1
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ Because $x^2|x|$ is piecewise defined, and that is such a form of $kx^3$. $\endgroup$
    – MH.Lee
    Jan 18, 2022 at 15:15
  • $\begingroup$ @Nightflight but it is not exactly equal to it . What is k ? $\endgroup$
    – neelkanth
    Jan 18, 2022 at 15:16
  • $\begingroup$ $k$ is constant, piecewise defined. $\endgroup$
    – MH.Lee
    Jan 18, 2022 at 15:17
  • $\begingroup$ 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 $x\ne0$. $\endgroup$
    – J.G.
    Jan 18, 2022 at 15:19
  • $\begingroup$ @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? $\endgroup$
    – neelkanth
    Jan 18, 2022 at 15:19

2 Answers 2

7
$\begingroup$

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 $\mathbb R$.

$\endgroup$
0
$\begingroup$

$$c_1x^3+c_2x^2$$ 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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .