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Example 25 Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes Solution Let C denote the family of circles in the second quadrant and touching the coordinate axes Let (–a, a) be the coordinate of the centre of any member of this family (see Fig 9 6)
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Solution Let C denote the family of circles in the second quadrant and touching the coordinate axes Let (–a, a) be the coordinate of the centre of any member of this family (see Fig 9 6) © NCERT not to be republished MATHEMATICS 416 X X’ Y Y’ (– , ) a a O Equation representing the family C is (x + a)2 + (y – a)2 = a2
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Let (–a, a) be the coordinate of the centre of any member of this family (see Fig 9 6) © NCERT not to be republished MATHEMATICS 416 X X’ Y Y’ (– , ) a a O Equation representing the family C is (x + a)2 + (y – a)2 = a2 (1) or x2 + y2 + 2ax – 2ay + a2 = 0
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6) © NCERT not to be republished MATHEMATICS 416 X X’ Y Y’ (– , ) a a O Equation representing the family C is (x + a)2 + (y – a)2 = a2 (1) or x2 + y2 + 2ax – 2ay + a2 = 0 (2) Differentiating equation (2) with respect to x, we get 2 2 2 2 dy dy x y a a dx dx + + − = 0 or dy x +y dx = 1 dy a dx ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ or a = 1 x y y y ′ + ′ − Substituting the value of a in equation (1), we get 2 2 1 1 x y y x y y x y y y ′ ′ + + ⎡ ⎤ ⎡ ⎤ + + − ⎢ ⎥ ⎢ ⎥ ′ ′ − − ⎣ ⎦ ⎣ ⎦ = 2 1 x yy y ′ + ⎡ ⎤ ⎢ ⎥ ′ − ⎣ ⎦ or [xy′ – x + x + y y′]2 + [y y′ – y – x – y y′]2 = [x + y y′]2 or (x + y)2 y′2 + [x + y]2 = [x + y y′]2 or (x + y)2 [(y′)2 + 1] = [x + y y′]2 which is the differential equation representing the given family of circles
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© NCERT not to be republished MATHEMATICS 416 X X’ Y Y’ (– , ) a a O Equation representing the family C is (x + a)2 + (y – a)2 = a2 (1) or x2 + y2 + 2ax – 2ay + a2 = 0 (2) Differentiating equation (2) with respect to x, we get 2 2 2 2 dy dy x y a a dx dx + + − = 0 or dy x +y dx = 1 dy a dx ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ or a = 1 x y y y ′ + ′ − Substituting the value of a in equation (1), we get 2 2 1 1 x y y x y y x y y y ′ ′ + + ⎡ ⎤ ⎡ ⎤ + + − ⎢ ⎥ ⎢ ⎥ ′ ′ − − ⎣ ⎦ ⎣ ⎦ = 2 1 x yy y ′ + ⎡ ⎤ ⎢ ⎥ ′ − ⎣ ⎦ or [xy′ – x + x + y y′]2 + [y y′ – y – x – y y′]2 = [x + y y′]2 or (x + y)2 y′2 + [x + y]2 = [x + y y′]2 or (x + y)2 [(y′)2 + 1] = [x + y y′]2 which is the differential equation representing the given family of circles Example 26 Find the particular solution of the differential equation log 3 4 dy x y ⎛dx ⎞ = + ⎜ ⎟ ⎝ ⎠ given that y = 0 when x = 0
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(1) or x2 + y2 + 2ax – 2ay + a2 = 0 (2) Differentiating equation (2) with respect to x, we get 2 2 2 2 dy dy x y a a dx dx + + − = 0 or dy x +y dx = 1 dy a dx ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ or a = 1 x y y y ′ + ′ − Substituting the value of a in equation (1), we get 2 2 1 1 x y y x y y x y y y ′ ′ + + ⎡ ⎤ ⎡ ⎤ + + − ⎢ ⎥ ⎢ ⎥ ′ ′ − − ⎣ ⎦ ⎣ ⎦ = 2 1 x yy y ′ + ⎡ ⎤ ⎢ ⎥ ′ − ⎣ ⎦ or [xy′ – x + x + y y′]2 + [y y′ – y – x – y y′]2 = [x + y y′]2 or (x + y)2 y′2 + [x + y]2 = [x + y y′]2 or (x + y)2 [(y′)2 + 1] = [x + y y′]2 which is the differential equation representing the given family of circles Example 26 Find the particular solution of the differential equation log 3 4 dy x y ⎛dx ⎞ = + ⎜ ⎟ ⎝ ⎠ given that y = 0 when x = 0 Solution The given differential equation can be written as dy dx = e(3x + 4y) or dy dx = e3x
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(2) Differentiating equation (2) with respect to x, we get 2 2 2 2 dy dy x y a a dx dx + + − = 0 or dy x +y dx = 1 dy a dx ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ or a = 1 x y y y ′ + ′ − Substituting the value of a in equation (1), we get 2 2 1 1 x y y x y y x y y y ′ ′ + + ⎡ ⎤ ⎡ ⎤ + + − ⎢ ⎥ ⎢ ⎥ ′ ′ − − ⎣ ⎦ ⎣ ⎦ = 2 1 x yy y ′ + ⎡ ⎤ ⎢ ⎥ ′ − ⎣ ⎦ or [xy′ – x + x + y y′]2 + [y y′ – y – x – y y′]2 = [x + y y′]2 or (x + y)2 y′2 + [x + y]2 = [x + y y′]2 or (x + y)2 [(y′)2 + 1] = [x + y y′]2 which is the differential equation representing the given family of circles Example 26 Find the particular solution of the differential equation log 3 4 dy x y ⎛dx ⎞ = + ⎜ ⎟ ⎝ ⎠ given that y = 0 when x = 0 Solution The given differential equation can be written as dy dx = e(3x + 4y) or dy dx = e3x e4y
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Example 26 Find the particular solution of the differential equation log 3 4 dy x y ⎛dx ⎞ = + ⎜ ⎟ ⎝ ⎠ given that y = 0 when x = 0 Solution The given differential equation can be written as dy dx = e(3x + 4y) or dy dx = e3x e4y (1) Separating the variables, we get edy4 y = e3x dx Therefore e4 y dy ∫− = 3x e dx ∫ Fig 9
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4926-4929
Solution The given differential equation can be written as dy dx = e(3x + 4y) or dy dx = e3x e4y (1) Separating the variables, we get edy4 y = e3x dx Therefore e4 y dy ∫− = 3x e dx ∫ Fig 9 6 © NCERT not to be republished DIFFERENTIAL EQUATIONS 417 or 4 4 y e− − = 3 C 3 ex + or 4 e3x + 3 e– 4y + 12 C = 0
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e4y (1) Separating the variables, we get edy4 y = e3x dx Therefore e4 y dy ∫− = 3x e dx ∫ Fig 9 6 © NCERT not to be republished DIFFERENTIAL EQUATIONS 417 or 4 4 y e− − = 3 C 3 ex + or 4 e3x + 3 e– 4y + 12 C = 0 (2) Substituting x = 0 and y = 0 in (2), we get 4 + 3 + 12 C = 0 or C = 127 − Substituting the value of C in equation (2), we get 4e3x + 3e– 4y – 7 = 0, which is a particular solution of the given differential equation
