Answer to Question #292774 in Differential Equations for luna

Question

Solve the first order linear inhomogeneous differential equation using the constant variation method

(2x+1)y^,=4x+2y

Answer 1

Given a first order inhomogeneous linear differential equation of the form;

$\displaystyle y\prime+p(x)y=f(x)\cdots\cdots\cdots\color{red}(1)$

using constant variation method, the general solution is given by;

$\displaystyle y(x)=v(x)e^{q(x)}+ce^{q(x)}$,

where $\displaystyle v\prime(x)=e^{-q(x)}f(x)$, $\displaystyle q(x)=\int[-p(x)]\ dx$, and $c$ is and arbitrary constant.

Now, the given DE is of the form;

$\displaystyle y\prime-\frac{2}{2x+1}y=\frac{4x}{2x+1}$

By comparing the given DE with $\color{red}(1)$;

$\displaystyle q(x)=\int[-p(x)]\ dx=\int\left[\frac{2}{2x+1}\right]\ dx=\ln(2x+1)$

$\displaystyle v\prime(x)=e^{-q(x)\ dx}\times f(x)=e^{-\ln(2x+1)}\times\frac{4x}{2x+1}=\frac{4x}{(2x+1)^2}$

$\displaystyle \Rightarrow v(x)=\int\frac{4x}{(2x+1)^2}\ dx=\int\left(\frac{2}{2x+1}-\frac{2}{(2x+1)^2}\right)\ dx$

$\displaystyle =\ln(2x+1)+\frac{1}{2x+1}$

Thus, the general solution is;

$\displaystyle y(x)=v(x)e^{q(x)}+Ae^{q(x)}=\left(\ln(2x+1)+\frac{1}{2x+1}\right)e^{\ln(2x+1)}+ce^{\ln(2x+1)}$

$\displaystyle =\left(\ln(2x+1)+\frac{1}{2x+1}\right)(2x+1)+c\ln(2x+1)$

$\displaystyle =(2x+1)\ln(2x+1)+1+c(2x+1)$

$=(2x+1)[c+\ln(2x+1)]+1$