Solve the following differential equations : (x+y) d y d x =1

Question

Solve the following differential equations : $(x+y) \frac{d y}{d x}=1$

✓

Answer

$
\begin{aligned}
& ( x + y ) \frac{d y}{d x}=1 \\
& \therefore \frac{d x}{d y}= x + y \\
& \therefore \frac{d x}{d y}- x = y \\
& \therefore \frac{d x}{d y}+(-1) x = y
\end{aligned}
$
This is the linear differential equation of the form
$
\begin{aligned}
& \frac{d x}{d y}+P \cdot x=Q, \text { where } P=-1 \text { and } Q=y \\
& \therefore \text { I.F. }=e^{\int P d y}=e^{\int-1 d y}=e^{-y}
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{array}{rl}
& x \cdot(\text { I.F. })=\int Q \cdot(\text { I.F. }) d y+c \\
\therefore x \cdot e^{-y} & =\int y \cdot e^{-y} d y+c \\
\therefore & e^{-y} \cdot x=y \int e^{-y} d y-\int\left[\frac{d}{d y}(y) \int e^{-y} d y\right] d y+c \\
= & y \cdot \frac{e^{-y}}{-1}-\int 1 \cdot \frac{e^{-y}}{-1} d y+c \\
& =-y e^{-y}+\int e^{-y} d y+c \\
\therefore & e^{-y} \cdot x=-y e^{-y}+\frac{e^{-y}}{-1}+c \\
\therefore & e^{-y} \cdot x+y e^{-y}+e^{-y}=c \\
\therefore & e^{-y}(x+y+1)=c \\
\therefore x+y & x=c e^y
\end{array}
$
This is the general solution.

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