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(1) Separating the variables, we get edy4 y = e3x dx Therefore e4 y dy ∫− = 3x e dx ∫ Fig 9 6 © NCERT not to be republished DIFFERENTIAL EQUATIONS 417 or 4 4 y e− − = 3 C 3 ex + or 4 e3x + 3 e– 4y + 12 C = 0 (2) Substituting x = 0 and y = 0 in (2), we get 4 + 3 + 12 C = 0 or C = 127 − Substituting the value of C in equation (2), we get 4e3x + 3e– 4y – 7 = 0, which is a particular solution of the given differential equation Example 27 Solve the differential equation (x dy – y dx) y sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = (y dx + x dy) x cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠
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4929-4932
6 © NCERT not to be republished DIFFERENTIAL EQUATIONS 417 or 4 4 y e− − = 3 C 3 ex + or 4 e3x + 3 e– 4y + 12 C = 0 (2) Substituting x = 0 and y = 0 in (2), we get 4 + 3 + 12 C = 0 or C = 127 − Substituting the value of C in equation (2), we get 4e3x + 3e– 4y – 7 = 0, which is a particular solution of the given differential equation Example 27 Solve the differential equation (x dy – y dx) y sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = (y dx + x dy) x cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Solution The given differential equation can be written as 2 2 sin cos cos sin y y y y x y x dy xy y dx x x x x ⎡ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − = + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ or dy dx = 2 2 cos sin sin cos y y xy y x x y y xy x x x ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Dividing numerator and denominator on RHS by x2, we get dy dx = 2 2 cos sin sin cos y y y y x x x x y y y x x x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
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(2) Substituting x = 0 and y = 0 in (2), we get 4 + 3 + 12 C = 0 or C = 127 − Substituting the value of C in equation (2), we get 4e3x + 3e– 4y – 7 = 0, which is a particular solution of the given differential equation Example 27 Solve the differential equation (x dy – y dx) y sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = (y dx + x dy) x cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Solution The given differential equation can be written as 2 2 sin cos cos sin y y y y x y x dy xy y dx x x x x ⎡ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − = + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ or dy dx = 2 2 cos sin sin cos y y xy y x x y y xy x x x ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Dividing numerator and denominator on RHS by x2, we get dy dx = 2 2 cos sin sin cos y y y y x x x x y y y x x x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (1) Clearly, equation (1) is a homogeneous differential equation of the form dy y g dx ⎛x ⎞ = ⎜ ⎟ ⎝ ⎠
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Example 27 Solve the differential equation (x dy – y dx) y sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = (y dx + x dy) x cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Solution The given differential equation can be written as 2 2 sin cos cos sin y y y y x y x dy xy y dx x x x x ⎡ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − = + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ or dy dx = 2 2 cos sin sin cos y y xy y x x y y xy x x x ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Dividing numerator and denominator on RHS by x2, we get dy dx = 2 2 cos sin sin cos y y y y x x x x y y y x x x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (1) Clearly, equation (1) is a homogeneous differential equation of the form dy y g dx ⎛x ⎞ = ⎜ ⎟ ⎝ ⎠ To solve it, we make the substitution y = vx
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Solution The given differential equation can be written as 2 2 sin cos cos sin y y y y x y x dy xy y dx x x x x ⎡ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − = + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ or dy dx = 2 2 cos sin sin cos y y xy y x x y y xy x x x ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Dividing numerator and denominator on RHS by x2, we get dy dx = 2 2 cos sin sin cos y y y y x x x x y y y x x x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (1) Clearly, equation (1) is a homogeneous differential equation of the form dy y g dx ⎛x ⎞ = ⎜ ⎟ ⎝ ⎠ To solve it, we make the substitution y = vx (2) or dy dx = dv v x dx + © NCERT not to be republished MATHEMATICS 418 or dv v +x dx = 2 cos sin sin cos v v v v v v v + − (using (1) and (2)) or dv x dx = sin2 cos cos v v v v v − or sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 2 dx x Therefore sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∫ = 1 2 xdx ∫ or 1 tanv dv vdv − ∫ ∫ = 1 2 xdx ∫ or log sec log | | v v − = 1 2log | | log | C | x + or logsec2 v xv = log |C1| or secv2 v x = ± C1
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4933-4936
(1) Clearly, equation (1) is a homogeneous differential equation of the form dy y g dx ⎛x ⎞ = ⎜ ⎟ ⎝ ⎠ To solve it, we make the substitution y = vx (2) or dy dx = dv v x dx + © NCERT not to be republished MATHEMATICS 418 or dv v +x dx = 2 cos sin sin cos v v v v v v v + − (using (1) and (2)) or dv x dx = sin2 cos cos v v v v v − or sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 2 dx x Therefore sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∫ = 1 2 xdx ∫ or 1 tanv dv vdv − ∫ ∫ = 1 2 xdx ∫ or log sec log | | v v − = 1 2log | | log | C | x + or logsec2 v xv = log |C1| or secv2 v x = ± C1 (3) Replacing v by y x in equation (3), we get 2 sec ( ) y x y x x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C where, C = ± C1 or sec y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C xy which is the general solution of the given differential equation
