Second order linear differential equation with non-constants coefficients $xy''+2(x+1)y'+((x+2)y=0$

Question

I am struggling with this equation on $\mathbb{R}$ $$xy''+2(x+1)y'+(x+2)y=0$$ I was searching for the right substitution but I didn"t find a good way! Any help, please?

Answer 1

By substitut $z=xy$ on this DE we get \begin{align} xy''+2(x+1)y'+(x+2)y&=(xy''+2y')+2(xy'+y)+xy\\ &=z"+2z'+z\\ &=0 \end{align}

now we use the characteristic equation $$r^2+2r+1=0$$ which has a unique solution $r=-1$ of multiplicity two, so the solution of this DE is given by

$$z(x)=(ax+b)e^{-x}$$ which mean that the solution of the initial problem on $]-\infty , 0[$ or $]0,\infty[$is of the form $$y(x)= \frac{ax+b}{x}e^{-x}$$ Now let $y$ be

$$ y(x)= \left\{ \begin{array}{l l} \frac{a_1x+b_1}{x}e^{-x} & x<0\\ \frac{a_2x+b_2}{x}e^{-x} &x>0 \end{array} \right. $$

So $y$ is a solution on $\mathbb{R}^*$, and to be a solution overall $\mathbb{R}$ it is easy to check that $a_1,a_2,b_1;b_2$ must verify $$ a_1=a_2 \text{ and } b_1=b_2=0$$ Here you are

Answer 2

A particular solution could be $y=e^{ax}$

Then we have that $$x(a^2+2a+1)+2(a+1)=0$$

$$\implies a=-1 \implies y_p=e^{-x}$$ Now you can use the reduction of order method $$y=v(x)e^{-x}$$ The equation becomes $$x(v''-2v')+2(x+1)v'=0$$ $$xv''+2v'=0$$

Which is separable ... $$(\ln |{v'}|)'=-\frac 2x$$ Integrate twice to get : $$\implies v(x)=K_1+\frac {K_2}x$$ $$\boxed {y(x)=e^{-x} \left (K_1+\frac {K_2}x \right)}$$