Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations2 Marks
Question
Solve the differential equation $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y(y \neq 0)$
✓
Answer
It is given that $\text { ye }^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y$ $\Rightarrow \text { ye }^{\frac{x}{y}} \frac{d x}{d y}=x e^{\frac{x}{y}}+y^{2}$ $\Rightarrow \mathrm{e}^{\frac{x}{y}}\left[y \cdot \frac{d x}{d y}-x\right]=y^{2}$ $\Rightarrow e^{\frac{x}{y}} \cdot \frac{\left[y \cdot \frac{d x}{d y}-x\right]}{y^{2}}=1$ .......(i) Let $e^{\frac{x}{y}}=z$ Differentiating it w.r.t. y, we get, $\frac{d}{d y}\left(e^{\frac{x}{y}}\right)=\frac{d z}{d y}$ $\Rightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y}\left(\frac{x}{y}\right)=\frac{d z}{d y}$ $\Rightarrow e^{\frac{x}{y}} \cdot\left[\frac{y \cdot \frac{d x}{d y}-x}{y^{2}}\right]=\frac{d z}{d y}$ ......(ii) From equation (i) and equation (ii), we get, $\frac{\mathrm{d} \mathrm{z}}{\mathrm{dy}}=1$ $\Rightarrow$ dz = dy On integrating both sides, we get, z = y + C $\Rightarrow \mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}}=\mathrm{y}+\mathrm{C}$
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