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4934-4937
To solve it, we make the substitution y = vx (2) or dy dx = dv v x dx + © NCERT not to be republished MATHEMATICS 418 or dv v +x dx = 2 cos sin sin cos v v v v v v v + − (using (1) and (2)) or dv x dx = sin2 cos cos v v v v v − or sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 2 dx x Therefore sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∫ = 1 2 xdx ∫ or 1 tanv dv vdv − ∫ ∫ = 1 2 xdx ∫ or log sec log | | v v − = 1 2log | | log | C | x + or logsec2 v xv = log |C1| or secv2 v x = ± C1 (3) Replacing v by y x in equation (3), we get 2 sec ( ) y x y x x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C where, C = ± C1 or sec y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C xy which is the general solution of the given differential equation Example 28 Solve the differential equation (tan–1y – x) dy = (1 + y2) dx
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(2) or dy dx = dv v x dx + © NCERT not to be republished MATHEMATICS 418 or dv v +x dx = 2 cos sin sin cos v v v v v v v + − (using (1) and (2)) or dv x dx = sin2 cos cos v v v v v − or sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 2 dx x Therefore sin coscos v v v dv v −v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∫ = 1 2 xdx ∫ or 1 tanv dv vdv − ∫ ∫ = 1 2 xdx ∫ or log sec log | | v v − = 1 2log | | log | C | x + or logsec2 v xv = log |C1| or secv2 v x = ± C1 (3) Replacing v by y x in equation (3), we get 2 sec ( ) y x y x x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C where, C = ± C1 or sec y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C xy which is the general solution of the given differential equation Example 28 Solve the differential equation (tan–1y – x) dy = (1 + y2) dx Solution The given differential equation can be written as 2 1 dx x dy y + + = 1 2 1tan y y − +
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(3) Replacing v by y x in equation (3), we get 2 sec ( ) y x y x x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C where, C = ± C1 or sec y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = C xy which is the general solution of the given differential equation Example 28 Solve the differential equation (tan–1y – x) dy = (1 + y2) dx Solution The given differential equation can be written as 2 1 dx x dy y + + = 1 2 1tan y y − + (1) © NCERT not to be republished DIFFERENTIAL EQUATIONS 419 Now (1) is a linear differential equation of the form 1P dx dy + x = Q1, where, P1 = 112 +y and 1 1 2 tan Q 1 y y − = +
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4937-4940
Example 28 Solve the differential equation (tan–1y – x) dy = (1 + y2) dx Solution The given differential equation can be written as 2 1 dx x dy y + + = 1 2 1tan y y − + (1) © NCERT not to be republished DIFFERENTIAL EQUATIONS 419 Now (1) is a linear differential equation of the form 1P dx dy + x = Q1, where, P1 = 112 +y and 1 1 2 tan Q 1 y y − = + Therefore, I
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4938-4941
Solution The given differential equation can be written as 2 1 dx x dy y + + = 1 2 1tan y y − + (1) © NCERT not to be republished DIFFERENTIAL EQUATIONS 419 Now (1) is a linear differential equation of the form 1P dx dy + x = Q1, where, P1 = 112 +y and 1 1 2 tan Q 1 y y − = + Therefore, I F = 1 12 tan 1 dy y y e e − ∫+ = Thus, the solution of the given differential equation is tan1 y xe − = 1 1 tan 2 tan C 1 y y e dy y − − ⎛ ⎞ + ⎜ ⎟ + ⎝ ⎠ ∫
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(1) © NCERT not to be republished DIFFERENTIAL EQUATIONS 419 Now (1) is a linear differential equation of the form 1P dx dy + x = Q1, where, P1 = 112 +y and 1 1 2 tan Q 1 y y − = + Therefore, I F = 1 12 tan 1 dy y y e e − ∫+ = Thus, the solution of the given differential equation is tan1 y xe − = 1 1 tan 2 tan C 1 y y e dy y − − ⎛ ⎞ + ⎜ ⎟ + ⎝ ⎠ ∫ (2) Let I = 1 1 tan 2 tan 1 y y e dy y − − ⎛ ⎞ ⎜ ⎟ + ⎝ ⎠ ∫ Substituting tan–1 y = t so that 112 dy dt y ⎛ ⎞ = ⎜ ⎟ ⎝+ ⎠ , we get I = t ∫t e dt = t et – ∫1
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4940-4943
Therefore, I F = 1 12 tan 1 dy y y e e − ∫+ = Thus, the solution of the given differential equation is tan1 y xe − = 1 1 tan 2 tan C 1 y y e dy y − − ⎛ ⎞ + ⎜ ⎟ + ⎝ ⎠ ∫ (2) Let I = 1 1 tan 2 tan 1 y y e dy y − − ⎛ ⎞ ⎜ ⎟ + ⎝ ⎠ ∫ Substituting tan–1 y = t so that 112 dy dt y ⎛ ⎞ = ⎜ ⎟ ⎝+ ⎠ , we get I = t ∫t e dt = t et – ∫1 et dt = t et – et = et (t – 1) or I = tan1 y e − (tan–1y –1) Substituting the value of I in equation (2), we get 1 1 tan tan 1
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4941-4944
F = 1 12 tan 1 dy y y e e − ∫+ = Thus, the solution of the given differential equation is tan1 y xe − = 1 1 tan 2 tan C 1 y y e dy y − − ⎛ ⎞ + ⎜ ⎟ + ⎝ ⎠ ∫ (2) Let I = 1 1 tan 2 tan 1 y y e dy y − − ⎛ ⎞ ⎜ ⎟ + ⎝ ⎠ ∫ Substituting tan–1 y = t so that 112 dy dt y ⎛ ⎞ = ⎜ ⎟ ⎝+ ⎠ , we get I = t ∫t e dt = t et – ∫1 et dt = t et – et = et (t – 1) or I = tan1 y e − (tan–1y –1) Substituting the value of I in equation (2), we get 1 1 tan tan 1 (tan 1) C y y x e e y − − − = − + or x = 1 1 tan (tan 1) C y y e − − − − + which is the general solution of the given differential equation
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4942-4945
(2) Let I = 1 1 tan 2 tan 1 y y e dy y − − ⎛ ⎞ ⎜ ⎟ + ⎝ ⎠ ∫ Substituting tan–1 y = t so that 112 dy dt y ⎛ ⎞ = ⎜ ⎟ ⎝+ ⎠ , we get I = t ∫t e dt = t et – ∫1 et dt = t et – et = et (t – 1) or I = tan1 y e − (tan–1y –1) Substituting the value of I in equation (2), we get 1 1 tan tan 1 (tan 1) C y y x e e y − − − = − + or x = 1 1 tan (tan 1) C y y e − − − − + which is the general solution of the given differential equation Miscellaneous Exercise on Chapter 9 1
