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1	4718-4721	16 For the differential equation ( 2) ( 2) xydy x y dx = + + , find the solution curve passing through the point (1, –1) 17 Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point
1	4719-4722	For the differential equation ( 2) ( 2) xydy x y dx = + + , find the solution curve passing through the point (1, –1) 17 Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point 18
1	4720-4723	17 Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point 18 At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3)
1	4721-4724	Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point 18 At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3) Find the equation of the curve given that it passes through (–2, 1)
1	4722-4725	18 At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3) Find the equation of the curve given that it passes through (–2, 1) 19
1	4723-4726	At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3) Find the equation of the curve given that it passes through (–2, 1) 19 The volume of spherical balloon being inflated changes at a constant rate
1	4724-4727	Find the equation of the curve given that it passes through (–2, 1) 19 The volume of spherical balloon being inflated changes at a constant rate If initially its radius is 3 units and after 3 seconds it is 6 units
1	4725-4728	19 The volume of spherical balloon being inflated changes at a constant rate If initially its radius is 3 units and after 3 seconds it is 6 units Find the radius of balloon after t seconds
1	4726-4729	The volume of spherical balloon being inflated changes at a constant rate If initially its radius is 3 units and after 3 seconds it is 6 units Find the radius of balloon after t seconds © NCERT not to be republished DIFFERENTIAL EQUATIONS 397 20
1	4727-4730	If initially its radius is 3 units and after 3 seconds it is 6 units Find the radius of balloon after t seconds © NCERT not to be republished DIFFERENTIAL EQUATIONS 397 20 In a bank, principal increases continuously at the rate of r% per year
1	4728-4731	Find the radius of balloon after t seconds © NCERT not to be republished DIFFERENTIAL EQUATIONS 397 20 In a bank, principal increases continuously at the rate of r% per year Find the value of r if Rs 100 double itself in 10 years (loge2 = 0
1	4729-4732	© NCERT not to be republished DIFFERENTIAL EQUATIONS 397 20 In a bank, principal increases continuously at the rate of r% per year Find the value of r if Rs 100 double itself in 10 years (loge2 = 0 6931)
1	4730-4733	In a bank, principal increases continuously at the rate of r% per year Find the value of r if Rs 100 double itself in 10 years (loge2 = 0 6931) 21
1	4731-4734	Find the value of r if Rs 100 double itself in 10 years (loge2 = 0 6931) 21 In a bank, principal increases continuously at the rate of 5% per year
1	4732-4735	6931) 21 In a bank, principal increases continuously at the rate of 5% per year An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0
1	4733-4736	21 In a bank, principal increases continuously at the rate of 5% per year An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0 5 = 1
1	4734-4737	In a bank, principal increases continuously at the rate of 5% per year An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0 5 = 1 648)
1	4735-4738	An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0 5 = 1 648) 22
1	4736-4739	5 = 1 648) 22 In a culture, the bacteria count is 1,00,000
1	4737-4740	648) 22 In a culture, the bacteria count is 1,00,000 The number is increased by 10% in 2 hours
1	4738-4741	22 In a culture, the bacteria count is 1,00,000 The number is increased by 10% in 2 hours In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present
1	4739-4742	In a culture, the bacteria count is 1,00,000 The number is increased by 10% in 2 hours In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present 23
1	4740-4743	The number is increased by 10% in 2 hours In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present 23 The general solution of the differential equation x y dy e dx + = is (A) ex + e–y = C (B) ex + ey = C (C) e–x + ey = C (D) e–x + e–y = C 9
1	4741-4744	In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present 23 The general solution of the differential equation x y dy e dx + = is (A) ex + e–y = C (B) ex + ey = C (C) e–x + ey = C (D) e–x + e–y = C 9 5
1	4742-4745	23 The general solution of the differential equation x y dy e dx + = is (A) ex + e–y = C (B) ex + ey = C (C) e–x + ey = C (D) e–x + e–y = C 9 5 2 Homogeneous differential equations Consider the following functions in x and y F1 (x, y) = y2 + 2xy, F2 (x, y) = 2x – 3y, F3 (x, y) = cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , F4 (x, y) = sin x + cos y If we replace x and y by λx and λy respectively in the above functions, for any nonzero constant λ, we get F1 (λx, λy) = λ2 (y2 + 2xy) = λ2 F1 (x, y) F2 (λx, λy) = λ (2x – 3y) = λ F2 (x, y) F3 (λx, λy) = cos cos y y x x ⎛λ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝λ ⎠ ⎝ ⎠ = λ0 F3 (x, y) F4 (λx, λy) = sin λx + cos λy ≠ λn F4 (x, y), for any n ∈ N Here, we observe that the functions F1, F2, F3 can be written in the form F(λx, λy) = λn F (x, y) but F4 can not be written in this form
1	4743-4746	The general solution of the differential equation x y dy e dx + = is (A) ex + e–y = C (B) ex + ey = C (C) e–x + ey = C (D) e–x + e–y = C 9 5 2 Homogeneous differential equations Consider the following functions in x and y F1 (x, y) = y2 + 2xy, F2 (x, y) = 2x – 3y, F3 (x, y) = cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , F4 (x, y) = sin x + cos y If we replace x and y by λx and λy respectively in the above functions, for any nonzero constant λ, we get F1 (λx, λy) = λ2 (y2 + 2xy) = λ2 F1 (x, y) F2 (λx, λy) = λ (2x – 3y) = λ F2 (x, y) F3 (λx, λy) = cos cos y y x x ⎛λ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝λ ⎠ ⎝ ⎠ = λ0 F3 (x, y) F4 (λx, λy) = sin λx + cos λy ≠ λn F4 (x, y), for any n ∈ N Here, we observe that the functions F1, F2, F3 can be written in the form F(λx, λy) = λn F (x, y) but F4 can not be written in this form This leads to the following definition: A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy) = λn F(x, y) for any nonzero constant λ
