ssh said:
How to solve \( (x+1) y'' - (2x+5) y' + 2y = (x+1) e^x\)
can we assume \(y_1 = (Ax+B) e^x \),
then \(y_2= vy_1\) Is this right? then solve for A and B
Finally \( y = c_1 y_1 + c_2 y_2\)
The solving procedure is the same used in...
http://www.mathhelpboards.com/f17/how-solve-differential-equation-second-order-linear-variable-coefficient-2089/
First You have to find the general solution of the incomplete equation...
$\displaystyle (x+1)\ y^{\ ''} - (2\ x + 5)\ y^{\ '} +2\ y=0$ (1)
... which has the form...
$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x)$ (2)
If $u(x)$ and $v(x)$ are solutions of (1) then...
$\displaystyle (x+1)\ u^{\ ''} - (2\ x + 5)\ u^{\ '} +2\ u=0$
$\displaystyle (x+1)\ v^{\ ''} - (2\ x + 5)\ v^{\ '} +2\ v=0$ (3)
Multiplying the first of (3) by v and the second by u and do the difference we obtain...
$\displaystyle (x+1)\ (v\ u^{\ ''}-u\ v^{\ ''}) - (2\ x + 5)\ (v\ u^{\ '} -u\ v^{\ '})=0$ (4)
Now we set $\displaystyle z= v\ u^{\ '} -u\ v^{\ '}$ so that (4) becomes...
$\displaystyle z^{\ '}= \frac{2\ x+5}{x+1}$ (5)
The (5) is a first order ODE one solution of which is...
$\displaystyle z= 2\ x + 3\ \ln (x+1)$ (6)
... so that is...$\displaystyle \frac{d}{d x} (\frac{u}{v})= \frac{z}{v^{2}} = \frac{2\ x + 3\ \ln (x+1) + c_{2}}{v^{2}} \implies u= v\ \int \frac{2\ x + 3\ \ln (x+1) }{v^{2}}\ dx$ (7)
Now it is easy enough to see that $\displaystyle v= x+ \frac{5}{2}$ is solution of (1) so that from (7) we derive that...
$\displaystyle u= \ln (2\ x +5) + 2\ (x+1)\ \ln (x+1) + (2\ x+5)\ \ln (2\ x +5)$ (8)
... and the general solution of (1) is...
$\displaystyle y(x)= c_{1}\ (2\ x+5) + c_{2}\ \{(x+1)\ \ln (x+1) + (x+3)\ \ln (2\ x +5)\}$ (9)
... and the first part is completed. Now You have to search a particular solution of the complete equation... $\displaystyle (x+1)\ y^{\ ''} - (2\ x + 5)\ y^{\ '} +2\ y= (x+1)\ e^{x}$ (10)
Kind regards
$\chi$ $\sigma$