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et dt = t et – et = et (t – 1) or I = tan1 y e − (tan–1y –1) Substituting the value of I in equation (2), we get 1 1 tan tan 1 (tan 1) C y y x e e y − − − = − + or x = 1 1 tan (tan 1) C y y e − − − − + which is the general solution of the given differential equation Miscellaneous Exercise on Chapter 9 1 For each of the differential equations given below, indicate its order and degree (if defined)
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4944-4947
(tan 1) C y y x e e y − − − = − + or x = 1 1 tan (tan 1) C y y e − − − − + which is the general solution of the given differential equation Miscellaneous Exercise on Chapter 9 1 For each of the differential equations given below, indicate its order and degree (if defined) (i) 2 2 2 5 6 log d y xdy y x dx dx ⎛ ⎞ + − = ⎜ ⎟ ⎝ ⎠ (ii) 3 2 4 7 sin dy dy y x dx dx ⎛ ⎞ ⎛ ⎞ − + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (iii) 4 3 4 3 sin 0 d y d y dx dx ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ © NCERT not to be republished MATHEMATICS 420 2
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Miscellaneous Exercise on Chapter 9 1 For each of the differential equations given below, indicate its order and degree (if defined) (i) 2 2 2 5 6 log d y xdy y x dx dx ⎛ ⎞ + − = ⎜ ⎟ ⎝ ⎠ (ii) 3 2 4 7 sin dy dy y x dx dx ⎛ ⎞ ⎛ ⎞ − + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (iii) 4 3 4 3 sin 0 d y d y dx dx ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ © NCERT not to be republished MATHEMATICS 420 2 For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation
1
4946-4949
For each of the differential equations given below, indicate its order and degree (if defined) (i) 2 2 2 5 6 log d y xdy y x dx dx ⎛ ⎞ + − = ⎜ ⎟ ⎝ ⎠ (ii) 3 2 4 7 sin dy dy y x dx dx ⎛ ⎞ ⎛ ⎞ − + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (iii) 4 3 4 3 sin 0 d y d y dx dx ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ © NCERT not to be republished MATHEMATICS 420 2 For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation (i) y = a ex + b e–x + x2 : 2 2 2 2 2 0 d y dy x xy x dx dx + − + − = (ii) y = ex (a cos x + b sin x) : 2 2 2 2 0 d y dy y dx dx − + = (iii) y = x sin 3x : 2 2 9 6cos3 0 d y y x dx + − = (iv) x2 = 2y2 log y : 2 2 ( ) 0 dy x y xy dx + − = 3
1
4947-4950
(i) 2 2 2 5 6 log d y xdy y x dx dx ⎛ ⎞ + − = ⎜ ⎟ ⎝ ⎠ (ii) 3 2 4 7 sin dy dy y x dx dx ⎛ ⎞ ⎛ ⎞ − + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (iii) 4 3 4 3 sin 0 d y d y dx dx ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ © NCERT not to be republished MATHEMATICS 420 2 For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation (i) y = a ex + b e–x + x2 : 2 2 2 2 2 0 d y dy x xy x dx dx + − + − = (ii) y = ex (a cos x + b sin x) : 2 2 2 2 0 d y dy y dx dx − + = (iii) y = x sin 3x : 2 2 9 6cos3 0 d y y x dx + − = (iv) x2 = 2y2 log y : 2 2 ( ) 0 dy x y xy dx + − = 3 Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant
1
4948-4951
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation (i) y = a ex + b e–x + x2 : 2 2 2 2 2 0 d y dy x xy x dx dx + − + − = (ii) y = ex (a cos x + b sin x) : 2 2 2 2 0 d y dy y dx dx − + = (iii) y = x sin 3x : 2 2 9 6cos3 0 d y y x dx + − = (iv) x2 = 2y2 log y : 2 2 ( ) 0 dy x y xy dx + − = 3 Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant 4
1
4949-4952
(i) y = a ex + b e–x + x2 : 2 2 2 2 2 0 d y dy x xy x dx dx + − + − = (ii) y = ex (a cos x + b sin x) : 2 2 2 2 0 d y dy y dx dx − + = (iii) y = x sin 3x : 2 2 9 6cos3 0 d y y x dx + − = (iv) x2 = 2y2 log y : 2 2 ( ) 0 dy x y xy dx + − = 3 Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant 4 Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter
1
4950-4953
Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant 4 Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter 5
1
4951-4954
4 Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter 5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes
1
4952-4955
Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter 5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes 6
1
4953-4956
5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes 6 Find the general solution of the differential equation 2 2 1 0 1 dy y dx −x + = −
1
4954-4957
Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes 6 Find the general solution of the differential equation 2 2 1 0 1 dy y dx −x + = − 7
1
4955-4958
6 Find the general solution of the differential equation 2 2 1 0 1 dy y dx −x + = − 7 Show that the general solution of the differential equation 2 2 1 0 1 dy y y dx x +x + + = + + is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter
1
4956-4959
Find the general solution of the differential equation 2 2 1 0 1 dy y dx −x + = − 7 Show that the general solution of the differential equation 2 2 1 0 1 dy y y dx x +x + + = + + is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter 8
1
4957-4960
7 Show that the general solution of the differential equation 2 2 1 0 1 dy y y dx x +x + + = + + is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter 8 Find the equation of the curve passing through the point 0, 4 π ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ whose differential equation is sin x cos y dx + cos x sin y dy = 0
1
4958-4961
Show that the general solution of the differential equation 2 2 1 0 1 dy y y dx x +x + + = + + is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter 8 Find the equation of the curve passing through the point 0, 4 π ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ whose differential equation is sin x cos y dx + cos x sin y dy = 0 9
1
4959-4962
8 Find the equation of the curve passing through the point 0, 4 π ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ whose differential equation is sin x cos y dx + cos x sin y dy = 0 9 Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0