1	4744-4747	5 2 Homogeneous differential equations Consider the following functions in x and y F1 (x, y) = y2 + 2xy, F2 (x, y) = 2x – 3y, F3 (x, y) = cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , F4 (x, y) = sin x + cos y If we replace x and y by λx and λy respectively in the above functions, for any nonzero constant λ, we get F1 (λx, λy) = λ2 (y2 + 2xy) = λ2 F1 (x, y) F2 (λx, λy) = λ (2x – 3y) = λ F2 (x, y) F3 (λx, λy) = cos cos y y x x ⎛λ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝λ ⎠ ⎝ ⎠ = λ0 F3 (x, y) F4 (λx, λy) = sin λx + cos λy ≠ λn F4 (x, y), for any n ∈ N Here, we observe that the functions F1, F2, F3 can be written in the form F(λx, λy) = λn F (x, y) but F4 can not be written in this form This leads to the following definition: A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy) = λn F(x, y) for any nonzero constant λ We note that in the above examples, F1, F2, F3 are homogeneous functions of degree 2, 1, 0 respectively but F4 is not a homogeneous function
1	4745-4748	2 Homogeneous differential equations Consider the following functions in x and y F1 (x, y) = y2 + 2xy, F2 (x, y) = 2x – 3y, F3 (x, y) = cos y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , F4 (x, y) = sin x + cos y If we replace x and y by λx and λy respectively in the above functions, for any nonzero constant λ, we get F1 (λx, λy) = λ2 (y2 + 2xy) = λ2 F1 (x, y) F2 (λx, λy) = λ (2x – 3y) = λ F2 (x, y) F3 (λx, λy) = cos cos y y x x ⎛λ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝λ ⎠ ⎝ ⎠ = λ0 F3 (x, y) F4 (λx, λy) = sin λx + cos λy ≠ λn F4 (x, y), for any n ∈ N Here, we observe that the functions F1, F2, F3 can be written in the form F(λx, λy) = λn F (x, y) but F4 can not be written in this form This leads to the following definition: A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy) = λn F(x, y) for any nonzero constant λ We note that in the above examples, F1, F2, F3 are homogeneous functions of degree 2, 1, 0 respectively but F4 is not a homogeneous function © NCERT not to be republished MATHEMATICS 398 We also observe that F1(x, y) = 2 2 2 1 2 2 y y y x x h x x x ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F1(x, y) = 2 2 2 2 1 x x y y h y y ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F2(x, y) = 1 1 3 3 2 y y x x h x x ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F2(x, y) = 1 1 4 2 3 x x y y h y y ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F3(x, y) = 0 0 5 cos y y x x h x x ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F4(x, y) ≠ 6 n y x h x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N or F4 (x, y) ≠ 7 n x y h ⎛y ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N Therefore, a function F (x, y) is a homogeneous function of degree n if F(x, y) = or n n y x x g y h x y ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ A differential equation of the form dy dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero
1	4746-4749	This leads to the following definition: A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy) = λn F(x, y) for any nonzero constant λ We note that in the above examples, F1, F2, F3 are homogeneous functions of degree 2, 1, 0 respectively but F4 is not a homogeneous function © NCERT not to be republished MATHEMATICS 398 We also observe that F1(x, y) = 2 2 2 1 2 2 y y y x x h x x x ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F1(x, y) = 2 2 2 2 1 x x y y h y y ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F2(x, y) = 1 1 3 3 2 y y x x h x x ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F2(x, y) = 1 1 4 2 3 x x y y h y y ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F3(x, y) = 0 0 5 cos y y x x h x x ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F4(x, y) ≠ 6 n y x h x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N or F4 (x, y) ≠ 7 n x y h ⎛y ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N Therefore, a function F (x, y) is a homogeneous function of degree n if F(x, y) = or n n y x x g y h x y ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ A differential equation of the form dy dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero To solve a homogeneous differential equation of the type ( ) F , dy x y dx = = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠
1	4747-4750	We note that in the above examples, F1, F2, F3 are homogeneous functions of degree 2, 1, 0 respectively but F4 is not a homogeneous function © NCERT not to be republished MATHEMATICS 398 We also observe that F1(x, y) = 2 2 2 1 2 2 y y y x x h x x x ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F1(x, y) = 2 2 2 2 1 x x y y h y y ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F2(x, y) = 1 1 3 3 2 y y x x h x x ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F2(x, y) = 1 1 4 2 3 x x y y h y y ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F3(x, y) = 0 0 5 cos y y x x h x x ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F4(x, y) ≠ 6 n y x h x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N or F4 (x, y) ≠ 7 n x y h ⎛y ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N Therefore, a function F (x, y) is a homogeneous function of degree n if F(x, y) = or n n y x x g y h x y ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ A differential equation of the form dy dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero To solve a homogeneous differential equation of the type ( ) F , dy x y dx = = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (1) We make the substitution y = v
