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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

$\displaystyle y^{\ ''} – \frac{2+x}{x}\ y^{\ ’}\ + \frac{2+x}{x^{2}}\ y = x\ e^{x}$ (1)

… and till now no satisfactory solution has been supplied. Well!... now we have the opportunity to test the procedure described in…

http://www.mathhelpboards.com/f17/h...-linear-variable-coefficient-2089/index2.html

The first step is to find the general solution of the incomplete ODE...

$\displaystyle y^{\ ''} – \frac{2+x}{x}\ y^{\ ’}\ + \frac{2+x}{x^{2}}\ y = 0$ (2)

If u and v are two independent solution of (2) then we arrive to write...

$\displaystyle (v\ u^{\ ''} - u\ v^{\ ''}) = \frac{2+x}{x}\ (v\ u^{\ '} - u\ v^{\ '})$ (3)

... and setting $z= v\ u^{\ '} - u\ v^{\ '}$ we obtain the first order ODE...

$\displaystyle z^{\ '}= \frac{2+x}{x}\ z$ (4)

... one solution of which is $\displaystyle z=x^{2}\ e^{x}$, so that is...

$\displaystyle \frac{z}{v^{2}} = \frac{d}{d x} (\frac{u}{v}) = \frac{x^{2}\ e^{x}}{v^{2}} \implies u= v\ \int \frac{x^{2}\ e^{x}}{v^{2}}\ dx$ (5)

It is easy enough to see that $v=x$ is solution of (2) so that…

$\displaystyle u= x\ \int e^{x}\ dx = x\ e^{x}$ (6)

Now that we have u and v we have to find the particular solution of (1) in the form...

$\displaystyle Y= C_{1}(x)\ u + C_{2} (x)\ v$ (7)

... where...

$\displaystyle C_{1}(x) = - \int \frac{ v\ \varphi(x)}{W_{u,v}(x)}\ dx$

$\displaystyle C_{2}(x) = \int \frac{ u\ \varphi(x)}{W_{u,v}(x)}\ dx$ (8)

The Wronskian is computed as $\displaystyle W_{u,v} (x)= u\ v^{\ '}-v\ u^{\ '} = - x^{2}\ e^{x}$ and is $\displaystyle \varphi(x)= x\ e^{x}$ so that we obtain...

$\displaystyle C_{1}(x)= \int dx = x$

$\displaystyle C_{2}(x)= - \int \frac{d x}{x}= \ln \frac{1}{|x|}$ (9)

... and the general solution of (1) is...

$\displaystyle y(x)= c_{1}\ x\ e^{x} + c_{2}\ x + x^{2}\ e^{x} + x\ \ln \frac{1}{|x|}$ (10)

Kind regards

$\chi$ $\sigma$