# Help with solving differential equations

I tried to solve this third order differential equation:

$$x^3y’’’+2x^2y’’-xy’+y=0$$

By substituting $$y=e^{kx}$$ and then finding $$k$$.

But I am not sure if I should solve this by using this method?

• Note: $y=x$ is a solution, so whatever ansatz you use, it should include that as a possibility – Wouter Jun 25 '19 at 18:38
• @user670302: This is a Euler-Cauchy equation and it is better to substitute $$y = x^m$$ Then solve for $m's$. You should of course find three of them, with a repeated one. – Moo Jun 25 '19 at 18:51
• @Moo I dont know what to do with the x before y? – user670302 Jun 25 '19 at 19:04
• – Hans Lundmark Jun 25 '19 at 19:08
• @user670302: For example, find the first, second and third derivative, plug those into the ODE and simplify to solve for the zeros of $m$. Try it! – Moo Jun 25 '19 at 19:20

Hint: $$x^3y'''+2x^2y''-xy'+y=(x^3y'''+3x^2y'')-(x^2y''+2xy')+(xy'+y)=(x^3y''-x^2y'+xy)'=0$$ then $$x^2y''-xy'+y=\dfrac{C}{x}$$ let $$x=e^t$$ then the DE reduces to $$Y''-2Y'+Y=Ce^{-t}$$

$$x^3y’’’+2x^2y’’-xy’+y=0\qquad . . . . . . . (1)$$

Let $$~x=e^z\implies z=\log x$$

Then $$\quad y'=\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{x}\frac{dy}{dz}\implies xy'= \frac{dy}{dz}\equiv Dy\qquad \text{where} \quad D\equiv \frac{d}{dz}$$

$$y''=\frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{dz}+\frac{1}{x}\frac{d^2y}{dz^2}\implies x^2y''=D(D-1)y$$

Similarly $$\quad x^3~y'''=D(D-1)(D-2)y$$

Now $$(1)$$ becomes, $$\{D(D-1)(D-2)+2D(D-1)-D+1\}y=0$$ $$\implies (D-1)\{D(D-2)+2D-1\}y=0$$ $$\implies (D-1)^2(D+1)y=0$$

Roots of the auxiliary equation are $$~1, ~1, ~-1$$.

Hence the general solution is $$y=(c_1+~c_2~ z)e^z+c_3e^{-z}$$i.e., $$y=(c_1+c_2 \log x)~x+c_3~x^{-1}$$where $$~c_1, ~c_2, ~c_3~$$ are arbitrary constants.

To solve the third order Cauchy Euler equation we start by making the substitution

$$y=x^r$$

then

$$y'=rx^{r-1}, y''=r(r-1)x^{r-2}$$ and $$y'''=r(r-1)(r-2)x^{r-3}$$

Substituting this back into the original differential equation yields,

$$x^3r(r-1)(r-2)x^{r-3}+2x^2r(r-1)x^{r-2}-xrx^{r-1}+x^r=0$$,

$$x^r(r(r-1)(r-2)+2r(r-1)-r+1)=0$$,

$$r^3-3r^2+2r+2r^2-2r-r+1=0$$,

$$r^3-r^2-r+1=0$$ or $$r^2(r-1)-(r-1)=0$$ or $$(r^2-1)(r-1)=0$$

so the roots are $$r_1=-1$$ of multiplicity $$1$$

and $$r_2=1$$ of multiplicity $$2$$. Hence the general equation is

$$y=c_1x^{-1}+c_2x + c_3x \ln{x}$$