1	4748-4751	© NCERT not to be republished MATHEMATICS 398 We also observe that F1(x, y) = 2 2 2 1 2 2 y y y x x h x x x ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F1(x, y) = 2 2 2 2 1 x x y y h y y ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F2(x, y) = 1 1 3 3 2 y y x x h x x ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or F2(x, y) = 1 1 4 2 3 x x y y h y y ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F3(x, y) = 0 0 5 cos y y x x h x x ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F4(x, y) ≠ 6 n y x h x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N or F4 (x, y) ≠ 7 n x y h ⎛y ⎞ ⎜ ⎟ ⎝ ⎠ , for any n ∈ N Therefore, a function F (x, y) is a homogeneous function of degree n if F(x, y) = or n n y x x g y h x y ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ A differential equation of the form dy dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero To solve a homogeneous differential equation of the type ( ) F , dy x y dx = = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (1) We make the substitution y = v x
1	4749-4752	To solve a homogeneous differential equation of the type ( ) F , dy x y dx = = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (1) We make the substitution y = v x (2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx
1	4750-4753	(1) We make the substitution y = v x (2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx (3) Substituting the value of dy dx from equation (3) in equation (1), we get © NCERT not to be republished DIFFERENTIAL EQUATIONS 399 dv v +x dx = g (v) or dv x dx = g (v) – v
1	4751-4754	x (2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx (3) Substituting the value of dy dx from equation (3) in equation (1), we get © NCERT not to be republished DIFFERENTIAL EQUATIONS 399 dv v +x dx = g (v) or dv x dx = g (v) – v (4) Separating the variables in equation (4), we get ( ) dv g v −v = dx x
1	4752-4755	(2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx (3) Substituting the value of dy dx from equation (3) in equation (1), we get © NCERT not to be republished DIFFERENTIAL EQUATIONS 399 dv v +x dx = g (v) or dv x dx = g (v) – v (4) Separating the variables in equation (4), we get ( ) dv g v −v = dx x (5) Integrating both sides of equation (5), we get ( ) dv g v −v ∫ = 1 C xdx + ∫
1	4753-4756	(3) Substituting the value of dy dx from equation (3) in equation (1), we get © NCERT not to be republished DIFFERENTIAL EQUATIONS 399 dv v +x dx = g (v) or dv x dx = g (v) – v (4) Separating the variables in equation (4), we get ( ) dv g v −v = dx x (5) Integrating both sides of equation (5), we get ( ) dv g v −v ∫ = 1 C xdx + ∫ (6) Equation (6) gives general solution (primitive) of the differential equation (1) when we replace v by y x
1	4754-4757	(4) Separating the variables in equation (4), we get ( ) dv g v −v = dx x (5) Integrating both sides of equation (5), we get ( ) dv g v −v ∫ = 1 C xdx + ∫ (6) Equation (6) gives general solution (primitive) of the differential equation (1) when we replace v by y x �Note If the homogeneous differential equation is in the form F( , ) dx x y dy = where, F (x, y) is homogenous function of degree zero, then we make substitution x y =v i
1	4755-4758	(5) Integrating both sides of equation (5), we get ( ) dv g v −v ∫ = 1 C xdx + ∫ (6) Equation (6) gives general solution (primitive) of the differential equation (1) when we replace v by y x �Note If the homogeneous differential equation is in the form F( , ) dx x y dy = where, F (x, y) is homogenous function of degree zero, then we make substitution x y =v i e
1	4756-4759	(6) Equation (6) gives general solution (primitive) of the differential equation (1) when we replace v by y x �Note If the homogeneous differential equation is in the form F( , ) dx x y dy = where, F (x, y) is homogenous function of degree zero, then we make substitution x y =v i e , x = vy and we proceed further to find the general solution as discussed above by writing F( , )
1	4757-4760	�Note If the homogeneous differential equation is in the form F( , ) dx x y dy = where, F (x, y) is homogenous function of degree zero, then we make substitution x y =v i e , x = vy and we proceed further to find the general solution as discussed above by writing F( , ) dx x x y h dy ⎛y ⎞ = = ⎜ ⎟ ⎝ ⎠ Example 15 Show that the differential equation (x – y) dy dx = x + 2y is homogeneous and solve it
1	4758-4761	e , x = vy and we proceed further to find the general solution as discussed above by writing F( , ) dx x x y h dy ⎛y ⎞ = = ⎜ ⎟ ⎝ ⎠ Example 15 Show that the differential equation (x – y) dy dx = x + 2y is homogeneous and solve it Solution The given differential equation can be expressed as dy dx = 2 x y x +y −
1	4759-4762	, x = vy and we proceed further to find the general solution as discussed above by writing F( , ) dx x x y h dy ⎛y ⎞ = = ⎜ ⎟ ⎝ ⎠ Example 15 Show that the differential equation (x – y) dy dx = x + 2y is homogeneous and solve it Solution The given differential equation can be expressed as dy dx = 2 x y x +y − (1) Let F(x, y) = 2 x y x y Now F(λx, λy) = 0 ( 2 ) ( , ) ( ) x y F x y x y © NCERT not to be republished MATHEMATICS 400 Therefore, F(x, y) is a homogenous function of degree zero
1	4760-4763	dx x x y h dy ⎛y ⎞ = = ⎜ ⎟ ⎝ ⎠ Example 15 Show that the differential equation (x – y) dy dx = x + 2y is homogeneous and solve it Solution The given differential equation can be expressed as dy dx = 2 x y x +y − (1) Let F(x, y) = 2 x y x y Now F(λx, λy) = 0 ( 2 ) ( , ) ( ) x y F x y x y © NCERT not to be republished MATHEMATICS 400 Therefore, F(x, y) is a homogenous function of degree zero So, the given differential equation is a homogenous differential equation
