Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Find the general solution of $(x+y) \frac{d y}{d x}=1$
✓
Answer
It is given that $(x+y) \frac{d y}{d x}=1$
$\Rightarrow \frac{d y}{d x}=\frac{1}{x+y}$
$\Rightarrow \frac{d x}{d y}=x+y$
$\Rightarrow \frac{d x}{d y}-x=y$
This is equation in the form of $\frac{d x}{d y}+p x=Q($where, $p = -1$ and $Q = y)$
Now, $ I.F. = \mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int-\mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}$
Thus, the solution of the given differential equation is given by the relation:
$x(I . F .)=\int(Q \times I . F .) d y+C$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=\int\left[\mathrm{y} \cdot \mathrm{e}^{-\mathrm{y}}\right] \mathrm{d} \mathrm{y}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=\mathrm{y} \int \mathrm{e}^{-\mathrm{y}} \mathrm{dy}-\int\left[\frac{\mathrm{d}}{\mathrm{dy}}(\mathrm{y}) \int \mathrm{e}^{-\mathrm{y}} \mathrm{dy}\right] \mathrm{dy}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=\mathrm{y}\left(-\mathrm{e}^{-\mathrm{y}}\right)-\int\left(-\mathrm{e}^{-\mathrm{y}}\right) \mathrm{d} \mathrm{y}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=-\mathrm{ye}^{-\mathrm{y}}+\int \mathrm{e}^{-\mathrm{y}} \mathrm{dy}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=-\mathrm{ye}^{-\mathrm{y}}-\mathrm{e}^{-\mathrm{y}}+\mathrm{C}$
$\Rightarrow x = - y – 1 + Ce^y$
$\Rightarrow x + y + 1 = Ce^y$
Therefore, the required general solution of the given differential equation is
$x + y + 1 = Ce^y$
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