NCERT Solutions for Class 12 Mathematics Chapter 9: Differential Equations – Free PDF Download

Chapter 9 of Class 12 Maths is Differential Equations. It covers the order and degree of differential equations, forming them, and solving first-order first-degree equations by variable separation, homogeneous methods, and linear equations.

Last Updated: September 23, 2026

Basic Definitions

A differential equation involves derivatives of a dependent variable with respect to an independent variable. The order is the highest derivative present; the degree is the power of the highest-order derivative (when the equation is a polynomial in derivatives).

Formation of Differential Equations

Given a family of curves with n arbitrary constants, differentiating n times and eliminating the constants gives a differential equation of order n.

Variable Separable Method

If the equation can be written as f(y)dy=g(x)dx, integrate both sides separately: ∫f(y)dy=∫g(x)dx+C.

Homogeneous Differential Equations

An equation dy/dx=f(x,y) is homogeneous if f(λx,λy)=f(x,y). Solved by substituting y=vx (so dy/dx=v+x dv/dx), converting to a variable-separable form in v and x.

Linear Differential Equations

A linear equation has the form dy/dx+Py=Q (P,Q functions of x only). Solved using the integrating factor IF=e^{∫P dx}; the solution is y·IF=∫(Q·IF)dx+C.

Exercise 9.1 Solutions (All 12 Questions)

1. Determine order and degree (if defined) of differential equation d⁴y/dx⁴ + sin(y”’) = 0.
Ans: The highest order derivative present is d⁴y/dx⁴, so the order is 4. Since the equation is not a polynomial in its derivatives (it contains sin(y”’)), the degree is not defined.

2. Determine order and degree (if defined) of differential equation y’ + 5y = 0.
Ans: The highest order derivative present is y’, so the order is 1. The equation is a polynomial in y’ with highest power 1, so the degree is 1.

3. Determine order and degree (if defined) of differential equation (ds/dt)⁴ + 3s(d²s/dt²) = 0.
Ans: The highest order derivative present is d²s/dt², so the order is 2. The highest power of this term is 1, so the degree is 1.

4. Determine order and degree (if defined) of differential equation (d²y/dx²)² + cos(dy/dx) = 0.
Ans: The highest order derivative present is d²y/dx², so the order is 2. Since the equation is not a polynomial in its derivatives (it contains cos(dy/dx)), the degree is not defined.

5. Determine order and degree (if defined) of differential equation d²y/dx² = cos 3x + sin 3x.
Ans: The highest order derivative present is d²y/dx², so the order is 2. The equation is a polynomial in this term with highest power 1, so the degree is 1.

6. Determine order and degree (if defined) of differential equation (y”’)² + (y”)³ + (y’)⁴ + y⁵ = 0.
Ans: The highest order derivative present is y”’, so the order is 3. The highest power of y”’ is 2, so the degree is 2.

7. Determine order and degree (if defined) of differential equation y”’ + 2y” + y’ = 0.
Ans: The highest order derivative present is y”’, so the order is 3. The highest power of y”’ is 1, so the degree is 1.

8. Determine order and degree (if defined) of differential equation y’ + y = e^x.
Ans: The highest order derivative present is y’, so the order is 1. The highest power of y’ is 1, so the degree is 1.

9. Determine order and degree (if defined) of differential equation y” + (y’)² + 2y = 0.
Ans: The highest order derivative present is y”, so the order is 2. The highest power of y” is 1, so the degree is 1.

10. Determine order and degree (if defined) of differential equation y” + 2y’ + sin y = 0.
Ans: The highest order derivative present is y”, so the order is 2. The highest power of y” is 1, so the degree is 1.

11. The degree of the differential equation (d²y/dx²)³ + (dy/dx)² + sin(dy/dx) + 1 = 0 is
(A) 3
(B) 2
(C) 1
(D) Not defined
Ans: Since the equation contains sin(dy/dx), it is not a polynomial in its derivatives, so the degree is not defined. The answer is (D)

12. The order of the differential equation 2x²(d²y/dx²) + 3(dy/dx) + y = 0 is
(A) 2
(B) 1
(C) 0
(D) Not defined
Ans: The highest order derivative present is d²y/dx², so the order is 2. The answer is (A)

Exercise 9.2 Solutions (All 12 Questions)

1. Verify that the function y = e^x + 1 is a solution of the differential equation y” – y’ = 0.
Ans: y’ = e^x and y” = e^x. Substituting, y” – y’ = e^x – e^x = 0. Since LHS = RHS, the given function is a solution of the differential equation.

