Answer to Question #171468 in Differential Equations for Jose

Question

y'=1+x-2y

Answer 1

y'=1+x-2y

Solution:

$y'+2y=1+x$

The complementary equation is:

$y'+2y=0$

The characteristic equation:

$r+2=0$

$r=-2$

The general solution for this homogeneous equation :

$y_0=Ce^{-2x}$

Now let's find the particular solution $y_p(x)$ of the nonhomogeneous equation. The particular solution might have the form:

$y_p(x)=Ax+B$

Then

$y_p'(x)=A$

For $y_p(x)$ to be a solution to the differential equation, we must find values for A and B such that

$A+2(Ax+B)=1+x$

Setting coefficients of like terms equal, we have

$A+2B=1$

$2A=1$

Then

$A=\frac12$ ; $B=\frac14$ .

$y_p(x)=\frac12 x+\frac14$ .

The general solution of a nonhomogeneous equation is:

$y(x)=y_0(x)+y_p(x)=Ce^{-2x}+\frac12x+\frac14$ .

Answer: $y(x)=Ce^{-2x}+\frac12x+\frac14$ .