1
4960-4963
Find the equation of the curve passing through the point 0, 4 π ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ whose differential equation is sin x cos y dx + cos x sin y dy = 0 9 Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0 10
1
4961-4964
9 Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0 10 Solve the differential equation 2 ( 0) x x y y y e dx x e y dy y ⎛ ⎞ ⎜ ⎟ = + ≠ ⎝ ⎠
1
4962-4965
Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0 10 Solve the differential equation 2 ( 0) x x y y y e dx x e y dy y ⎛ ⎞ ⎜ ⎟ = + ≠ ⎝ ⎠ 11
1
4963-4966
10 Solve the differential equation 2 ( 0) x x y y y e dx x e y dy y ⎛ ⎞ ⎜ ⎟ = + ≠ ⎝ ⎠ 11 Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0
1
4964-4967
Solve the differential equation 2 ( 0) x x y y y e dx x e y dy y ⎛ ⎞ ⎜ ⎟ = + ≠ ⎝ ⎠ 11 Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0 (Hint: put x – y = t) © NCERT not to be republished DIFFERENTIAL EQUATIONS 421 12
1
4965-4968
11 Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0 (Hint: put x – y = t) © NCERT not to be republished DIFFERENTIAL EQUATIONS 421 12 Solve the differential equation 2 1( 0) x e y dx x dy x x ⎡− ⎤ − = ≠ ⎢ ⎥ ⎣ ⎦
1
4966-4969
Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0 (Hint: put x – y = t) © NCERT not to be republished DIFFERENTIAL EQUATIONS 421 12 Solve the differential equation 2 1( 0) x e y dx x dy x x ⎡− ⎤ − = ≠ ⎢ ⎥ ⎣ ⎦ 13
1
4967-4970
(Hint: put x – y = t) © NCERT not to be republished DIFFERENTIAL EQUATIONS 421 12 Solve the differential equation 2 1( 0) x e y dx x dy x x ⎡− ⎤ − = ≠ ⎢ ⎥ ⎣ ⎦ 13 Find a particular solution of the differential equation cot dy y x dx + = 4x cosec x (x ≠ 0), given that y = 0 when 2 x =π
1
4968-4971
Solve the differential equation 2 1( 0) x e y dx x dy x x ⎡− ⎤ − = ≠ ⎢ ⎥ ⎣ ⎦ 13 Find a particular solution of the differential equation cot dy y x dx + = 4x cosec x (x ≠ 0), given that y = 0 when 2 x =π 14
1
4969-4972
13 Find a particular solution of the differential equation cot dy y x dx + = 4x cosec x (x ≠ 0), given that y = 0 when 2 x =π 14 Find a particular solution of the differential equation (x + 1) dy dx = 2 e–y – 1, given that y = 0 when x = 0
1
4970-4973
Find a particular solution of the differential equation cot dy y x dx + = 4x cosec x (x ≠ 0), given that y = 0 when 2 x =π 14 Find a particular solution of the differential equation (x + 1) dy dx = 2 e–y – 1, given that y = 0 when x = 0 15
1
4971-4974
14 Find a particular solution of the differential equation (x + 1) dy dx = 2 e–y – 1, given that y = 0 when x = 0 15 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time
1
4972-4975
Find a particular solution of the differential equation (x + 1) dy dx = 2 e–y – 1, given that y = 0 when x = 0 15 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009
1
4973-4976
15 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009 16
1
4974-4977
The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009 16 The general solution of the differential equation 0 y dx −yx dy = is (A) xy = C (B) x = Cy2 (C) y = Cx (D) y = Cx2 17
1
4975-4978
If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009 16 The general solution of the differential equation 0 y dx −yx dy = is (A) xy = C (B) x = Cy2 (C) y = Cx (D) y = Cx2 17 The general solution of a differential equation of the type 1 1 P Q dx x dy + = is (A) ( ) 1 1 P P Q1 C dy dy y e e dy ∫ ∫ = + ∫ (B) ( ) 1 1 P P 1
1
4976-4979
16 The general solution of the differential equation 0 y dx −yx dy = is (A) xy = C (B) x = Cy2 (C) y = Cx (D) y = Cx2 17 The general solution of a differential equation of the type 1 1 P Q dx x dy + = is (A) ( ) 1 1 P P Q1 C dy dy y e e dy ∫ ∫ = + ∫ (B) ( ) 1 1 P P 1 Q C dx dx y e e dx ∫ ∫ = + ∫ (C) ( ) 1 1 P P Q1 C dy dy x e e dy ∫ ∫ = + ∫ (D) ( ) 1 1 P P Q1 C dx dx x e e dx ∫ ∫ = + ∫ 18
1
4977-4980
The general solution of the differential equation 0 y dx −yx dy = is (A) xy = C (B) x = Cy2 (C) y = Cx (D) y = Cx2 17 The general solution of a differential equation of the type 1 1 P Q dx x dy + = is (A) ( ) 1 1 P P Q1 C dy dy y e e dy ∫ ∫ = + ∫ (B) ( ) 1 1 P P 1 Q C dx dx y e e dx ∫ ∫ = + ∫ (C) ( ) 1 1 P P Q1 C dy dy x e e dy ∫ ∫ = + ∫ (D) ( ) 1 1 P P Q1 C dx dx x e e dx ∫ ∫ = + ∫ 18 The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is (A) x ey + x2 = C (B) x ey + y2 = C (C) y ex + x2 = C (D) y ey + x2 = C © NCERT not to be republished MATHEMATICS 422 Summary � An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation
1
4978-4981
The general solution of a differential equation of the type 1 1 P Q dx x dy + = is (A) ( ) 1 1 P P Q1 C dy dy y e e dy ∫ ∫ = + ∫ (B) ( ) 1 1 P P 1 Q C dx dx y e e dx ∫ ∫ = + ∫ (C) ( ) 1 1 P P Q1 C dy dy x e e dy ∫ ∫ = + ∫ (D) ( ) 1 1 P P Q1 C dx dx x e e dx ∫ ∫ = + ∫ 18 The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is (A) x ey + x2 = C (B) x ey + y2 = C (C) y ex + x2 = C (D) y ey + x2 = C © NCERT not to be republished MATHEMATICS 422 Summary � An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation � Order of a differential equation is the order of the highest order derivative occurring in the differential equation
1
4979-4982
Q C dx dx y e e dx ∫ ∫ = + ∫ (C) ( ) 1 1 P P Q1 C dy dy x e e dy ∫ ∫ = + ∫ (D) ( ) 1 1 P P Q1 C dx dx x e e dx ∫ ∫ = + ∫ 18 The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is (A) x ey + x2 = C (B) x ey + y2 = C (C) y ex + x2 = C (D) y ey + x2 = C © NCERT not to be republished MATHEMATICS 422 Summary � An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation � Order of a differential equation is the order of the highest order derivative occurring in the differential equation � Degree of a differential equation is defined if it is a polynomial equation in its derivatives