1	4761-4764	Solution The given differential equation can be expressed as dy dx = 2 x y x +y − (1) Let F(x, y) = 2 x y x y Now F(λx, λy) = 0 ( 2 ) ( , ) ( ) x y F x y x y © NCERT not to be republished MATHEMATICS 400 Therefore, F(x, y) is a homogenous function of degree zero So, the given differential equation is a homogenous differential equation Alternatively, 2 1 1 y dy yx dx x ⎛ ⎞ ⎜+ ⎟ = ⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠
1	4762-4765	(1) Let F(x, y) = 2 x y x y Now F(λx, λy) = 0 ( 2 ) ( , ) ( ) x y F x y x y © NCERT not to be republished MATHEMATICS 400 Therefore, F(x, y) is a homogenous function of degree zero So, the given differential equation is a homogenous differential equation Alternatively, 2 1 1 y dy yx dx x ⎛ ⎞ ⎜+ ⎟ = ⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (2) R
1	4763-4766	So, the given differential equation is a homogenous differential equation Alternatively, 2 1 1 y dy yx dx x ⎛ ⎞ ⎜+ ⎟ = ⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (2) R H
1	4764-4767	Alternatively, 2 1 1 y dy yx dx x ⎛ ⎞ ⎜+ ⎟ = ⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ = y g x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (2) R H S
1	4765-4768	(2) R H S of differential equation (2) is of the form y g x and so it is a homogeneous function of degree zero
1	4766-4769	H S of differential equation (2) is of the form y g x and so it is a homogeneous function of degree zero Therefore, equation (1) is a homogeneous differential equation
1	4767-4770	S of differential equation (2) is of the form y g x and so it is a homogeneous function of degree zero Therefore, equation (1) is a homogeneous differential equation To solve it we make the substitution y = vx
1	4768-4771	of differential equation (2) is of the form y g x and so it is a homogeneous function of degree zero Therefore, equation (1) is a homogeneous differential equation To solve it we make the substitution y = vx (3) Differentiating equation (3) with respect to, x we get dy dx = dv v +x dx
1	4769-4772	Therefore, equation (1) is a homogeneous differential equation To solve it we make the substitution y = vx (3) Differentiating equation (3) with respect to, x we get dy dx = dv v +x dx (4) Substituting the value of y and dy dx in equation (1) we get dv v +x dx = 1 12 +vv − or dv x dx = 1 12 v v +v − − or dv x dx = 2 1 v1 vv or 2 1 1 v dv v v = dx x Integrating both sides of equation (5), we get 2 1 1 v dv v v = xdx or 2 1 2 1 3 2 1 v dv v v = – log \|x\| + C1 © NCERT not to be republished DIFFERENTIAL EQUATIONS 401 or 1 2 2 1 2 1 3 1 log C 2 2 1 1 v dv dv x v v v v or 2 1 2 2 1 3 1 log 1 log C 2 2 1 3 2 2 v v dv x v or 2 1 1 1 3 2 2 1 log 1
1	4770-4773	To solve it we make the substitution y = vx (3) Differentiating equation (3) with respect to, x we get dy dx = dv v +x dx (4) Substituting the value of y and dy dx in equation (1) we get dv v +x dx = 1 12 +vv − or dv x dx = 1 12 v v +v − − or dv x dx = 2 1 v1 vv or 2 1 1 v dv v v = dx x Integrating both sides of equation (5), we get 2 1 1 v dv v v = xdx or 2 1 2 1 3 2 1 v dv v v = – log \|x\| + C1 © NCERT not to be republished DIFFERENTIAL EQUATIONS 401 or 1 2 2 1 2 1 3 1 log C 2 2 1 1 v dv dv x v v v v or 2 1 2 2 1 3 1 log 1 log C 2 2 1 3 2 2 v v dv x v or 2 1 1 1 3 2 2 1 log 1 tan log C 2 2 3 3 v v v x or 2 2 1 1 1 1 2 1 log 1 log 3 tan C 2 2 3 v v v x (Why
1	4771-4774	(3) Differentiating equation (3) with respect to, x we get dy dx = dv v +x dx (4) Substituting the value of y and dy dx in equation (1) we get dv v +x dx = 1 12 +vv − or dv x dx = 1 12 v v +v − − or dv x dx = 2 1 v1 vv or 2 1 1 v dv v v = dx x Integrating both sides of equation (5), we get 2 1 1 v dv v v = xdx or 2 1 2 1 3 2 1 v dv v v = – log \|x\| + C1 © NCERT not to be republished DIFFERENTIAL EQUATIONS 401 or 1 2 2 1 2 1 3 1 log C 2 2 1 1 v dv dv x v v v v or 2 1 2 2 1 3 1 log 1 log C 2 2 1 3 2 2 v v dv x v or 2 1 1 1 3 2 2 1 log 1 tan log C 2 2 3 3 v v v x or 2 2 1 1 1 1 2 1 log 1 log 3 tan C 2 2 3 v v v x (Why ) Replacing v by y x , we get or 2 2 1 1 2 1 1 2 log 1 log 3 tan C 2 2 3 y y y x x x x x or 2 2 1 1 2 1 2 log 1 3 tan C 2 3 y y y x x x x x − ⎛ ⎞ + ⎛ ⎞ + + = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or 2 2 1 1 2 log ( ) 2 3 tan 2C y3 x y xy x x − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ or 2 2 1 2 log ( ) 2 3 tan C 3 − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ x y x xy y x which is the general solution of the differential equation (1) Example 16 Show that the differential equation cos cos y dy y x y x x dx x ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ is homogeneous and solve it
1	4772-4775	(4) Substituting the value of y and dy dx in equation (1) we get dv v +x dx = 1 12 +vv − or dv x dx = 1 12 v v +v − − or dv x dx = 2 1 v1 vv or 2 1 1 v dv v v = dx x Integrating both sides of equation (5), we get 2 1 1 v dv v v = xdx or 2 1 2 1 3 2 1 v dv v v = – log \|x\| + C1 © NCERT not to be republished DIFFERENTIAL EQUATIONS 401 or 1 2 2 1 2 1 3 1 log C 2 2 1 1 v dv dv x v v v v or 2 1 2 2 1 3 1 log 1 log C 2 2 1 3 2 2 v v dv x v or 2 1 1 1 3 2 2 1 log 1 tan log C 2 2 3 3 v v v x or 2 2 1 1 1 1 2 1 log 1 log 3 tan C 2 2 3 v v v x (Why ) Replacing v by y x , we get or 2 2 1 1 2 1 1 2 log 1 log 3 tan C 2 2 3 y y y x x x x x or 2 2 1 1 2 1 2 log 1 3 tan C 2 3 y y y x x x x x − ⎛ ⎞ + ⎛ ⎞ + + = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or 2 2 1 1 2 log ( ) 2 3 tan 2C y3 x y xy x x − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ or 2 2 1 2 log ( ) 2 3 tan C 3 − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ x y x xy y x which is the general solution of the differential equation (1) Example 16 Show that the differential equation cos cos y dy y x y x x dx x ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ is homogeneous and solve it Solution The given differential equation can be written as dy dx = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠
