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

[MarkFL]

Here is the question:

The differential equation

$$y''+2y'+y=0$$

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:

$$f(x)=xe^{-x}$$

So, let's set:

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

$$y'(x)=v(x)f'(x)+v'(x)f(x)$$

$$y''(x)=v''(x)f(x)+2v'(x)f'(x)+v(x)f''(x)$$

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

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

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

And so we find:

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

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

Substituting into the original ODE, we find:

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

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

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

Distribute:

$$xv''(x)+2v'(x)-2xv'(x)+xv(x)-2v(x)+2v(x)-2xv(x)+2xv'(x)+xv(x)=0$$

Combine like terms:

$$xv''(x)+2v'(x)=0$$

Multiply through by $x$:

$$x^2v''(x)+2xv'(x)=0$$

Now the left side is the differentiation of a product:

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

Integrate with respect to $x$ to obtain:

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

$$x^2v'(x)=c_1$$

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

Integrate again with respect to $x$:

$$\int\,dv=c_1\int x^{-2}\,dx$$

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

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

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

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

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

Shown as required.