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

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

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

Solution:

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