1	4773-4776	tan log C 2 2 3 3 v v v x or 2 2 1 1 1 1 2 1 log 1 log 3 tan C 2 2 3 v v v x (Why ) Replacing v by y x , we get or 2 2 1 1 2 1 1 2 log 1 log 3 tan C 2 2 3 y y y x x x x x or 2 2 1 1 2 1 2 log 1 3 tan C 2 3 y y y x x x x x − ⎛ ⎞ + ⎛ ⎞ + + = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or 2 2 1 1 2 log ( ) 2 3 tan 2C y3 x y xy x x − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ or 2 2 1 2 log ( ) 2 3 tan C 3 − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ x y x xy y x which is the general solution of the differential equation (1) Example 16 Show that the differential equation cos cos y dy y x y x x dx x ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ is homogeneous and solve it Solution The given differential equation can be written as dy dx = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠ (1) © NCERT not to be republished MATHEMATICS 402 It is a differential equation of the form F( , ) dy x y dx =
1	4774-4777	) Replacing v by y x , we get or 2 2 1 1 2 1 1 2 log 1 log 3 tan C 2 2 3 y y y x x x x x or 2 2 1 1 2 1 2 log 1 3 tan C 2 3 y y y x x x x x − ⎛ ⎞ + ⎛ ⎞ + + = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or 2 2 1 1 2 log ( ) 2 3 tan 2C y3 x y xy x x − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ or 2 2 1 2 log ( ) 2 3 tan C 3 − + ⎛ ⎞ + + = + ⎜ ⎟ ⎝ ⎠ x y x xy y x which is the general solution of the differential equation (1) Example 16 Show that the differential equation cos cos y dy y x y x x dx x ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ is homogeneous and solve it Solution The given differential equation can be written as dy dx = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠ (1) © NCERT not to be republished MATHEMATICS 402 It is a differential equation of the form F( , ) dy x y dx = Here F(x, y) = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠ Replacing x by λx and y by λy, we get F(λx, λy) = 0 [ cos ] [F( , )] cos y y x x x y y x x ⎛ ⎞ λ + ⎜ ⎟ ⎝ ⎠ = λ ⎛ ⎞ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero
1	4775-4778	Solution The given differential equation can be written as dy dx = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠ (1) © NCERT not to be republished MATHEMATICS 402 It is a differential equation of the form F( , ) dy x y dx = Here F(x, y) = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠ Replacing x by λx and y by λy, we get F(λx, λy) = 0 [ cos ] [F( , )] cos y y x x x y y x x ⎛ ⎞ λ + ⎜ ⎟ ⎝ ⎠ = λ ⎛ ⎞ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero Therefore, the given differential equation is a homogeneous differential equation
1	4776-4779	(1) © NCERT not to be republished MATHEMATICS 402 It is a differential equation of the form F( , ) dy x y dx = Here F(x, y) = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠ Replacing x by λx and y by λy, we get F(λx, λy) = 0 [ cos ] [F( , )] cos y y x x x y y x x ⎛ ⎞ λ + ⎜ ⎟ ⎝ ⎠ = λ ⎛ ⎞ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero Therefore, the given differential equation is a homogeneous differential equation To solve it we make the substitution y = vx
1	4777-4780	Here F(x, y) = cos cos y y x x y x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎛⎠ ⎞ ⎜ ⎟ ⎝ ⎠ Replacing x by λx and y by λy, we get F(λx, λy) = 0 [ cos ] [F( , )] cos y y x x x y y x x ⎛ ⎞ λ + ⎜ ⎟ ⎝ ⎠ = λ ⎛ ⎞ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero Therefore, the given differential equation is a homogeneous differential equation To solve it we make the substitution y = vx (2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx
1	4778-4781	Therefore, the given differential equation is a homogeneous differential equation To solve it we make the substitution y = vx (2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx (3) Substituting the value of y and dy dx in equation (1), we get dv v +x dx = cos 1 vcos vv + or dv x dx = cos 1 vcos v v v + − or dv x dx = 1 cosv or cosv dv = dx x Therefore ∫cosv dv = 1 dx x∫ © NCERT not to be republished DIFFERENTIAL EQUATIONS 403 or sin v = log \|x\| + log \|C\| or sin v = log \|Cx\| Replacing v by y x , we get sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = log \|Cx\| which is the general solution of the differential equation (1)
1	4779-4782	To solve it we make the substitution y = vx (2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx (3) Substituting the value of y and dy dx in equation (1), we get dv v +x dx = cos 1 vcos vv + or dv x dx = cos 1 vcos v v v + − or dv x dx = 1 cosv or cosv dv = dx x Therefore ∫cosv dv = 1 dx x∫ © NCERT not to be republished DIFFERENTIAL EQUATIONS 403 or sin v = log \|x\| + log \|C\| or sin v = log \|Cx\| Replacing v by y x , we get sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = log \|Cx\| which is the general solution of the differential equation (1) Example 17 Show that the differential equation 2 2 0 x x y y y e dx y x e dy ⎛ ⎞ ⎜ ⎟ + − = ⎝ ⎠ is homogeneous and find its particular solution, given that, x = 0 when y = 1
1	4780-4783	(2) Differentiating equation (2) with respect to x, we get dy dx = dv v +x dx (3) Substituting the value of y and dy dx in equation (1), we get dv v +x dx = cos 1 vcos vv + or dv x dx = cos 1 vcos v v v + − or dv x dx = 1 cosv or cosv dv = dx x Therefore ∫cosv dv = 1 dx x∫ © NCERT not to be republished DIFFERENTIAL EQUATIONS 403 or sin v = log \|x\| + log \|C\| or sin v = log \|Cx\| Replacing v by y x , we get sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = log \|Cx\| which is the general solution of the differential equation (1) Example 17 Show that the differential equation 2 2 0 x x y y y e dx y x e dy ⎛ ⎞ ⎜ ⎟ + − = ⎝ ⎠ is homogeneous and find its particular solution, given that, x = 0 when y = 1 Solution The given differential equation can be written as dx dy = 2 2 x y x y x e y y e −