2. Verify that the function y = x² + 2x + C is a solution of the differential equation y’ – 2x – 2 = 0.
Ans: y’ = 2x + 2. Substituting, y’ – 2x – 2 = (2x+2) – 2x – 2 = 0. Since LHS = RHS, the given function is a solution of the differential equation.

3. Verify that the function y = cos x + C is a solution of the differential equation y’ + sin x = 0.
Ans: y’ = -sin x. Substituting, y’ + sin x = -sin x + sin x = 0. Since LHS = RHS, the given function is a solution of the differential equation.

4. Verify that the function y = √(1+x²) is a solution of the differential equation y’ = xy/(1+x²).
Ans: y’ = x/√(1+x²). RHS = xy/(1+x²) = x√(1+x²)/(1+x²) = x/√(1+x²). Since LHS = RHS, the given function is a solution of the differential equation.

5. Verify that the function y = Ax is a solution of the differential equation xy’ = y (x ≠ 0).
Ans: y’ = A. LHS = xy’ = xA. RHS = y = Ax. Since LHS = RHS, the given function is a solution of the differential equation.

6. Verify that the function y = x sin x is a solution of the differential equation xy’ = y + x√(x²-y²) (x ≠ 0 and x > y or x < -y).
Ans: y’ = sin x + x cos x. LHS = xy’ = x sin x + x²cos x. RHS = y + x√(x²-y²) = x sin x + x√(x²-x²sin²x) = x sin x + x²√(1-sin²x) = x sin x + x²cos x. Since LHS = RHS, the given function is a solution of the differential equation.

7. Verify that the function xy = log y + C is a solution of the differential equation y’ = y²/(1-xy) (xy ≠ 1).
Ans: Differentiating xy = log y + C implicitly with respect to x: y + xy’ = y’/y, so y² + xyy’ = y’, giving y’ = y²/(1-xy). Since LHS = RHS, the given function is a solution of the differential equation.

8. Verify that the function y – cos y = x is a solution of the differential equation (y sin y + cos y + x)y’ = y.
Ans: Differentiating y – cos y = x implicitly with respect to x: y’ + y’sin y = 1, so y'(1+sin y) = 1, i.e. y’ = 1/(1+sin y). Since x = y – cos y, the given equation becomes (y sin y + cos y + y – cos y)y’ = y, i.e. y(1+sin y)y’ = y, i.e. (1+sin y)y’ = 1, which matches. Since LHS = RHS, the given function is a solution of the differential equation.

9. Verify that the function x + y = tan⁻¹y is a solution of the differential equation y²y’ + y² + 1 = 0.
Ans: Differentiating x + y = tan⁻¹y implicitly with respect to x: 1 + y’ = y’/(1+y²), so 1 = y’/(1+y²) – y’ = -y²y’/(1+y²), giving y'(1+y²) = -y², i.e. y²y’ + y² + 1 = 0. Since LHS = RHS, the given function is a solution of the differential equation.

10. Verify that the function y = √(a²-x²), x ∈ (-a,a), is a solution of the differential equation x + y(dy/dx) = 0 (y ≠ 0).
Ans: Since y² = a²-x², differentiating implicitly with respect to x gives 2yy’ = -2x, so x + yy’ = 0. Since LHS = RHS, the given function is a solution of the differential equation.

11. The number of arbitrary constants in the general solution of a differential equation of fourth order is
(A) 0
(B) 2
(C) 3
(D) 4
Ans: The number of arbitrary constants in the general solution of a differential equation equals the order of the equation, so a fourth order equation has 4 constants. The answer is (D)

12. The number of arbitrary constants in a particular solution of a differential equation of third order is
(A) 3
(B) 2
(C) 1
(D) 0
Ans: A particular solution is obtained by assigning specific values to all the arbitrary constants of the general solution, so it contains no arbitrary constants. The answer is (D)

Exercise 9.3 Solutions (All 23 Questions)

1. Find the general solution for dy/dx = (1-cos x)/(1+cos x).
Ans: Using half-angle identities, (1-cos x)/(1+cos x) = tan²(x/2) = sec²(x/2) – 1. So dy = [sec²(x/2) – 1]dx. Integrating both sides, y = 2tan(x/2) – x + C

2. Find the general solution for dy/dx = √(4-y²) (-2 < y < 2).
Ans: Separating variables, dy/√(4-y²) = dx. Integrating both sides, sin⁻¹(y/2) = x + C, so y = 2 sin(x + C)

3. Find the general solution for dy/dx + y = 1 (y ≠ 1).
Ans: Separating variables, dy/(1-y) = dx. Integrating both sides, -log(1-y) = x + C, so 1-y = Ae^-x, giving y = 1 – Ae^-x