1
4980-4983
The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is (A) x ey + x2 = C (B) x ey + y2 = C (C) y ex + x2 = C (D) y ey + x2 = C © NCERT not to be republished MATHEMATICS 422 Summary � An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation � Order of a differential equation is the order of the highest order derivative occurring in the differential equation � Degree of a differential equation is defined if it is a polynomial equation in its derivatives � Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it
1
4981-4984
� Order of a differential equation is the order of the highest order derivative occurring in the differential equation � Degree of a differential equation is defined if it is a polynomial equation in its derivatives � Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it � A function which satisfies the given differential equation is called its solution
1
4982-4985
� Degree of a differential equation is defined if it is a polynomial equation in its derivatives � Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it � A function which satisfies the given differential equation is called its solution The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution
1
4983-4986
� Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it � A function which satisfies the given differential equation is called its solution The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution � To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants
1
4984-4987
� A function which satisfies the given differential equation is called its solution The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution � To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants � Variable separable method is used to solve such an equation in which variables can be separated completely i
1
4985-4988
The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution � To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants � Variable separable method is used to solve such an equation in which variables can be separated completely i e
1
4986-4989
� To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants � Variable separable method is used to solve such an equation in which variables can be separated completely i e terms containing y should remain with dy and terms containing x should remain with dx
1
4987-4990
� Variable separable method is used to solve such an equation in which variables can be separated completely i e terms containing y should remain with dy and terms containing x should remain with dx � A differential equation which can be expressed in the form ( , ) or ( , ) dy dx f x y g x y dx dy where, f (x, y) and g(x, y) are homogenous functions of degree zero is called a homogeneous differential equation
1
4988-4991
e terms containing y should remain with dy and terms containing x should remain with dx � A differential equation which can be expressed in the form ( , ) or ( , ) dy dx f x y g x y dx dy where, f (x, y) and g(x, y) are homogenous functions of degree zero is called a homogeneous differential equation � A differential equation of the form +P Q dy y dx , where P and Q are constants or functions of x only is called a first order linear differential equation
1
4989-4992
terms containing y should remain with dy and terms containing x should remain with dx � A differential equation which can be expressed in the form ( , ) or ( , ) dy dx f x y g x y dx dy where, f (x, y) and g(x, y) are homogenous functions of degree zero is called a homogeneous differential equation � A differential equation of the form +P Q dy y dx , where P and Q are constants or functions of x only is called a first order linear differential equation Historical Note One of the principal languages of Science is that of differential equations
1
4990-4993
� A differential equation which can be expressed in the form ( , ) or ( , ) dy dx f x y g x y dx dy where, f (x, y) and g(x, y) are homogenous functions of degree zero is called a homogeneous differential equation � A differential equation of the form +P Q dy y dx , where P and Q are constants or functions of x only is called a first order linear differential equation Historical Note One of the principal languages of Science is that of differential equations Interestingly, the date of birth of differential equations is taken to be November, 11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black and white the identity 212 y dy y = ∫ , thereby introducing both the symbols ∫ and dy
1
4991-4994
� A differential equation of the form +P Q dy y dx , where P and Q are constants or functions of x only is called a first order linear differential equation Historical Note One of the principal languages of Science is that of differential equations Interestingly, the date of birth of differential equations is taken to be November, 11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black and white the identity 212 y dy y = ∫ , thereby introducing both the symbols ∫ and dy © NCERT not to be republished DIFFERENTIAL EQUATIONS 423 Leibnitz was actually interested in the problem of finding a curve whose tangents were prescribed
1
4992-4995
Historical Note One of the principal languages of Science is that of differential equations Interestingly, the date of birth of differential equations is taken to be November, 11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black and white the identity 212 y dy y = ∫ , thereby introducing both the symbols ∫ and dy © NCERT not to be republished DIFFERENTIAL EQUATIONS 423 Leibnitz was actually interested in the problem of finding a curve whose tangents were prescribed This led him to discover the ‘method of separation of variables’ 1691
