Mike's question at Yahoo! Answers regarding reduction of order

[MarkFL]

Here is the question:

The differential equation

\(\displaystyle y''+2y'+y=0\)

has linear independent solutions $e^{-x}$ and $xe^{-x}$. Pretend you only know that $xe^{-x}$ is a solution, and use reduction of order to obtain a second linearly independent solution.

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Mike,

We are given the nontrivial solution:

\(\displaystyle f(x)=xe^{-x}\)

So, let's set:

\(\displaystyle y(x)=v(x)f(x)\) and substitute for the derivatives of $y(x)$ into the given ODE. Thus, computing the needed derivatives, we find:

\(\displaystyle y'(x)=v(x)f'(x)+v'(x)f(x)\)

\(\displaystyle y''(x)=v''(x)f(x)+2v'(x)f'(x)+v(x)f''(x)\)

Now we need to compute the derivatives of $f(x)$:

\(\displaystyle f'(x)=-xe^{-x}+e^{-x}=e^{-x}(1-x)\)

\(\displaystyle f''(x)=-e^{-x}+(x-1)e^{-x}=e^{-x}(x-2)\)

And so we find:

\(\displaystyle y'(x)=v(x)e^{-x}(1-x)+v'(x)xe^{-x}=e^{-x}\left(v(x)(1-x)+xv'(x) \right)\)

\(\displaystyle y''(x)=v''(x)xe^{-x}+2v'(x)e^{-x}(1-x)+v(x)e^{-x}(x-2)=e^{-x}\left(xv''(x)+2v'(x)(1-x)+v(x)(x-2) \right)\)

Substituting into the original ODE, we find:

\(\displaystyle e^{-x}\left(xv''(x)+2v'(x)(1-x)+v(x)(x-2) \right)+2\left(e^{-x}\left(v(x)(1-x)+xv'(x) \right) \right)+v(x)xe^{-x}=0\)

Since \(\displaystyle e^{-x}\ne0\) we may divide through by this factor to obtain:

\(\displaystyle xv''(x)+2v'(x)(1-x)+v(x)(x-2)+2v(x)(1-x)+2xv'(x)+v(x)x=0\)

Distribute:

\(\displaystyle xv''(x)+2v'(x)-2xv'(x)+xv(x)-2v(x)+2v(x)-2xv(x)+2xv'(x)+xv(x)=0\)

Combine like terms:

\(\displaystyle xv''(x)+2v'(x)=0\)

Multiply through by $x$:

\(\displaystyle x^2v''(x)+2xv'(x)=0\)

Now the left side is the differentiation of a product:

\(\displaystyle \frac{d}{dx}\left(x^2v'(x) \right)=0\)

Integrate with respect to $x$ to obtain:

\(\displaystyle \int\,d\left(x^2v'(x) \right)=\int\,dx\)

\(\displaystyle x^2v'(x)=c_1\)

\(\displaystyle v'(x)=c_1x^{-2}\)

Integrate again with respect to $x$:

\(\displaystyle \int\,dv=c_1\int x^{-2}\,dx\)

\(\displaystyle v(x)=-c_1x^{-1}+c_2\)

For simplicity, let $c_1=-1$ and $c_2=0$ and so:

\(\displaystyle v(x)=\frac{1}{x}\)

And hence, we find our second linearly independent solution is:

\(\displaystyle y(x)=v(x)f(x)=\frac{1}{x}xe^{-x}=e^{-x}\)

Shown as required.