Solve the following differential equation: dy dx =(x+y+1)^ 2 - Vidyadip

Question

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=(\text{x}+\text{y}+1)^{2}$

✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}=(\text{x}+\text{y}+1)^{2}$
Putting $\text{x}+\text{y}+1=\text{v}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}-1$
$\Rightarrow \frac{\text{dv}}{\text{dx}}-1=\text{v}^{2}$
$\Rightarrow \frac{\text{dv}}{\text{dx}}=\text{v}^{2}+1$
$\Rightarrow \frac{1}{\text{v}^{2}+1}\text{dv}=\text{dx}$
Integrating both sides, we get
$\int \frac{1}{\text{v}^{2}+1}\text{dv}=\int\text{dx}$
$\Rightarrow \tan^{-1}\text{v}=\text{x}+\text{C}$
$\Rightarrow \tan^{-1}(\text{x}+\text{y}+1)=\text{x}+\text{C}$

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