Find the general solution for each of the following differential equations. $(x+y) \frac{d y}{d x}=1$

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CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal

Find the general solution for each of the following differential equations. $(x+y) \frac{d y}{d x}=1$

Solution:

$(\mathrm{x}+\mathrm{y}) \frac{d y}{d x}=1-\ldots(1)$
General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ is given by,
$y \cdot\left(I . F_{.}\right)=\int Q \cdot(I . F .) d x+c$
Where, integrating factor,
$I . F_{1}=e \int^{p d x}$
$\frac{d x}{d y}-x=y \ldots \ldots \ldots(1)$
$(x+y) \frac{d y}{d x}=1$
In equation (1)P $=-1$ and $Q=y$
Therefore, integrating factor is
$\text { I.F. }=e \int^{p d x}=e^{-y}$
General solution is
$\begin{array}{r} y \cdot(I . F \cdot)=\int Q \cdot(I . F .) d x+c \\ x \cdot e^{-y}=\int y \cdot e^{-y} d y+c \ldots \ldots(2) \\ I=\int y \cdot e^{-y} d y \end{array}$
Let, $u=y$ and $v=e^{-y}$
$\begin{array}{l} \mathrm{I}=\mathrm{y} \int \mathrm{e}^{-y} \mathrm{~d} \mathrm{y}-\int \frac{d y}{d y} \cdot \int e^{-y} d y \mathrm{dy} \\ \mathrm{I}=-\mathrm{ye}^{-\mathrm{y}}-\int 1 \cdot\left(-\mathrm{e}^{-y}\right) \mathrm{dy} \\ \mathrm{I}=-\mathrm{y} \mathrm{e}^{-\mathrm{y}} \mathrm{e}^{-\mathrm{y}} \end{array}$
Substituting I in (2),
$\begin{array}{l} x \cdot\left(e^{-y}\right)=-y e^{-y}-e^{-y}+c \\ e^{-y}(x+y+1)=c \text { (general solution) } \end{array}$