Question

The solution of the differential equation ${\left(x+y\right)}^2\left(\frac{dy}{dx}\right)=a^2$ is

• A. ${\left(x+y\right)}^2\left(\frac{dy}{dx}\right)=\frac{a^{2}x}{2}+c$
• B. ${\left(x+y\right)}^2\left(\frac{dy}{dx}\right)=a^{2}x+c$
• C. ${\left(x+y\right)}^2\left(\frac{dy}{dx}\right)=2a^{2}x+c$
• D. None of these.

Answer 1

The correct option is D

None of these.

Explanation of the correct option.

Compute the required value.

Given :${\left(x+y\right)}^2\left(\frac{dy}{dx}\right)=a^2$

Let $x+y=u$

Differentiate with respect to $x$,

$1+\frac{dy}{dx}=\frac{du}{dx}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{du}{dx}-1$

Substitute the values,

$u^2\left(\frac{du}{dx}-1\right)=a^2$

$\Rightarrow$ $u^2\frac{du}{dx}-u^2=a^2$

$\Rightarrow$ $u^2\frac{du}{dx}=a^2+u^2$

$\Rightarrow$$\left(\frac{u^2}{a^2+u^2}\right)du=dx$

Integrate both side,

$\Rightarrow$ $\int \frac{u^2+a^2-a^2}{u^2+a^2}du=\int dx$

$\Rightarrow$ $\int du-\int \frac{a^2}{u^2+a^2}du=x+c$

$\Rightarrow$ $u-a^2.\frac{1}{a}{\tan }^{-1}\left(\frac{u}{a}\right)=x+c$

$\Rightarrow$$x+y-a{\tan }^{-1}\left(\frac{x+y}{a}\right)=x+c$

$\Rightarrow$ $y=a{\tan }^{-1}\left(\frac{x+y}{a}\right)+c$

Hence option $\left(D\right)$ is the correct option.