1	4781-4784	(3) Substituting the value of y and dy dx in equation (1), we get dv v +x dx = cos 1 vcos vv + or dv x dx = cos 1 vcos v v v + − or dv x dx = 1 cosv or cosv dv = dx x Therefore ∫cosv dv = 1 dx x∫ © NCERT not to be republished DIFFERENTIAL EQUATIONS 403 or sin v = log \|x\| + log \|C\| or sin v = log \|Cx\| Replacing v by y x , we get sin y x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = log \|Cx\| which is the general solution of the differential equation (1) Example 17 Show that the differential equation 2 2 0 x x y y y e dx y x e dy ⎛ ⎞ ⎜ ⎟ + − = ⎝ ⎠ is homogeneous and find its particular solution, given that, x = 0 when y = 1 Solution The given differential equation can be written as dx dy = 2 2 x y x y x e y y e − (1) Let F(x, y) = 2 2 x y x y xe y ye − Then F(λx, λy) = 0 2 [F( , )] 2 x y x y xe y x y ye ⎛ ⎞ ⎜ ⎟ λ − ⎜ ⎟ ⎝ ⎠ =λ ⎛ ⎞ ⎜ ⎟ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero
1	4782-4785	Example 17 Show that the differential equation 2 2 0 x x y y y e dx y x e dy ⎛ ⎞ ⎜ ⎟ + − = ⎝ ⎠ is homogeneous and find its particular solution, given that, x = 0 when y = 1 Solution The given differential equation can be written as dx dy = 2 2 x y x y x e y y e − (1) Let F(x, y) = 2 2 x y x y xe y ye − Then F(λx, λy) = 0 2 [F( , )] 2 x y x y xe y x y ye ⎛ ⎞ ⎜ ⎟ λ − ⎜ ⎟ ⎝ ⎠ =λ ⎛ ⎞ ⎜ ⎟ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero Therefore, the given differential equation is a homogeneous differential equation
1	4783-4786	Solution The given differential equation can be written as dx dy = 2 2 x y x y x e y y e − (1) Let F(x, y) = 2 2 x y x y xe y ye − Then F(λx, λy) = 0 2 [F( , )] 2 x y x y xe y x y ye ⎛ ⎞ ⎜ ⎟ λ − ⎜ ⎟ ⎝ ⎠ =λ ⎛ ⎞ ⎜ ⎟ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero Therefore, the given differential equation is a homogeneous differential equation To solve it, we make the substitution x = vy
1	4784-4787	(1) Let F(x, y) = 2 2 x y x y xe y ye − Then F(λx, λy) = 0 2 [F( , )] 2 x y x y xe y x y ye ⎛ ⎞ ⎜ ⎟ λ − ⎜ ⎟ ⎝ ⎠ =λ ⎛ ⎞ ⎜ ⎟ λ⎜ ⎟ ⎝ ⎠ Thus, F(x, y) is a homogeneous function of degree zero Therefore, the given differential equation is a homogeneous differential equation To solve it, we make the substitution x = vy (2) Differentiating equation (2) with respect to y, we get dx dy = + dv v y dy © NCERT not to be republished MATHEMATICS 404 Substituting the value of xand dx dy in equation (1), we get dv v +y dy = 2 1 2 v v ev e − or dv y dy = 2 1 2 v v ev v e − − or dv y dy = 2 ve1 − or 2ev dv = dy y − or 2 ve ⋅dv ∫ = ydy −∫ or 2 ev = – log \|y\| + C and replacing v by x y , we get 2 yx e + log \|y\| = C
1	4785-4788	Therefore, the given differential equation is a homogeneous differential equation To solve it, we make the substitution x = vy (2) Differentiating equation (2) with respect to y, we get dx dy = + dv v y dy © NCERT not to be republished MATHEMATICS 404 Substituting the value of xand dx dy in equation (1), we get dv v +y dy = 2 1 2 v v ev e − or dv y dy = 2 1 2 v v ev v e − − or dv y dy = 2 ve1 − or 2ev dv = dy y − or 2 ve ⋅dv ∫ = ydy −∫ or 2 ev = – log \|y\| + C and replacing v by x y , we get 2 yx e + log \|y\| = C (3) Substituting x = 0 and y = 1 in equation (3), we get 2 e0 + log \|1\| = C ⇒ C = 2 Substituting the value of C in equation (3), we get 2 yx e + log \|y \| = 2 which is the particular solution of the given differential equation
1	4786-4789	To solve it, we make the substitution x = vy (2) Differentiating equation (2) with respect to y, we get dx dy = + dv v y dy © NCERT not to be republished MATHEMATICS 404 Substituting the value of xand dx dy in equation (1), we get dv v +y dy = 2 1 2 v v ev e − or dv y dy = 2 1 2 v v ev v e − − or dv y dy = 2 ve1 − or 2ev dv = dy y − or 2 ve ⋅dv ∫ = ydy −∫ or 2 ev = – log \|y\| + C and replacing v by x y , we get 2 yx e + log \|y\| = C (3) Substituting x = 0 and y = 1 in equation (3), we get 2 e0 + log \|1\| = C ⇒ C = 2 Substituting the value of C in equation (3), we get 2 yx e + log \|y \| = 2 which is the particular solution of the given differential equation Example 18 Show that the family of curves for which the slope of the tangent at any point (x, y) on it is 2 2 2 x y +xy , is given by x2 – y2 = cx
1	4787-4790	(2) Differentiating equation (2) with respect to y, we get dx dy = + dv v y dy © NCERT not to be republished MATHEMATICS 404 Substituting the value of xand dx dy in equation (1), we get dv v +y dy = 2 1 2 v v ev e − or dv y dy = 2 1 2 v v ev v e − − or dv y dy = 2 ve1 − or 2ev dv = dy y − or 2 ve ⋅dv ∫ = ydy −∫ or 2 ev = – log \|y\| + C and replacing v by x y , we get 2 yx e + log \|y\| = C (3) Substituting x = 0 and y = 1 in equation (3), we get 2 e0 + log \|1\| = C ⇒ C = 2 Substituting the value of C in equation (3), we get 2 yx e + log \|y \| = 2 which is the particular solution of the given differential equation Example 18 Show that the family of curves for which the slope of the tangent at any point (x, y) on it is 2 2 2 x y +xy , is given by x2 – y2 = cx Solution We know that the slope of the tangent at any point on a curve is dy dx
1	4788-4791	(3) Substituting x = 0 and y = 1 in equation (3), we get 2 e0 + log \|1\| = C ⇒ C = 2 Substituting the value of C in equation (3), we get 2 yx e + log \|y \| = 2 which is the particular solution of the given differential equation Example 18 Show that the family of curves for which the slope of the tangent at any point (x, y) on it is 2 2 2 x y +xy , is given by x2 – y2 = cx Solution We know that the slope of the tangent at any point on a curve is dy dx Therefore, dy dx = 2 2 2 x y xy + © NCERT not to be republished DIFFERENTIAL EQUATIONS 405 or dy dx = 2 2 1 2 yxy x +
