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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.

neelkanth
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  • Because $x^2|x|$ is piecewise defined, and that is such a form of $kx^3$. – MH.Lee Jan 18 '22 at 15:15
  • @Nightflight but it is not exactly equal to it . What is k ? – neelkanth Jan 18 '22 at 15:16
  • $k$ is constant, piecewise defined. – MH.Lee Jan 18 '22 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 '22 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? – neelkanth Jan 18 '22 at 15:19
  • @J.G. in solving Cauchy Euler we put $x=e^z$ So that $z=ln( x)$ so no role of zero ? Am I correct now ? – neelkanth Jan 18 '22 at 15:22
  • It seems that Cauchy-Euler equation may have many non-smooth solutions. Why did no textbook talk about this? – Zerox Jan 18 '22 at 15:24
  • @neelkanth Your definition of $z$ diverges at $x=0$ (in fact, the case $x<0$ requires either complex $z$ or the alternative substitution $x=\pm e^z$), so behaviour either side of that can be unrelated. This is a similar example. – J.G. Jan 18 '22 at 15:33

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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 $\mathbb R$.

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