1
4993-4996
Interestingly, the date of birth of differential equations is taken to be November, 11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black and white the identity 212 y dy y = ∫ , thereby introducing both the symbols ∫ and dy © NCERT not to be republished DIFFERENTIAL EQUATIONS 423 Leibnitz was actually interested in the problem of finding a curve whose tangents were prescribed This led him to discover the ‘method of separation of variables’ 1691 A year later he formulated the ‘method of solving the homogeneous differential equations of the first order’
1
4994-4997
© NCERT not to be republished DIFFERENTIAL EQUATIONS 423 Leibnitz was actually interested in the problem of finding a curve whose tangents were prescribed This led him to discover the ‘method of separation of variables’ 1691 A year later he formulated the ‘method of solving the homogeneous differential equations of the first order’ He went further in a very short time to the discovery of the ‘method of solving a linear differential equation of the first-order’
1
4995-4998
This led him to discover the ‘method of separation of variables’ 1691 A year later he formulated the ‘method of solving the homogeneous differential equations of the first order’ He went further in a very short time to the discovery of the ‘method of solving a linear differential equation of the first-order’ How surprising is it that all these methods came from a single man and that too within 25 years of the birth of differential equations
1
4996-4999
A year later he formulated the ‘method of solving the homogeneous differential equations of the first order’ He went further in a very short time to the discovery of the ‘method of solving a linear differential equation of the first-order’ How surprising is it that all these methods came from a single man and that too within 25 years of the birth of differential equations In the old days, what we now call the ‘solution’ of a differential equation, was used to be referred to as ‘integral’ of the differential equation, the word being coined by James Bernoulli (1654 - 1705) in 1690
1
4997-5000
He went further in a very short time to the discovery of the ‘method of solving a linear differential equation of the first-order’ How surprising is it that all these methods came from a single man and that too within 25 years of the birth of differential equations In the old days, what we now call the ‘solution’ of a differential equation, was used to be referred to as ‘integral’ of the differential equation, the word being coined by James Bernoulli (1654 - 1705) in 1690 The word ‘solution was first used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost hundred years since the birth of differential equations
1
4998-5001
How surprising is it that all these methods came from a single man and that too within 25 years of the birth of differential equations In the old days, what we now call the ‘solution’ of a differential equation, was used to be referred to as ‘integral’ of the differential equation, the word being coined by James Bernoulli (1654 - 1705) in 1690 The word ‘solution was first used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost hundred years since the birth of differential equations It was Jules Henri Poincare (1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the word ‘solution’ has found its deserved place in modern terminology
1
4999-5002
In the old days, what we now call the ‘solution’ of a differential equation, was used to be referred to as ‘integral’ of the differential equation, the word being coined by James Bernoulli (1654 - 1705) in 1690 The word ‘solution was first used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost hundred years since the birth of differential equations It was Jules Henri Poincare (1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the word ‘solution’ has found its deserved place in modern terminology The name of the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748), a younger brother of James Bernoulli
1
5000-5003
The word ‘solution was first used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost hundred years since the birth of differential equations It was Jules Henri Poincare (1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the word ‘solution’ has found its deserved place in modern terminology The name of the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748), a younger brother of James Bernoulli Application to geometric problems were also considered
1
5001-5004
It was Jules Henri Poincare (1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the word ‘solution’ has found its deserved place in modern terminology The name of the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748), a younger brother of James Bernoulli Application to geometric problems were also considered It was again John Bernoulli who first brought into light the intricate nature of differential equations
1
5002-5005
The name of the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748), a younger brother of James Bernoulli Application to geometric problems were also considered It was again John Bernoulli who first brought into light the intricate nature of differential equations In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the differential equation x2 y″ = 2y, which led to three types of curves, viz
1
5003-5006
Application to geometric problems were also considered It was again John Bernoulli who first brought into light the intricate nature of differential equations In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the differential equation x2 y″ = 2y, which led to three types of curves, viz , parabolas, hyperbolas and a class of cubic curves