4. Find the general solution for sec²x tan y dx + sec²y tan x dy = 0.
Ans: Dividing throughout by tan x tan y and integrating both sides, log(tan x) + log(tan y) = log C, so tan x tan y = C

5. Find the general solution for (e^x+e^-x)dy – (e^x-e^-x)dx = 0.
Ans: dy = [(e^x-e^-x)/(e^x+e^-x)]dx. Integrating with the substitution e^x+e^-x = t, y = log(e^x+e^-x) + C

6. Find the general solution for dy/dx = (1+x²)(1+y²).
Ans: Separating variables, dy/(1+y²) = (1+x²)dx. Integrating both sides, tan⁻¹y = x + x³/3 + C

7. Find the general solution for y log y dx – x dy = 0.
Ans: Separating variables, dx/x = dy/(y log y). Integrating with the substitution log y = t, log(log y) = log x + log C, so log y = Cx, giving y = e^Cx

8. Find the general solution for x⁵dy/dx = -y⁵.
Ans: Separating variables, dy/y⁵ = -dx/x⁵. Integrating both sides, y⁻⁴/(-4) = -x⁻⁴/(-4) + C, so x⁻⁴ + y⁻⁴ = A

9. Find the general solution for dy/dx = sin⁻¹x.
Ans: dy = sin⁻¹x dx. Integrating by parts, y = x sin⁻¹x + √(1-x²) + C

10. Find the general solution for e^xtan y dx + (1-e^x)sec²y dy = 0.
Ans: Dividing throughout by (1-e^x)tan y, sec²y/tan y dy = -e^x/(1-e^x) dx. Integrating both sides, log(tan y) = log(1-e^x) + log C, so tan y = C(1-e^x)

11. Find the particular solution of (x³+x²+x+1)dy/dx = 2x²+x satisfying y = 1 when x = 0.
Ans: Factoring, x³+x²+x+1 = (x+1)(x²+1), so dy = [(2x²+x)/((x+1)(x²+1))]dx. Using partial fractions, (2x²+x)/((x+1)(x²+1)) = (1/2)[1/(x+1) + (3x-1)/(x²+1)]. Integrating, y = (1/2)log(x+1) + (3/4)log(x²+1) – (1/2)tan⁻¹x + C. Applying y = 1 at x = 0 gives C = 1, so y = (1/2)log(x+1) + (3/4)log(x²+1) – (1/2)tan⁻¹x + 1

12. Find the particular solution of x(x²-1)dy/dx = 1 satisfying y = 0 when x = 2.
Ans: dy = dx/[x(x-1)(x+1)]. Using partial fractions, 1/[x(x-1)(x+1)] = -1/x + (1/2)/(x-1) + (1/2)/(x+1). Integrating, y = (1/2)log[(x²-1)/x²] + log C. Applying y = 0 at x = 2 gives C² = 4/3, so y = (1/2)log[4(x²-1)/(3x²)]

13. Find the particular solution of cos(dy/dx) = a (a ∈ R) satisfying y = 1 when x = 0.
Ans: dy/dx = cos⁻¹a, a constant, so dy = cos⁻¹a dx. Integrating, y = x cos⁻¹a + C. Applying y = 1 at x = 0 gives C = 1, so y = x cos⁻¹a + 1, which can also be written as cos[(y-1)/x] = a

14. Find the particular solution of dy/dx = y tan x satisfying y = 1 when x = 0.
Ans: Separating variables, dy/y = tan x dx. Integrating, log y = log(sec x) + log C, so y = C sec x. Applying y = 1 at x = 0 gives C = 1, so y = sec x

15. Find the equation of a curve passing through the point (0, 0) whose differential equation is y’ = e^xsin x.
Ans: dy = e^xsin x dx. Using integration by parts twice, ∫e^xsin x dx = (e^x/2)(sin x – cos x) + C. So y = (e^x/2)(sin x – cos x) + C. Applying the curve through (0,0) gives 0 = (1/2)(0-1) + C, so C = 1/2, giving y = (e^x/2)(sin x – cos x) + 1/2

16. For the differential equation xy(dy/dx) = (x+2)(y+2), find the solution curve passing through the point (1,-1).
Ans: Rewriting, [1 – 2/(y+2)]dy = [(x+2)/x]dx. Integrating both sides, y – 2log(y+2) = x + 2log x + C, i.e. y – x = log[x²(y+2)²] + C. Applying the curve through (1,-1) gives C = -2, so y – x + 2 = log[x²(y+2)²]

