Solve the differential equation given below $$ \frac{x d y}{d x}=y+\sqrt{x^2+y^2} $$
Solve the differential equation given below
$$
\frac{x d y}{d x}=y+\sqrt{x^2+y^2}
$$
- $x^2=c\left[y+\sqrt{y^2+x^2}\right]$
- $y^2=c\left[x+\sqrt{y^2-x^2}\right]$
- $y^2=c\left[x+\tan ^{-1}\left(\sqrt{1+y^2}\right)\right]$
- $y^2=c\left[x-\sqrt{y^2+x^2}\right]$
Solution
Given differential equation
$
\begin{aligned}
\quad x \frac{d y}{d x} & =y+\sqrt{x^2+y^2} \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{y+\sqrt{x^2+y^2}}{x}
\end{aligned}
$
Now, put $y=v \cdot x$
$
\therefore \quad \frac{d y}{d x}=v+x \frac{d v}{d x}
$
On doing substitution, we get
$
\begin{aligned}
& v+x \frac{d v}{d x}=v+\sqrt{1+v^2} \\
\Rightarrow & \int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x} \\
\Rightarrow & \log _{\ell} C+\log _{\ell}\left(\sqrt{1+v^2}+v\right)=\log _{\ell} x \\
\Rightarrow & \log \left[\left(y+\sqrt{x^2+y^2}\right) C\right]=2 \log _{\ell} x \\
\Rightarrow & x^2=C\left[y+\sqrt{y^2+x^2}\right]
\end{aligned}
$
Hence, option (1) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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