Differential Equations: 5 marks Questions (CBSE Class 12)

2 different 5 marks questions on Differential Equations from CBSE Class 12 Maths board exams 2026, newest first.

Solve the differential equation $y e^{y} d x = (y^{3} + 2 x e^{y}) d y$ , when $y (0) = 1$ .

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Answer:

x = y^{2} (e^{- 1} - e^{- y})

, i.e.

x = \frac{y ^{2}}{e} - y^{2} e^{- y}

Write as $\frac{d x}{d y} = \frac{y ^{3} + 2 x e ^{y}}{y e ^{y}}$ , i.e. $\frac{d x}{d y} - \frac{2}{y} x = y^{2} e^{- y}$ , linear in x.
I.F. $= e^{\int - \frac{2}{y} d y} = e^{- 2 l o g y} = \frac{1}{y ^{2}}$ .
$\frac{x}{y ^{2}} = \int y^{2} e^{- y} \cdot \frac{1}{y ^{2}} d y = \int e^{- y} d y = - e^{- y} + C$ .
$y (0) = 1$ : $x = 0$ , $y = 1$ gives $0 = - e^{- 1} + C$ , so $C = \frac{1}{e}$ .
Solution: $x = y^{2} (\frac{1}{e} - e^{- y})$ .

Find the general solution of the differential equation
$(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ .

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Answer:

x^{2} - y^{2} = C (x^{2} + y^{2})^{2}

$\frac{d y}{d x} = \frac{x ^{3} - 3 x y ^{2}}{y ^{3} - 3 x ^{2} y}$ is homogeneous. Put $y = v x$ : $v + x \frac{d v}{d x} = \frac{1 - 3 v ^{2}}{v ^{3} - 3 v}$ .
$x \frac{d v}{d x} = \frac{1 - 3 v ^{2} - v ^{4} + 3 v ^{2}}{v ^{3} - 3 v} = \frac{1 - v ^{4}}{v ^{3} - 3 v}$ , so $\int \frac{v ^{3} - 3 v}{1 - v ^{4}} d v = \int \frac{d x}{x}$ .
Put $v^{2} = t$ : $\frac{1}{2} \int \frac{t - 3}{( 1 - t ) ( 1 + t )} d t = \frac{1}{2} \int (\frac{- 1}{1 - t} - \frac{2}{1 + t}) d t = \frac{1}{2} lo g ∣1 - t ∣ - lo g (1 + t)$ .
So $\frac{1}{2} lo g ∣1 - v^{2} ∣ - lo g (1 + v^{2}) = lo g ∣ x ∣ + c$ , i.e. $\frac{∣1 - v ^{2} ∣}{( 1 + v ^{2} ) ^{2} x ^{2}} =$ constant.
With $v = \frac{y}{x}$ : $\frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = C$ , i.e. $x^{2} - y^{2} = C (x^{2} + y^{2})^{2}$ .

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