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 equals the x-coordinate.
Ans: The condition gives y(dy/dx) = x, so y dy = x dx. Integrating, y²/2 = x²/2 + C, i.e. y² – x² = 2C. Applying the curve through (0,-2) gives C = 2, so y² – x² = 4

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).
Ans: The slope of the segment is (y+3)/(x+4), so dy/dx = 2(y+3)/(x+4). Separating variables and integrating, log(y+3) = 2log(x+4) + log C, so y+3 = C(x+4)². Applying the curve through (-2,1) gives C = 1, so y + 3 = (x+4)²

19. The volume of a 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 the balloon after t seconds.
Ans: Let V = (4/3)πr³ and dV/dt = k. Then (4/3)πr³ = kt + C. At t=0, r=3 gives C = 36π. At t=3, r=6 gives k = 84π. So (4/3)πr³ = 84πt + 36π, giving r³ = 63t + 27, i.e. r = (63t+27)^1/3

20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 doubles itself in 10 years (log_e2 = 0.6931).
Ans: dp/dt = (r/100)p. Integrating, p = 100e^rt/100 (using p=100 at t=0). At t=10, p=200 gives e^r/10 = 2, so r/10 = 0.6931, giving r = 6.931%

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 be worth after 10 years (e^0.5 = 1.648)?
Ans: dp/dt = p/20. Integrating, p = 1000e^t/20 (using p=1000 at t=0). At t=10, p = 1000e^0.5 = 1000 × 1.648 = Rs. 1648

22. In a culture, the bacteria count is 100000. The number increases by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?
Ans: dy/dt = cy. Integrating, y = 100000e^ct. At t=2, y=110000 gives c = (1/2)log(11/10). Setting y=200000 gives t = log 2 / c = 2log2 / log(11/10) hours

23. Find the general solution of the differential equation dy/dx = e^x+y.
(A) e^x+e^-y=C
(B) e^x+e^y=C
(C) e^-x+e^y=C
(D) e^-x+e^-y=C
Ans: dy/dx = e^xe^y, so e^-ydy = e^xdx. Integrating both sides, -e^-y = e^x + D, so e^x+e^-y=C. The answer is (A)

Exercise 9.4 Solutions (All 17 Questions)

1. Solve the differential equation (x²+xy)dy = (x²+y²)dx.
Ans: Rearranging, dy/dx = (x²+y²)/(x²+xy), which is homogeneous. Substituting y = vx, dy/dx = v + x(dv/dx), the equation becomes v + x(dv/dx) = (1+v²)/(1+v), so x(dv/dx) = (1-v)/(1+v). Separating variables, [(1+v)/(1-v)]dv = dx/x. Integrating both sides, -v – 2log(1-v) = log x + C. Substituting v = y/x and simplifying gives the general solution (x-y)²e^y/x = Cx

2. Solve the differential equation y’ = (x+y)/x.
Ans: Rearranging, dy/dx = 1 + y/x, which is homogeneous. Substituting y = vx, the equation becomes v + x(dv/dx) = 1 + v, so dv = dx/x. Integrating both sides, v = log x + C. Substituting v = y/x, y = x log x + Cx

3. Solve the differential equation (x-y)dy = (x+y)dx.
Ans: Rearranging, dy/dx = (x+y)/(x-y), which is homogeneous. Substituting y = vx, the equation becomes v + x(dv/dx) = (1+v)/(1-v), so x(dv/dx) = (1+v²)/(1-v). Separating variables, [(1-v)/(1+v²)]dv = dx/x. Integrating both sides, tan⁻¹v – (1/2)log(1+v²) = log x + C. Substituting v = y/x and simplifying gives the general solution tan⁻¹(y/x) = (1/2)log(x²+y²) + C

4. Solve the differential equation (x²-y²)dx + 2xy dy = 0.
Ans: Rearranging, dy/dx = (y²-x²)/(2xy), which is homogeneous. Substituting y = vx, the equation becomes v + x(dv/dx) = (v²-1)/(2v), so x(dv/dx) = -(1+v²)/(2v). Separating variables and integrating with the substitution 1+v²=t, log(1+v²) = -log x + C. Substituting v = y/x and simplifying gives x²+y² = Kx

5. Solve the differential equation x²(dy/dx) = x²-2y²+xy.
Ans: Rearranging, dy/dx = 1 – 2v² + v where v = y/x, which is homogeneous. Substituting y = vx, x(dv/dx) = 1-2v². Separating variables, dv/(1-2v²) = dx/x. Integrating both sides using standard log-form integration, the general solution is (1/(2√2))log|(x+√2 y)/(x-√2 y)| = log|x| + C

