# ODE 2nd order non constant coefficient

I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$.

Given $y_1=x$ is a solution $$(2x^2 + 1)y'' − 4xy' + 4y = 0$$ Find the second solution $y_2$ for equation.

• Could you show what you did when attempting reduction of order? That should work fine. – user170231 Jul 11 '16 at 17:53
• I don't have the work any more I threw it out but at the end I got after substituting y = VX in, V'''+V=0.I could of made a mistake I'm unsure. – user353452 Jul 11 '16 at 17:55

Via reduction of order, assume $y_2=y_1v=xv$ is a solution. Then you have derivatives $$\begin{cases}y_2=xv\\{y_2}'=xv'+v\\{y_2}''=xv''+2v'\end{cases}$$ Substitute into the ODE: $$(2x^2+1)\left(xv''+2v'\right)-4x\left(xv'+v\right)+4xv$$ or, grouping together the same-order derivatives, $$x(2x^2+1)v''+\left(2\left(2x^2+1\right)-4x^2\right)v'+(4x-4x)v=0$$ and simplifying a bit gives $$x(2x^2+1)v''+2v'=0$$ As you can see, the order of the ODE has been reduced. Substituting $z=v'$, you have a separable ODE linear in $z$. $$x\left(2x^2+1\right)z'+2z=0$$ (I can add more details on finding the solution upon request, but this should hopefully address whatever previous mistake you made.)

• Oh, okay thanks but how would you go about getting the third solution then. – user353452 Jul 11 '16 at 18:08
• Ah I see, I made a mistake somewhere along the line thanks. – user353452 Jul 11 '16 at 18:21

Hint

By the reduction of order method, if you know a solution $y_1(x)$ of the homogeneous equation

$$y^{''}(x)+p(x)y^{'}(x)+q(x)=0$$

then the other solution is given by

$$y_2(x)=y_1(x)\int \frac{e^{-\int{p(x)dx}}}{y_1^2(x)}dx$$

Solution

So in your example $p(x)=-\frac{4x}{2x^2+1}$ and $y_1(x)=x$. Hence

\begin{align} y_2(x) &= x \int \frac{e^{\int{\frac{4x}{2x^2+1}dx}}}{x^2}dx \\ &= x \int \frac{e^{\ln(2x^2+1)}}{x^2}dx \\ &= x \int \frac{2x^2+1}{x^2}dx \\ &= x \int (2+\frac{1}{x^2})dx \\ &= x (2x-\frac{1}{x}) \\ &= 2x^2-1 \end{align}

• @user353452: I added the final answer too. Hope this helps. :) – H. R. Jul 11 '16 at 18:17
• @user353452: You are welcome. :) – H. R. Jul 11 '16 at 18:21