Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations3 Marks
Question
Find the general solution of $y d x+\left(x-y^{2}\right) d y=0$
✓
Answer
It is given that $\mathrm{ydx}+\left(\mathrm{x}-\mathrm{y}^{2}\right) \mathrm{dy}=0$ $\Rightarrow \mathrm{ydx}=\left(\mathrm{y}^{2}-\mathrm{x}\right) \mathrm{dy}$ $\Rightarrow \frac{d x}{d y}=\frac{\left(y^{2}-x\right)}{y}=y-\frac{x}{y}$ $\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y$ This is equation in the form of $\frac{d x}{d y}+p x=Q$ (where, p = $\frac{1}{y}$ and Q = y) Now, I.F. = $\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log \mathrm{y}}=\mathrm{y}$ Thus, the solution of the given differential equation is given by the relation: $x(I . F .)=\int(Q \times I . F .) d y+C$ $\Rightarrow \mathrm{x} \cdot \mathrm{y}=\int[\mathrm{y} \cdot \mathrm{y}] \mathrm{d} \mathrm{y}+\mathrm{C}$ $\Rightarrow x . y=\int y^{2} d y+C$ $\Rightarrow x \cdot y=\frac{y^{3}}{3}+C$ $\Rightarrow x =\frac{y^{2}}{3}+\frac{C}{y}$ Therefore, the required general solution of the given differential equation is $x =\frac{y^{2}}{3}+\frac{c}{y}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.