6. Solve the differential equation xdy – ydx = √(x²+y²)dx.
Ans: Rearranging, dy/dx = y/x + √(1+(y/x)²), which is homogeneous. Substituting y = vx, x(dv/dx) = √(1+v²). Separating variables and integrating, log|v+√(1+v²)| = log x + C. Substituting v = y/x and simplifying gives y + √(x²+y²) = Cx²

7. Solve the differential equation {x cos(y/x) + y sin(y/x)}y dx = {y sin(y/x) – x cos(y/x)}x dy.
Ans: Dividing throughout by x²y and substituting y = vx, the equation reduces to a variable-separable form in v and x. Integrating and substituting back v = y/x, the general solution is xy cos(y/x) = K

8. Solve the differential equation x(dy/dx) – y + x sin(y/x) = 0.
Ans: Rearranging, dy/dx = y/x – sin(y/x), which is homogeneous. Substituting y = vx, x(dv/dx) = -sin v. Separating variables, dv/sin v = -dx/x, i.e. cosec v dv = -dx/x. Integrating both sides and substituting v = y/x, the general solution is x[1 – cos(y/x)] = C sin(y/x)

9. Solve the differential equation y dx + x log(y/x) dy – 2x dy = 0.
Ans: Rearranging, dx/dy = x[2 – log(y/x)]/y, which is homogeneous in x and y. Substituting x = vy, integrating, and substituting back v = x/y, the general solution is log(y/x) – 1 = Cy

10. Solve the differential equation (1+e^x/y)dx + e^x/y(1-x/y)dy = 0.
Ans: This is homogeneous in x and y. Substituting x = vy, dx = v dy + y dv, and simplifying, the equation reduces to a variable-separable form. Integrating and substituting back v = x/y, the general solution is x + ye^x/y = C

11. Solve the differential equation (x+y)dy + (x-y)dx = 0, given y = 1 when x = 1.
Ans: Rearranging, dy/dx = (y-x)/(x+y), which is homogeneous. Substituting y = vx and integrating, tan⁻¹v + (1/2)log(1+v²) = -log x + C. Substituting v = y/x and simplifying gives 2tan⁻¹(y/x) + log(x²+y²) = K. Applying y=1 at x=1 gives K = π/2 + log 2, so 2tan⁻¹(y/x) + log(x²+y²) = π/2 + log 2

12. Solve the differential equation x²dy + (xy+y²)dx = 0, given y = 1 when x = 1.
Ans: Rearranging, dy/dx = -(xy+y²)/x², which is homogeneous. Substituting y = vx and integrating, the general solution is y/(2x+y) = K/x². Applying y=1 at x=1 gives K = 1/3, so 2x + y = 3x²y

13. Solve the differential equation [x sin²(y/x) – y]dx + x dy = 0, given y = π/4 when x = 1.
Ans: Rearranging, dy/dx = y/x – sin²(y/x), which is homogeneous. Substituting y = vx, integrating with the substitution cot v, and applying y=π/4 at x=1 gives the particular solution cot(y/x) = log|xe|

14. Solve the differential equation dy/dx – y/x + cosec(y/x) = 0, given y = 0 when x = 1.
Ans: Rearranging, dy/dx = y/x – cosec(y/x), which is homogeneous. Substituting y = vx, integrating, and applying y=0 at x=1 gives the particular solution cos(y/x) = log|xe|

15. Solve the differential equation 2xy + y² – 2x²(dy/dx) = 0, given y = 2 when x = 1.
Ans: Rearranging, dy/dx = (2xy+y²)/(2x²), which is homogeneous. Substituting y = vx, integrating, and applying y=2 at x=1 gives the particular solution y = 2x/(1-log x)

16. A homogeneous differential equation of the form dx/dy = h(x/y) can be solved by making the substitution
(A) y = vx
(B) v = yx
(C) x = vy
(D) x = v
Ans: Since the right side is a function of x/y, the substitution x = vy is used to make the equation variable-separable in v and y. The answer is (C)

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) y²dx + (x²-xy-y²)dy = 0
Ans: Writing dx/dy = -(x²-xy-y²)/y² = f(x,y) for option (D), replacing x by kx and y by ky gives f(kx,ky) = k⁰f(x,y), so f is homogeneous of degree zero. The equation in option (D) is homogeneous. The answer is (D)

Exercise 9.5 Solutions (All 18 Questions)

1. Solve the differential equation dy/dx + 2y = sin x.
Ans: This is linear with P=2, Q=sin x. Integrating factor I.F. = e^2x. The general solution is ye^2x = ∫e^2xsin x dx + C = (e^2x/5)(2sin x – cos x) + C, so y = (1/5)(2sin x – cos x) + Ce^-2x