1	4789-4792	Example 18 Show that the family of curves for which the slope of the tangent at any point (x, y) on it is 2 2 2 x y +xy , is given by x2 – y2 = cx Solution We know that the slope of the tangent at any point on a curve is dy dx Therefore, dy dx = 2 2 2 x y xy + © NCERT not to be republished DIFFERENTIAL EQUATIONS 405 or dy dx = 2 2 1 2 yxy x + (1) Clearly, (1) is a homogenous differential equation
1	4790-4793	Solution We know that the slope of the tangent at any point on a curve is dy dx Therefore, dy dx = 2 2 2 x y xy + © NCERT not to be republished DIFFERENTIAL EQUATIONS 405 or dy dx = 2 2 1 2 yxy x + (1) Clearly, (1) is a homogenous differential equation To solve it we make substitution y = vx Differentiating y = vx with respect to x, we get dy dx = dv v x dx + or dv v +x dx = 2 1 2 vv + or dv x dx = 2 1 2 vv − 122 v −vdv = dx x or 22 v1 v −dv = xdx − Therefore 22 v1 ∫v −dv = x1 dx −∫ or log \|v2 – 1 \| = – log \|x\| + log \|C1\| or log \|(v2 – 1) (x)\| = log \|C1\| or (v2 – 1) x = ± C1 Replacing v by y x , we get 2 2 1 y x x ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ = ± C1 or (y2 – x2) = ± C1 x or x2 – y2 = Cx © NCERT not to be republished MATHEMATICS 406 EXERCISE 9
1	4791-4794	Therefore, dy dx = 2 2 2 x y xy + © NCERT not to be republished DIFFERENTIAL EQUATIONS 405 or dy dx = 2 2 1 2 yxy x + (1) Clearly, (1) is a homogenous differential equation To solve it we make substitution y = vx Differentiating y = vx with respect to x, we get dy dx = dv v x dx + or dv v +x dx = 2 1 2 vv + or dv x dx = 2 1 2 vv − 122 v −vdv = dx x or 22 v1 v −dv = xdx − Therefore 22 v1 ∫v −dv = x1 dx −∫ or log \|v2 – 1 \| = – log \|x\| + log \|C1\| or log \|(v2 – 1) (x)\| = log \|C1\| or (v2 – 1) x = ± C1 Replacing v by y x , we get 2 2 1 y x x ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ = ± C1 or (y2 – x2) = ± C1 x or x2 – y2 = Cx © NCERT not to be republished MATHEMATICS 406 EXERCISE 9 5 In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them
1	4792-4795	(1) Clearly, (1) is a homogenous differential equation To solve it we make substitution y = vx Differentiating y = vx with respect to x, we get dy dx = dv v x dx + or dv v +x dx = 2 1 2 vv + or dv x dx = 2 1 2 vv − 122 v −vdv = dx x or 22 v1 v −dv = xdx − Therefore 22 v1 ∫v −dv = x1 dx −∫ or log \|v2 – 1 \| = – log \|x\| + log \|C1\| or log \|(v2 – 1) (x)\| = log \|C1\| or (v2 – 1) x = ± C1 Replacing v by y x , we get 2 2 1 y x x ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ = ± C1 or (y2 – x2) = ± C1 x or x2 – y2 = Cx © NCERT not to be republished MATHEMATICS 406 EXERCISE 9 5 In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them 1
1	4793-4796	To solve it we make substitution y = vx Differentiating y = vx with respect to x, we get dy dx = dv v x dx + or dv v +x dx = 2 1 2 vv + or dv x dx = 2 1 2 vv − 122 v −vdv = dx x or 22 v1 v −dv = xdx − Therefore 22 v1 ∫v −dv = x1 dx −∫ or log \|v2 – 1 \| = – log \|x\| + log \|C1\| or log \|(v2 – 1) (x)\| = log \|C1\| or (v2 – 1) x = ± C1 Replacing v by y x , we get 2 2 1 y x x ⎛ −⎞ ⎜ ⎟ ⎝ ⎠ = ± C1 or (y2 – x2) = ± C1 x or x2 – y2 = Cx © NCERT not to be republished MATHEMATICS 406 EXERCISE 9 5 In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them 1 (x2 + xy) dy = (x2 + y2) dx 2
1	4794-4797	5 In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them 1 (x2 + xy) dy = (x2 + y2) dx 2 x y y +x ′ = 3
1	4795-4798	1 (x2 + xy) dy = (x2 + y2) dx 2 x y y +x ′ = 3 (x – y) dy – (x + y) dx = 0 4
1	4796-4799	(x2 + xy) dy = (x2 + y2) dx 2 x y y +x ′ = 3 (x – y) dy – (x + y) dx = 0 4 (x2 – y2) dx + 2xy dy = 0 5
1	4797-4800	x y y +x ′ = 3 (x – y) dy – (x + y) dx = 0 4 (x2 – y2) dx + 2xy dy = 0 5 2 2 22 xdy x y xy dx = − + 6
1	4798-4801	(x – y) dy – (x + y) dx = 0 4 (x2 – y2) dx + 2xy dy = 0 5 2 2 22 xdy x y xy dx = − + 6 x dy – y dx = 2 2 x y dx + 7
1	4799-4802	(x2 – y2) dx + 2xy dy = 0 5 2 2 22 xdy x y xy dx = − + 6 x dy – y dx = 2 2 x y dx + 7 cos sin sin cos y y y y x y y dx y x x dy x x x x ⎧ ⎫ ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + = − ⎨ ⎬ ⎨ ⎬ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎩ ⎭ ⎩ ⎭ 8
1	4800-4803	2 2 22 xdy x y xy dx = − + 6 x dy – y dx = 2 2 x y dx + 7 cos sin sin cos y y y y x y y dx y x x dy x x x x ⎧ ⎫ ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + = − ⎨ ⎬ ⎨ ⎬ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎩ ⎭ ⎩ ⎭ 8 sin 0 dy y x y x dx ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ 9
1	4801-4804	x dy – y dx = 2 2 x y dx + 7 cos sin sin cos y y y y x y y dx y x x dy x x x x ⎧ ⎫ ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + = − ⎨ ⎬ ⎨ ⎬ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎩ ⎭ ⎩ ⎭ 8 sin 0 dy y x y x dx ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ 9 log 2 0 y y dx x dy x dy ⎛x ⎞ + − = ⎜ ⎟ ⎝ ⎠ 10
1	4802-4805	cos sin sin cos y y y y x y y dx y x x dy x x x x ⎧ ⎫ ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + = − ⎨ ⎬ ⎨ ⎬ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎩ ⎭ ⎩ ⎭ 8 sin 0 dy y x y x dx ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ 9 log 2 0 y y dx x dy x dy ⎛x ⎞ + − = ⎜ ⎟ ⎝ ⎠ 10 1 1 0 x x y y x e dx e dy y For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition: 11
1	4803-4806	sin 0 dy y x y x dx ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ 9 log 2 0 y y dx x dy x dy ⎛x ⎞ + − = ⎜ ⎟ ⎝ ⎠ 10 1 1 0 x x y y x e dx e dy y For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition: 11 (x + y) dy + (x – y) dx = 0; y = 1 when x = 1 12
1	4804-4807	log 2 0 y y dx x dy x dy ⎛x ⎞ + − = ⎜ ⎟ ⎝ ⎠ 10 1 1 0 x x y y x e dx e dy y For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition: 11 (x + y) dy + (x – y) dx = 0; y = 1 when x = 1 12 x2 dy + (xy + y2) dx = 0; y = 1 when x = 1 13
