Previous year questions: Differential Equations

Q1 (CBSE Board 2026, 3 marks)

Find the general solution of three x squared dy by dx equal to x squared plus x y plus y squared.

Answer: Divide by three x squared to get dy by dx equal one third of one plus y over x plus y over x whole squared, which is homogeneous. Put y equal to v x, so dy by dx equal v plus x dv by dx. The equation becomes x dv by dx equal v minus one whole squared divided by three. Separate and integrate: three dv over v minus one whole squared equal dx over x, which gives minus three over v minus one equal log x plus C. Replacing v gives minus three x over y minus x equal log x plus C, and rearranging gives the general solution y equal x minus three x over C plus log x.

Q2 (CBSE Board 2026, 3 marks)

Find the general solution of two x squared dy by dx equal to y squared plus two x y.

Answer: Divide by two x squared to get dy by dx equal half of y over x squared plus y over x, which is homogeneous. Put y equal to v x and dy by dx equal v plus x dv by dx. The terms v cancel and x dv by dx equals v squared over two. Separate: two dv over v squared equal dx over x, which integrates to minus two over v equal log x plus C. Replacing v by y over x gives minus two x over y equal log x plus C, so the general solution is y equal minus two x over C plus log x.

Q3 (CBSE Board 2026, 3 marks)

Find the general solution of three times y squared minus x squared dx equal two x y dy.

Answer: Rewrite as dy by dx equal three y squared minus x squared over two x y, which is homogeneous. With y equal to v x the equation becomes x dv by dx equal v squared minus three over two v after simplification. Separate: two v dv over v squared minus three equal dx over x, which integrates to log of v squared minus three equal log x plus a constant. Replacing v gives y squared minus three x squared equal C x cubed, which is the general solution in implicit form.

Q4 (CBSE Board 2026, 5 marks)

Find the general solution of x cubed minus three x y squared dx equal y cubed minus three x squared y dy.

Answer: Divide by x cubed to write dy by dx as a homogeneous function of y over x. With y equal to v x, the equation becomes x dv by dx equal one minus v to the fourth divided by v cubed minus three v. Separate and integrate using the substitution u equal v squared on the left. The partial fraction split gives one half log of v squared minus one minus log of v squared plus one equal log x plus a constant. Replacing v by y over x and simplifying gives y squared minus x squared equal C times x squared plus y squared whole squared, which is the general solution.

Q5 (CBSE Board 2026, 5 marks)

Solve x squared dy plus y times x plus y dx equal zero, given that y equals one when x equals one.

Answer: Rewrite as dy by dx equal minus y over x minus y over x whole squared, which is homogeneous. With y equal to v x, the equation becomes x dv by dx equal minus v squared minus two v. Separate: dv over v squared plus two v equal minus dx over x, and the partial fraction split integrates to one half log of v over v plus two equal minus log x plus a constant. Replacing v gives y over y plus two x equal C over x squared. The initial condition at one one gives one third equal C, so the particular solution is three x squared y equal y plus two x.

Q6 (CBSE Board 2025, 3 marks)

Find the particular solution of dy by dx minus three y cotangent x equal sine two x, given that y equals two when x equals pi by two.

Answer: The equation is linear with P equal minus three cotangent x, so the integrating factor is cosecant cubed x. Multiplying gives the derivative of y cosecant cubed x equal to sine two x times cosecant cubed x, which simplifies to two cotangent x cosecant x. Integrating gives minus two cosecant x plus C, so y equal minus two sine squared x plus C sine cubed x. The condition at pi by two gives two equal minus two plus C, so C is four and the particular solution is y equal four sine cubed x minus two sine squared x.

Q7 (CBSE Board 2025, 3 marks)

Find the general solution of two x squared plus y dx equal x dy.

Answer: Divide by x to get dy by dx equal two x plus y over x, that is dy by dx minus y over x equal two x, which is linear with P equal minus one over x and Q equal two x. The integrating factor is one over x. Multiplying gives the derivative of y over x equal two, so y over x equal two x plus C. The general solution is y equal two x squared plus C x. Differentiating confirms the left side of the original equation.

Q8 (CBSE Board 2024, 3 marks)

Find the particular solution of x cosine y dy equal x e to the x log x plus e to the x dx, given that y equals pi by two when x equals one.

Answer: Divide by x to get cosine y dy equal e to the x log x plus e to the x over x dx. The right side is the derivative of e to the x log x, so integrating gives sine y equal e to the x log x plus C. Using y equal pi by two at x equal one gives one equal zero plus C, so C is one. The particular solution is sine y equal one plus e to the x log x.

Q9 (CBSE Board 2024, 3 marks)

Solve dy by dx plus two y tangent x equal sine x, given that y equals zero when x equals pi by three.

Answer: The equation is linear with P equal two tangent x, so the integrating factor is secant squared x. Multiplying gives the derivative of y secant squared x equal to sine x over cosine squared x, whose integral is secant x. Therefore y secant squared x equal secant x plus C. The condition at pi by three gives zero equal two plus C, so C is minus two. The particular solution is y equal cosine x minus two cosine squared x.

Q10 (CBSE Board 2023, 3 marks)

Find the particular solution of dy by dx plus y cotangent x equal cosine squared x, given that y equals zero when x equals pi by two.

Answer: The integrating factor is sine x, so multiplying gives the derivative of y sine x equal to sine x cosine squared x. Integrating gives y sine x equal minus cosine cubed x over three plus C. The condition at pi by two gives zero equal zero plus C, so the particular solution is y sine x equal minus cosine cubed x over three, that is y equal minus cosine cubed x over three sine x.

How the board answers are marked

The marking schemes award the first mark for identifying the type of equation and writing the standard form, then marks for the substitution or the integrating factor, then marks for the integration, and the final mark for the particular solution with the constant evaluated. That is why every answer on this page starts with the method and shows the substitution or the factor as a separate line. A student who jumps to the integrated relation loses the method marks even when the relation is correct.

A last check before writing the answer

Differentiate the general solution once and compare it with the original equation. For a homogeneous equation, the check is usually quick because the derivative of the implicit relation returns the two coefficients. For a linear equation, the check confirms both the factor and the sign of the constant. Keep the check as the last line of the answer when time permits.