Another second order non homogeneous ODE....

[chisigma]

Four days ago on mathhelpforum.com the user ssh [I don’t know if he the same as in MHB…] has proposed the following second order complete linear ODE…

$\displaystyle y^{\ ''} – \frac{2+x}{x}\ y^{\ ’}\ + \frac{2+x}{x^{2}}\ y = x\ e^{x}$ (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/how-solve-differential-equation-second-order-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

$\chi$ $\sigma$

 

[chisigma]

For completeness sake now we describe the general solving procedure of an ODE of the type...

$\displaystyle y^{\ ''} + p(x)\ y^{\ ’}\ + q(x)\ y = \varphi(x)$ (1)

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

$\displaystyle y^{\ ''} + p(x)\ y^{\ ’}\ + q(x)\ y = 0$ (2)... which can be written as... $y(x)= c_{1}\ u(x) + c_{2}\ v(x)$ (3)... where u(*) and v(*) are two independent solutions of (2), $c_{1}$ and $c_{2}$ are constants. If u and v are both solution of (2) then ...

$\displaystyle u^{\ ''} + p(x)\ u^{\ ’}\ + q(x)\ u = 0$

$\displaystyle v^{\ ''} + p(x)\ v^{\ ’}\ + q(x)\ v = 0$ (4)

Multiplying the first of the (4) by v, the second by u and subtracting is...

$\displaystyle (v\ u^{\ ''} - u\ v^{\ ''}) = - p(x)\ (v\ u^{\ '} - u\ v^{\ '})$ (5)

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

$\displaystyle z^{\ '}= - p(x)\ z$ (6)

... one solution of which is $\displaystyle z = e^{- \int p(x)\ dx}$, so that is...

$\displaystyle \frac{d}{d x} (\frac{u}{v}) = \frac{z}{v^{2}} \implies u= v\ \int \frac{z}{v^{2}}\ dx$ (7)

... and the (7) allows to obtain, if we are able to find a solution v of (2), another solution u independent from it.

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

$\displaystyle Y= C_{1}(x)\ u + C_{2} (x)\ v$ (8)

... where...

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

$\displaystyle C_{2}(x) = \int \frac{ u\ \varphi(x)}{W_{u,v}(x)}\ dx$ (9)

... the Wronskian of u and v being...

$\displaystyle W_{u,v} (x)= u\ v^{\ '}-v\ u^{\ '}$ (9)

The general solution of (1) is...

$\displaystyle y(x)= \{c_{1} + C_{1} (x)\}\ u(x) + \{c_{2} + C_{2}(x)\}\ v(x) $ (10)

The procedure we have described permits a comfortable solution of an ODE like (1). The only 'weak point' is that it can start only if a particular solution of the incomplete ODE is known and that sometime can be an hard task...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

An interesting example of second order linear incomplete ODE can be found in mathhelpforum.com...

series solution!

The ODE is...

$\displaystyle y^{\ ''} + (1+\frac{1}{x})\ y^{\ '} - \frac{1}{x^{2}}=0$ (1)

... and for that a 'series solution' is explicity required. Applying the procedure we have described in previous post we arrive at the first order ODE...

$\displaystyle z^{\ '} = -(1+\frac{1}{x})\ z$ (2)

... one solution of which is...

$\displaystyle z= \frac{e^{-x}}{x}$ (3)

... so that if u and v are two independent solutions of (1) then is...

$\displaystyle u= v\ \int \frac{e^{-x}}{x\ v^{2}}\ dx$ (4)

Now $v= \frac{e^{-x}}{x}$ is solution of (1) so that we obtain...

$\displaystyle u= \frac{e^{-x}}{x}\ \int x\ e^{x}\ dx = \frac{x-1}{x}$ (5)... so that the general solution of (1) is...

$\displaystyle y(x)=c_{1}\ \frac{x-1}{x} + c_{2}\ \frac{e^{-x}}{x}$ (6)

Observing (6) it is obvious that any non zero solution of (1) has a singularity in x=0 and therefore the standard Mc Laurin series solution attempt necessarly fails...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

... so that the general solution of (1) is...

$\displaystyle y(x)=c_{1}\ \frac{x-1}{x} + c_{2}\ \frac{e^{-x}}{x}$ (6)

Observing (6) it is obvious that any non zero solution of (1) has a singularity in x=0 and therefore the standard Mc Laurin series solution attempt necessarly fails...

Of course that is not fully true, because in the particular case $\displaystyle c_{2}=c_{1}=c$ the series solution do exist and is...

$\displaystyle y(x)= c\ \sum_{n=1}^{\infty} (-1)^{n+1}\ \frac{x^{n}}{(n+1)!}$

In this case however is $y(0)=0$ and You can only impose a value to $y^{\ '}(0)$...

Kind regards

$\chi$ $\sigma$

 

[CellCoree]

Dear user ssh,

Thank you for sharing this second order non-homogeneous ODE and the proposed solution on the mathhelpforum.com. It is always interesting to see new problems and different approaches to solving them.

Upon reviewing your solution, I would like to offer some feedback. First, it is important to note that the method used in your solution, known as the method of variation of parameters, can only be applied to linear differential equations with constant coefficients. Therefore, it may not be applicable to all second order ODEs.

Additionally, in your solution, you mention that $v=x$ is a solution of the homogeneous ODE (2), but it is actually a solution of the non-homogeneous ODE (1). This is because the right side of (2) is equal to 0, while the right side of (1) is not. Therefore, the particular solution $Y$ in (7) should be in the form of $Y=C_1(x)v+C_2(x)u$.

Lastly, I would like to point out that the solution to the homogeneous ODE (2) is actually $y_h(x)=c_1x+c_2x^2$. This can be easily verified by plugging it into (2) and observing that the left side is equal to the right side.

I hope this feedback is helpful in further refining your solution. Keep up the good work and keep exploring new problems in mathematics.

Best regards,
Scientist