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Mathematics · 2025 · 5 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations5 marksApplylong answer
CBSE 2025 · Region 6 · Set 1 · Q34
(a) Solve the differential equation: x^2 y d x -( x^3+ y^3) d y =0.
(a)
Solve the differential equation : $\displaystyle \mathrm{x}^{2} \mathrm{y} d \mathrm{x}-\left(\mathrm{x}^{3}+\mathrm{y}^{3}\right) d \mathrm{y}=0$.
OR
(b)
Solve the differential equation $\displaystyle \left(1+\mathrm{x}^{2}\right) \frac{d \mathrm{y}}{d \mathrm{x}}+2 \mathrm{x} \mathrm{y}-4 \mathrm{x}^{2}=0$ subject to initial condition $\displaystyle \mathrm{y}(0)=0$.
Marking-scheme solution
(a)
Given differential equation can be written as $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\mathrm{y} \mathrm{x}^{2}}{\mathrm{x}^{3}+\mathrm{y}^{3}}$
Put $\displaystyle \mathrm{y}=\mathrm{vx}$, so $\displaystyle \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}$
Therefore, $\displaystyle \mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}^{3}}{\mathrm{x}^{3}+\mathrm{v}^{3} \mathrm{x}^{3}}=\frac{\mathrm{v}}{1+\mathrm{v}^{3}}$
\[\mathrm{x} \frac{d \mathrm{v}}{d \mathrm{x}}=\frac{-\mathrm{v}^{4}}{1+\mathrm{v}^{3}}
\]
\[\left(\frac{1}{\mathrm{v}^{4}}+\frac{1}{\mathrm{v}}\right) d \mathrm{v}=\frac{-d \mathrm{x}}{\mathrm{x}}
\]
Integrating we get
\[\frac{-1}{3 \mathrm{v}^{3}}+\log |\mathrm{v}|=-\log |\mathrm{x}|+\mathrm{C}
\]
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.