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I want to find another solution for equation $$ y'' + \frac{y'}{x} - \frac{y}{4x^2} = 0,\ x\in\ (0, \infty) $$ when one solution is $y_1(x) = \sqrt x$.

Another linearly independent solution for equations in the format $y'' + p(x)y' + q(x)y = 0$ should be $$ y_2(x)=y_1(x)\int\frac{e^{-\int p(x)dx}}{y_1(x)^2}dx = \dots = -\frac{1}{\sqrt x}, $$ but now calculating Wronskian determinant gives $$ W = y_1y_2'-y_2y_1' = \frac{1}{2\sqrt x}-\frac{1}{2\sqrt x}=0. $$ So now universal solution isn't $y_h=C_1y_1+C_2y_2$. Is that a wrong approach for finding another solution for this differential equation?

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    $\begingroup$ Are you sure you did the wronskian properly? $\endgroup$ Commented Dec 11, 2019 at 21:03

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I think your Wronskian is off.

$$y_1 = \sqrt{x}$$ $$y'_1 = \frac{1}{2\sqrt{x}}$$ $$y_2 = -\frac{1}{\sqrt{x}}$$ $$y'_2 = \frac{1}{2 x \sqrt{x}}$$ $$y_1 y'_2 = \frac{1}{2x}$$ $$y'_1 y_2 = -\frac{1}{2x}$$ $$y_1 y'_2 - y'_1 y_2 = \frac{1}{2x}-(-\frac{1}{2x})=\frac{1}{x}$$


Addendum

It's been suggested that the OP's solution is inaccurate. Let's try it out: $$ y_1 = \sqrt{x}$$ $$y'_1 = \frac{1}{2\sqrt{x}}$$ $$y'_1 = -\frac{1}{4x\sqrt{x}}$$ $$ \frac{y_1}{4 x^2} = \frac{1}{4 x \sqrt{x}}$$ $$ \frac{y'_1}{ x} = \frac{1}{2 x \sqrt{x}}$$ $$ y''_1 = -\frac{1}{4 x \sqrt{x}}$$ my understanding is that this forms:

$$-\frac{1}{4 x \sqrt{x}}+\frac{1}{2 x \sqrt{x}}-\frac{1}{4 x \sqrt{x}} = (-\frac{1}{4}+\frac{1}{2}-\frac{1}{4})\frac{1}{x\sqrt{x}}=0$$

OP's proposed answer is accurate.


Alternative approach

One can come upon the OP's answer using an ansatz that $y=Cx^m$ We then find based on the derivatives $C$ is a free parameter and $m(m-1)+m-1/4=0$ which leaves $m = \pm \frac{1}{2}$. These are the solutions found by OP

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  • $\begingroup$ No way! Double checked all my calculations literally million times and it turns out that I can't derive $\sqrt x$. Embarrasing. $\endgroup$
    – jte
    Commented Dec 11, 2019 at 21:08
  • $\begingroup$ @jte it happens :) glad I could help $\endgroup$ Commented Dec 11, 2019 at 21:09
  • $\begingroup$ @Isham How come $y = \sqrt x$ is not a solution? I get $y' = \frac{1}{2\sqrt x}$ and $y'' = -\frac{1}{4\sqrt x x}$ and placing them in to equation gives 0. $\endgroup$
    – jte
    Commented Dec 11, 2019 at 21:20
  • $\begingroup$ @jte : seems that taking derivatives is a cursed operation all around today :) $\endgroup$ Commented Dec 11, 2019 at 21:27
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    $\begingroup$ The derivative of $-x^{-1/2}$ is $\frac12x^{-3/2}$, you have the coefficient wrong. $\endgroup$ Commented Dec 12, 2019 at 0:25

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