Question

Solve the given differential equation

$\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Short answer

The general solution of the given differential equation is$2y+2-\ln |4x+2y+3|=c_1$

Step-by-step solution

Do the derivative of the given function with respect to X

Let$u=2x+y+1$

Take the derivative of u and find dydx

$\frac{du}{dx}=2+\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{du}{dx}-2$

Substitute $\mathit{u}$and $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$ and simplify

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

$\frac{du}{dx}=\frac{2u+1}{u}$

$\frac{u}{2u+1}du=dx$

Integrating both sides

$\begin{array}{rcl}\left(\frac{1}{2}-\frac{1}{2}\frac{1}{2u+1}\right)du& =& dx\\ \frac{1}{2}u-\frac{1}{4}\ln |2u+1|& =& x+c\end{array}$

Multiply both sides by 4

$2u-\ln |2u+1|=4x+c_1$

Resubstituting for $\mathit{u}$ and simplify

$4x+2y+2-\ln |4x+2y+3|=4x+c_1$

$2y+2-\ln |4x+2y+3|=c_1$