1	4805-4808	1 1 0 x x y y x e dx e dy y For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition: 11 (x + y) dy + (x – y) dx = 0; y = 1 when x = 1 12 x2 dy + (xy + y2) dx = 0; y = 1 when x = 1 13 sin2 0; 4 y x y dx x dy y x when x = 1 14
1	4806-4809	(x + y) dy + (x – y) dx = 0; y = 1 when x = 1 12 x2 dy + (xy + y2) dx = 0; y = 1 when x = 1 13 sin2 0; 4 y x y dx x dy y x when x = 1 14 cosec 0 dy y y dx x ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ ; y = 0 when x = 1 15
1	4807-4810	x2 dy + (xy + y2) dx = 0; y = 1 when x = 1 13 sin2 0; 4 y x y dx x dy y x when x = 1 14 cosec 0 dy y y dx x ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ ; y = 0 when x = 1 15 2 2 2 2 0 dy xy y x dx + − = ; y = 2 when x = 1 16
1	4808-4811	sin2 0; 4 y x y dx x dy y x when x = 1 14 cosec 0 dy y y dx x ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ ; y = 0 when x = 1 15 2 2 2 2 0 dy xy y x dx + − = ; y = 2 when x = 1 16 A homogeneous differential equation of the from dx hx dy ⎛y ⎞ = ⎜ ⎟ ⎝ ⎠ can be solved by making the substitution
1	4809-4812	cosec 0 dy y y dx x ⎛x ⎞ − + = ⎜ ⎟ ⎝ ⎠ ; y = 0 when x = 1 15 2 2 2 2 0 dy xy y x dx + − = ; y = 2 when x = 1 16 A homogeneous differential equation of the from dx hx dy ⎛y ⎞ = ⎜ ⎟ ⎝ ⎠ can be solved by making the substitution (A) y = vx (B) v = yx (C) x = vy (D) x = v © NCERT not to be republished DIFFERENTIAL EQUATIONS 407 17
1	4810-4813	2 2 2 2 0 dy xy y x dx + − = ; y = 2 when x = 1 16 A homogeneous differential equation of the from dx hx dy ⎛y ⎞ = ⎜ ⎟ ⎝ ⎠ can be solved by making the substitution (A) y = vx (B) v = yx (C) x = vy (D) x = v © NCERT not to be republished DIFFERENTIAL EQUATIONS 407 17 Which of the following is a homogeneous differential equation
1	4811-4814	A homogeneous differential equation of the from dx hx dy ⎛y ⎞ = ⎜ ⎟ ⎝ ⎠ can be solved by making the substitution (A) y = vx (B) v = yx (C) x = vy (D) x = v © NCERT not to be republished DIFFERENTIAL EQUATIONS 407 17 Which of the following is a homogeneous differential equation (A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (B) (xy) dx – (x3 + y3) dy = 0 (C) (x3 + 2y2) dx + 2xy dy = 0 (D) y2 dx + (x2 – xy – y2) dy = 0 9
1	4812-4815	(A) y = vx (B) v = yx (C) x = vy (D) x = v © NCERT not to be republished DIFFERENTIAL EQUATIONS 407 17 Which of the following is a homogeneous differential equation (A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (B) (xy) dx – (x3 + y3) dy = 0 (C) (x3 + 2y2) dx + 2xy dy = 0 (D) y2 dx + (x2 – xy – y2) dy = 0 9 5
1	4813-4816	Which of the following is a homogeneous differential equation (A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (B) (xy) dx – (x3 + y3) dy = 0 (C) (x3 + 2y2) dx + 2xy dy = 0 (D) y2 dx + (x2 – xy – y2) dy = 0 9 5 3 Linear differential equations A differential equation of the from P dy dx +y = Q where, P and Q are constants or functions of x only, is known as a first order linear differential equation
1	4814-4817	(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (B) (xy) dx – (x3 + y3) dy = 0 (C) (x3 + 2y2) dx + 2xy dy = 0 (D) y2 dx + (x2 – xy – y2) dy = 0 9 5 3 Linear differential equations A differential equation of the from P dy dx +y = Q where, P and Q are constants or functions of x only, is known as a first order linear differential equation Some examples of the first order linear differential equation are dy dx +y = sin x 1 dy y dx ⎛x ⎞ + ⎜ ⎟ ⎝ ⎠ = ex log dy y dx x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ = 1 x Another form of first order linear differential equation is 1P dx x dy + = Q1 where, P1 and Q1 are constants or functions of y only
1	4815-4818	5 3 Linear differential equations A differential equation of the from P dy dx +y = Q where, P and Q are constants or functions of x only, is known as a first order linear differential equation Some examples of the first order linear differential equation are dy dx +y = sin x 1 dy y dx ⎛x ⎞ + ⎜ ⎟ ⎝ ⎠ = ex log dy y dx x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ = 1 x Another form of first order linear differential equation is 1P dx x dy + = Q1 where, P1 and Q1 are constants or functions of y only Some examples of this type of differential equation are dx dy +x = cos y 2 dx x dy +−y = y2e – y To solve the first order linear differential equation of the type P dy y dx = Q
1	4816-4819	3 Linear differential equations A differential equation of the from P dy dx +y = Q where, P and Q are constants or functions of x only, is known as a first order linear differential equation Some examples of the first order linear differential equation are dy dx +y = sin x 1 dy y dx ⎛x ⎞ + ⎜ ⎟ ⎝ ⎠ = ex log dy y dx x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ = 1 x Another form of first order linear differential equation is 1P dx x dy + = Q1 where, P1 and Q1 are constants or functions of y only Some examples of this type of differential equation are dx dy +x = cos y 2 dx x dy +−y = y2e – y To solve the first order linear differential equation of the type P dy y dx = Q (1) Multiply both sides of the equation by a function of x say g (x) to get g(x) dy dx + P
1	4817-4820	Some examples of the first order linear differential equation are dy dx +y = sin x 1 dy y dx ⎛x ⎞ + ⎜ ⎟ ⎝ ⎠ = ex log dy y dx x x ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ = 1 x Another form of first order linear differential equation is 1P dx x dy + = Q1 where, P1 and Q1 are constants or functions of y only Some examples of this type of differential equation are dx dy +x = cos y 2 dx x dy +−y = y2e – y To solve the first order linear differential equation of the type P dy y dx = Q (1) Multiply both sides of the equation by a function of x say g (x) to get g(x) dy dx + P (g(x)) y = Q

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