Question

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

• A. $\frac{1}{x+y}=c$
• B. ${\sin }^{-1}(x+y)=x+c$
• C. ${\tan }^{-1}(x+y)=c$
• D. ${\tan }^{-1}(x+y)=x+c$

Answer 1

The correct option is D ${\tan }^{-1}(x+y)=x+c$

Given DE is $\frac{dy}{dx}=(x+y{)}^2$

Let's substitute $t=x+y$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$

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

$\Rightarrow \frac{dt}{dx}-1=t^2$

$\Rightarrow \frac{dt}{dx}=t^2+1$

$\Rightarrow \frac{dt}{t^2+1}=dx$

On integrating both sides we get

$\int \frac{dt}{t^2+1}=\int dx$

$\Rightarrow {\tan }^{-1}t=x+c$

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