solving Second order non - homogeneous Differential Equation

[ssh]

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 \( y = c_1 y_1 + c_2 y_2\)

 

[tkhunny]

Are you sure you don't want to include x^2 or x^3?

 

[chisigma]

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/h...econd-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

$\chi$ $\sigma$

 

[chisigma]

The procedure described in post #12 of...

http://www.mathhelpboards.com/f17/how-solve-differential-equation-second-order-linear-variable-coefficient-2089/index2.html

... may be used to find the particular solution Y(x) of the ODE proposed in this thread. The computation however is very unpleasant and the task is left as exercize to the readers ...

Kind regards

$\chi$ $\sigma$