2. Solve the differential equation dy/dx + 3y = e^-2x.
Ans: This is linear with P=3, Q=e^-2x. Integrating factor I.F. = e^3x. The general solution is ye^3x = ∫e^3xe^-2xdx + C = e^x + C, so y = e^-2x + Ce^-3x

3. Solve the differential equation dy/dx + y/x = x².
Ans: This is linear with P=1/x, Q=x². Integrating factor I.F. = x. The general solution is yx = ∫x³dx + C = x⁴/4 + C

4. Solve the differential equation dy/dx + (sec x)y = tan x (0 ≤ x ≤ π/2).
Ans: This is linear with P=sec x, Q=tan x. Integrating factor I.F. = sec x + tan x. The general solution is y(sec x + tan x) = ∫tan x(sec x + tan x)dx + C = sec x + tan x – x + C

5. Solve the differential equation cos²x(dy/dx) + y = tan x (0 ≤ x ≤ π/2).
Ans: Dividing by cos²x, dy/dx + y sec²x = tan x sec²x, linear with P=sec²x, Q=tan x sec²x. Integrating factor I.F. = e^{tan x}. Using the substitution t=tan x, the general solution is ye^{tan x} = tan x · e^{tan x} – e^{tan x} + C, so y = tan x – 1 + Ce^{-tan x}

6. Solve the differential equation x(dy/dx) + 2y = x²log x.
Ans: Dividing by x, dy/dx + 2y/x = x log x, linear with P=2/x, Q=x log x. Integrating factor I.F. = x². The general solution is yx² = ∫x³log x dx + C = (x⁴log x)/4 – x⁴/16 + C, so y = (x²/16)(4log x – 1) + Cx⁻²

7. Solve the differential equation x log x(dy/dx) + y = (2/x)log x.
Ans: Dividing by x log x, dy/dx + y/(x log x) = 2/x², linear with P=1/(x log x), Q=2/x². Integrating factor I.F. = log x. The general solution is y log x = -(2/x)(1+log x) + C

8. Solve the differential equation (1+x²)dy + 2xy dx = cot x dx.
Ans: Dividing by (1+x²), dy/dx + 2xy/(1+x²) = cot x/(1+x²), linear with integrating factor I.F. = 1+x². The general solution is y(1+x²) = ∫cot x dx + C = log(sin x) + C

9. Solve the differential equation x(dy/dx) + y – x + xy cot x = 0.
Ans: Dividing by x, dy/dx + y(1/x + cot x) = 1, linear with integrating factor I.F. = x sin x. The general solution is yx sin x = -x cos x + sin x + C, so y = -cot x + 1/x + C/(x sin x)

10. Solve the differential equation (x+y)dy/dx = 1.
Ans: Treating x as a function of y, dx/dy = x+y, i.e. dx/dy – x = y, linear in x with integrating factor I.F. = e^-y. The general solution is xe^-y = ∫ye^-ydy + C = -ye^-y – e^-y + C, so x + y + 1 = Ce^y

11. Solve the differential equation ydx + (x-y²)dy = 0.
Ans: Treating x as a function of y, dx/dy + x/y = y, linear in x with integrating factor I.F. = y. The general solution is xy = ∫y²dy + C = y³/3 + C

12. Solve the differential equation (x+3y²)dy/dx = y (y > 0).
Ans: Treating x as a function of y, dx/dy – x/y = 3y, linear in x with integrating factor I.F. = 1/y. The general solution is x/y = ∫3dy + C = 3y + C, so x = 3y² + Cy

13. Solve the differential equation dy/dx + 2y tan x = sin x, given y = 0 when x = π/3.
Ans: Linear with integrating factor I.F. = sec²x. The general solution is y sec²x = ∫sin x sec²x dx + C = sec x + C. Applying y=0 at x=π/3 gives C = -2, so y sec²x = sec x – 2, i.e. y = cos x – 2cos²x

14. Solve the differential equation (1+x²)dy/dx + 2xy = 1/(1+x²), given y = 0 when x = 1.
Ans: Linear with integrating factor I.F. = 1+x². The general solution is y(1+x²) = ∫dx/(1+x²) + C = tan⁻¹x + C. Applying y=0 at x=1 gives C = -π/4, so y(1+x²) = tan⁻¹x – π/4

15. Solve the differential equation dy/dx – 3y cot x = sin 2x, given y = 2 when x = π/2.
Ans: Linear with integrating factor I.F. = (sin x)⁻³. The general solution is y(sin x)⁻³ = ∫sin 2x(sin x)⁻³dx + C. Applying y=2 at x=π/2 gives C = 4, so y = 4sin³x – 2sin²x

16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x,y) equals the sum of the coordinates of the point.
Ans: dy/dx = x+y, linear with integrating factor I.F. = e^-x. The general solution is ye^-x = -xe^-x – e^-x + C, so x+y+1 = Ce^x. Applying the curve through the origin gives C = 1, so x + y + 1 = e^x

17. Find the equation of a curve passing through the point (0,2), given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Ans: x+y = dy/dx + 5, i.e. dy/dx – y = x – 5, linear with integrating factor I.F. = e^-x. The general solution is ye^-x = ∫(x-5)e^-xdx + C = (4-x)e^-x + C, so y + x – 4 = Ce^x. Applying the curve through (0,2) gives C = -2, so y + x – 4 = -2e^x

18. The integrating factor of the differential equation x(dy/dx) – y = 2x² is
(A) e^-x
(B) e^-y
(C) 1/x
(D) x
Ans: Rearranging, dy/dx – y/x = 2x, linear with P = -1/x. The integrating factor is I.F. = e^∫Pdx = e^{-log x} = 1/x. The answer is (C)

Miscellaneous Exercise Solutions (All 15 Questions)

1. For each of the differential equations given below, indicate its order and degree (if defined):
(i) d²y/dx² + 5x(dy/dx)² – 6y = log x
(ii) (dy/dx)³ – 4(dy/dx)² + 7y = sin x
(iii) d⁴y/dx⁴ – sin(d³y/dx³) = 0
Ans: (i) The highest order derivative is d²y/dx², so order = 2; it is a polynomial with highest power 1, so degree = 1.
(ii) The highest order derivative is dy/dx, so order = 1; its highest power is 3, so degree = 3.
(iii) The highest order derivative is d⁴y/dx⁴, so order = 4; since the equation contains sin(d³y/dx³), it is not a polynomial in its derivatives, so degree is not defined.

2. For each of the following, verify that the given function is a solution of the corresponding differential equation:
(i) xy = ae^x+be^-x+x² : x(d²y/dx²) + 2(dy/dx) – xy + x² – 2 = 0
(ii) y = e^x(a cos x + b sin x) : d²y/dx² – 2(dy/dx) + 2y = 0
(iii) y = x sin 3x : d²y/dx² + 9y – 6cos 3x = 0
(iv) x² = 2y²log y : (x²+y²)(dy/dx) – xy = 0
Ans: (i) Differentiating xy = ae^x+be^-x+x² twice gives x(d²y/dx²)+2(dy/dx) = ae^x+be^-x+2. Substituting into the LHS of the differential equation: (ae^x+be^-x+2) – (ae^x+be^-x+x²) + x² – 2 = 0. Verified.
(ii) Differentiating y = e^x(a cos x + b sin x) twice and substituting into the LHS gives 0. Verified.
(iii) Differentiating y = x sin 3x twice gives d²y/dx² = 6cos 3x – 9x sin 3x. Substituting into the LHS: (6cos 3x – 9x sin 3x) + 9(x sin 3x) – 6cos 3x = 0. Verified.
(iv) Differentiating x² = 2y²log y gives dy/dx = xy/(x²+y²). Substituting into the LHS: (x²+y²)·xy/(x²+y²) – xy = 0. Verified.

3. Prove that x²-y² = c(x²+y²)² is the general solution of the differential equation (x³-3xy²)dx = (y³-3x²y)dy, where c is a parameter.
Ans: Rearranging, dy/dx = (x³-3xy²)/(y³-3x²y), which is homogeneous. Substituting y = vx and integrating using partial fractions, (1/2)log[(1-v²)/(1+v²)²] = log x + log C. Substituting v = y/x and simplifying gives x²-y² = c(x²+y²)², as required.

4. Find the general solution of the differential equation dy/dx + √[(1-y²)/(1-x²)] = 0.
Ans: Separating variables, dy/√(1-y²) = -dx/√(1-x²). Integrating both sides, sin⁻¹y = -sin⁻¹x + C, so sin⁻¹y + sin⁻¹x = C

5. Show that the general solution of the differential equation dy/dx + (y²+y+1)/(x²+x+1) = 0 is given by (x+y+1) = A(1-x-y-2xy), where A is a parameter.
Ans: Separating variables, dy/(y²+y+1) = -dx/(x²+x+1). Completing the square and integrating both sides using the standard arctan form, (2/√3)[tan⁻¹((2y+1)/√3) + tan⁻¹((2x+1)/√3)] = C. This is equivalent, on taking the tangent of both sides and simplifying, to (x+y+1) = A(1-x-y-2xy).

6. Find the equation of the curve passing through the point (0, π/4) whose differential equation is sin x cos y dx + cos x sin y dy = 0.
Ans: Dividing throughout by cos x cos y, tan x dx + tan y dy = 0. Integrating both sides, log(sec x) + log(sec y) = C, so sec x sec y = k. Applying the curve through (0,π/4) gives k = √2, so sec x sec y = √2

7. Find the particular solution of the differential equation (1+e^2x)dy + (1+y²)e^xdx = 0, given that y = 1 when x = 0.
Ans: Separating variables, dy/(1+y²) = -e^xdx/(1+e^2x). Integrating with the substitution t = e^x, tan⁻¹y = -tan⁻¹(e^x) + C, so tan⁻¹y + tan⁻¹(e^x) = C. Applying y=1 at x=0 gives C = π/2, so tan⁻¹y + tan⁻¹(e^x) = π/2

8. Solve the differential equation ye^x/ydx = (xe^x/y+y²)dy (y ≠ 0).
Ans: Rearranging as dx/dy and substituting z = e^x/y, the equation simplifies to dz/dy = 1. Integrating, z = y + C, so e^x/y = y + C

9. 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)
Ans: Rearranging, (x-y+1)dy = (1-x+y)dx, so dy/dx = (1-x+y)/(x-y+1). Substituting t = x-y, the equation reduces to dt/dx = 2t/(1+t). Separating variables and integrating, log|t| + t = 2x + C, so log|x-y| + (x-y) = 2x + C, i.e. log|x-y| – y = x + C. Applying y=-1 at x=0 gives C = 1, so log|x-y| – y = x + 1

10. Solve the differential equation [e^-2√x/√x – y/√x](dx/dy) = 1 (x ≠ 0).
Ans: Rearranging, dy/dx + y/√x = e^-2√x/√x, linear with integrating factor I.F. = e^2√x. The general solution is ye^2√x = ∫(1/√x)dx + C = 2√x + C

11. Find a particular solution of the differential equation dy/dx + y cot x = 4x cosec x (x ≠ 0), given that y = 0 when x = π/2.
Ans: Linear with integrating factor I.F. = sin x. The general solution is y sin x = ∫4x dx + C = 2x² + C. Applying y=0 at x=π/2 gives C = -π²/2, so y sin x = 2x² – π²/2

12. Find a particular solution of the differential equation (x+1)dy/dx = 2e^-y-1, given that y = 0 when x = 0.
Ans: Separating variables, dy/(2e^-y-1) = dx/(x+1). Integrating with the substitution t = 2-e^y, (2-e^y) = 1/[C(x+1)]. Applying y=0 at x=0 gives C = 1, so e^y = (2x+1)/(x+1), i.e. y = log[(2x+1)/(x+1)]

13. The general solution of the differential equation (y dx – x dy)/y = 0 is
(A) xy = C
(B) x = Cy²
(C) y = Cx
(D) y = Cx²
Ans: Dividing by xy, dx/x – dy/y = 0. Integrating both sides, log|x| – log|y| = log k, so x/y = k, giving y = Cx. The answer is (C)

14. The general solution of a differential equation of the type dx/dy + P₁x = Q₁ is
(A) ye^∫P₁dy = ∫(Q₁e^∫P₁dy)dy + C
(B) ye^∫P₁dx = ∫(Q₁e^∫P₁dx)dy + C
(C) xe^∫P₁dy = ∫(Q₁e^∫P₁dy)dy + C
(D) xe^∫P₁dy = ∫(Q₁e^∫P₁dx)dy + C
Ans: This is a linear equation in x, with integrating factor I.F. = e^∫P₁dy, and its general solution is xe^∫P₁dy = ∫(Q₁e^∫P₁dy)dy + C. The answer is (C)

15. The general solution of the differential equation e^xdy + (ye^x+2x)dx = 0 is
(A) xe^y+x²=C
(B) xe^y+y²=C
(C) ye^x+x²=C
(D) ye^y+x²=C
Ans: Rearranging, dy/dx + y = -2xe^-x, linear with integrating factor I.F. = e^x. The general solution is ye^x = -2∫x dx + C = -x²+C, so ye^x+x²=C. The answer is (C)

Frequently Asked Questions

What is the order and degree of a differential equation?
Order is the highest derivative present; degree is the power of that highest derivative (equation must be polynomial in derivatives).

How do you solve a homogeneous differential equation?
Substitute y=vx to convert to a variable-separable equation in v and x.

What is the integrating factor for dy/dx+Py=Q?
IF=e^{∫P dx}.

Written by Satish

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