# Solutions of the differential equation $x^2y’’-4xy’+6y=0$.

In one of my test it given to prove that $$x^3$$ and $$x^2|x|$$ are linear independent solutions of the differential equation $$x^2y’’-4xy’+6y=0$$ on $$\mathbb R$$( here $$x$$ is independent variable).

But according to me it’s Cauchy Euler equation having general solution as $$y=c_1x^3+c_2x^2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants. How can be $$x^2|x|$$ a solution of given ODE as I am unable to find its by giving particular values of constants $$c_1$$ and $$c_2$$? Please help me to solve it . Thank you.

• Because $x^2|x|$ is piecewise defined, and that is such a form of $kx^3$. Jan 18 at 15:15
• @Nightflight but it is not exactly equal to it . What is k ? Jan 18 at 15:16
• $k$ is constant, piecewise defined. Jan 18 at 15:17
• Writing $y=x^2z$ so $x^4z^{\prime\prime}=0$, the real question is why $z=|x|$, rather than just $z=c_1x+c_2$, solves $z^{\prime\prime}=0$ for $x\ne0$.
– J.G.
Jan 18 at 15:19
• @Nightflight but as I know every particular solution must be obtained by general solution by giving particular values of constants. Is it not a true statement? Jan 18 at 15:19

The differential equation has a singularity at $$x=0$$, so the Existence and Uniqueness Theorem doesn't apply there. On each of the intervals $$(-\infty, 0)$$ and $$(0,\infty)$$ where the theorem does apply, you have two-parameter families of solutions. But it turns out any solution on $$(-\infty, 0)$$ and any solution on $$(0,\infty)$$ with the same $$c_2$$ can be put together to make a solution on $$\mathbb R$$.
$$c_1x^3+c_2x^2$$ is not the general solution of the Cauchy-Euler equation.
If you set $$x=e^u$$, you indeed linearize the original equation to
$$\ddot y-5\dot y+6y=0$$ with the obvious solution $$y=c_3e^{3u}+c_2e^{2u}=c_3x^3+c_2x^2.$$
But recall that by our change of variable, $$x$$ was assumed positive. Now we can solve again by setting $$x=-e^u$$ and get a solution of a similar shape, but the constants have no reason to be equal on both sides.