CBSE 2026 · Region 5 · Set 2 · Q28 · 3 marks
Solve the differential equation $\displaystyle \mathrm{y} \mathrm{dx}+\left(\mathrm{x}-\mathrm{y}^{3}\right) \mathrm{dy}=0$.
Marking-scheme solution
Given differential equation can be written as $\displaystyle \dfrac{dx}{dy}+\dfrac{x}{y}=y^{2}$
Integrating factor $\displaystyle =\mathrm{e}^{\int \frac{1}{y} dy}=\mathrm{e}^{\log y}=y$
Solution is $\displaystyle x \cdot y=\int y^{3} dy+C$
i.e., $\displaystyle x \cdot y=\dfrac{y^{4}}{4}+C$ or $\displaystyle x=\dfrac{y^{3}}{4}+\dfrac{C}{y}$
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.