Solve (x+y) d y=a^2 d x

Question

Solve $(x+y) d y=a^2 d x$

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Answer

$
\begin{aligned}
& \quad(x+y) d y=a^2 d x \\
& \therefore \frac{d y}{d x}=\frac{a^2}{x+y}
\end{aligned}
$
Put $x+y=v \quad \therefore 1+\frac{d y}{d x}=\frac{d v}{d x}$
$\therefore \frac{d y}{d x}=\frac{d v}{d x}-1$
$\therefore$ (1) becomes, $\frac{d v}{d x}-1=\frac{a^2}{v}$
$
\begin{aligned}
& \therefore \frac{d v}{d x}=\frac{a^2}{v}+1=\frac{a^2+v}{v} \\
& \therefore \frac{v}{a^2+v} d v=d x
\end{aligned}
$
Integrating, we get
$
\begin{aligned}
& \int \frac{v}{a^2+v} d v=\int d x \\
\therefore & \int \frac{\left(a^2+v\right)-a^2}{a^2+v} d v=\int d x \\
\therefore & \int\left(1-\frac{a^2}{a^2+v}\right) d v=\int d x
\end{aligned}
$
$
\begin{aligned}
& \therefore \int 1 d v-a^2 \int \frac{1}{a^2+v} d v=\int d x \\
& \therefore v-a^2 \log \left|a^2+v\right|=x+c_1 \\
& \therefore x+y-a^2 \log \left|a^2+x+y\right|=x+c_1 \\
& \therefore a^2 \log \left|x+y+a^2\right|=y-c_1 \\
& \therefore \log \left|x+y+a^2\right|=\frac{y-c_1}{a^2}=\frac{y}{a^2}-\frac{c_1}{a^2}
\end{aligned}
$
$
\therefore x+y+a^2=e^{\left(\frac{y}{a^2}-\frac{c_1}{a^2}\right)}=e^{\frac{y}{a^2}} \cdot e^{\frac{-c_1}{a^2}}
$
$\therefore x+y+a^2=c \cdot e^{y / a^2}$, where $c=e^{-c_1 / a^2}$
This is the general solution.

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