1
5004-5007
It was again John Bernoulli who first brought into light the intricate nature of differential equations In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the differential equation x2 y″ = 2y, which led to three types of curves, viz , parabolas, hyperbolas and a class of cubic curves This shows how varied the solutions of such innocent looking differential equation can be
1
5005-5008
In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the differential equation x2 y″ = 2y, which led to three types of curves, viz , parabolas, hyperbolas and a class of cubic curves This shows how varied the solutions of such innocent looking differential equation can be From the second half of the twentieth century attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading ‘qualitative analysis of differential equations’
1
5006-5009
, parabolas, hyperbolas and a class of cubic curves This shows how varied the solutions of such innocent looking differential equation can be From the second half of the twentieth century attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading ‘qualitative analysis of differential equations’ Now-a-days, this has acquired prime importance being absolutely necessary in almost all investigations
1
5007-5010
This shows how varied the solutions of such innocent looking differential equation can be From the second half of the twentieth century attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading ‘qualitative analysis of differential equations’ Now-a-days, this has acquired prime importance being absolutely necessary in almost all investigations —� � � � �— © NCERT not to be republished MATHEMATICS 424 �In most sciences one generation tears down what another has built and what one has established another undoes
1
5008-5011
From the second half of the twentieth century attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading ‘qualitative analysis of differential equations’ Now-a-days, this has acquired prime importance being absolutely necessary in almost all investigations —� � � � �— © NCERT not to be republished MATHEMATICS 424 �In most sciences one generation tears down what another has built and what one has established another undoes In Mathematics alone each generation builds a new story to the old structure
1
5009-5012
Now-a-days, this has acquired prime importance being absolutely necessary in almost all investigations —� � � � �— © NCERT not to be republished MATHEMATICS 424 �In most sciences one generation tears down what another has built and what one has established another undoes In Mathematics alone each generation builds a new story to the old structure – HERMAN HANKEL � 10
1
5010-5013
—� � � � �— © NCERT not to be republished MATHEMATICS 424 �In most sciences one generation tears down what another has built and what one has established another undoes In Mathematics alone each generation builds a new story to the old structure – HERMAN HANKEL � 10 1 Introduction In our day to day life, we come across many queries such as – What is your height
1
5011-5014
In Mathematics alone each generation builds a new story to the old structure – HERMAN HANKEL � 10 1 Introduction In our day to day life, we come across many queries such as – What is your height How should a football player hit the ball to give a pass to another player of his team
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5012-5015
– HERMAN HANKEL � 10 1 Introduction In our day to day life, we come across many queries such as – What is your height How should a football player hit the ball to give a pass to another player of his team Observe that a possible answer to the first query may be 1
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5013-5016
1 Introduction In our day to day life, we come across many queries such as – What is your height How should a football player hit the ball to give a pass to another player of his team Observe that a possible answer to the first query may be 1 6 meters, a quantity that involves only one value (magnitude) which is a real number
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5014-5017
How should a football player hit the ball to give a pass to another player of his team Observe that a possible answer to the first query may be 1 6 meters, a quantity that involves only one value (magnitude) which is a real number Such quantities are called scalars
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5015-5018
Observe that a possible answer to the first query may be 1 6 meters, a quantity that involves only one value (magnitude) which is a real number Such quantities are called scalars However, an answer to the second query is a quantity (called force) which involves muscular strength (magnitude) and direction (in which another player is positioned)
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5016-5019
6 meters, a quantity that involves only one value (magnitude) which is a real number Such quantities are called scalars However, an answer to the second query is a quantity (called force) which involves muscular strength (magnitude) and direction (in which another player is positioned) Such quantities are called vectors
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5017-5020
Such quantities are called scalars However, an answer to the second query is a quantity (called force) which involves muscular strength (magnitude) and direction (in which another player is positioned) Such quantities are called vectors In mathematics, physics and engineering, we frequently come across with both types of quantities, namely, scalar quantities such as length, mass, time, distance, speed, area, volume, temperature, work, money, voltage, density, resistance etc