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I feel like I'm missing a crucial step but I cannot seem to solve the following differential equation.

It is given to us that:

$$ \\ 2x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}-1=0 \\ $$

where $y(1)=1$ and $y(2)=2$

How would I solve this? Using which method? I have only been taught at university the method for solving this type of equation using the complementary function method.

Any help would be really appreciated!

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  • $\begingroup$ This is a Cauchy-Euler equation. See en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation $\endgroup$ – Chee Han Mar 8 '17 at 19:05
  • $\begingroup$ Try the substitution: $x=e^{z}$ $\endgroup$ – mrnovice Mar 8 '17 at 19:05
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    $\begingroup$ hint: $$y=-\frac{1}{x}$$ is one solution $\endgroup$ – Dr. Sonnhard Graubner Mar 8 '17 at 19:12
  • $\begingroup$ It's not a Cauchy-Euler equation. $2x^2 y'' +4x y' - y = 0$ is one though. $\endgroup$ – BobaFret Mar 8 '17 at 19:30
  • $\begingroup$ @Dr. Sonnhard Graubner You should check twice what you write before writing it : this solution is erroneous. $\endgroup$ – Jean Marie Mar 8 '17 at 19:32
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Considering $$\\ 2x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}-1=0 \\$$ first reduce the order using $z=\frac{dy}{dx}$. This gives $$2x^2\frac{dz}{dx}+4xz-1=0$$ The homogneous equation is then $$x\frac{dz}{dx}+2z=0$$ which is easy to integrate (since separable). So, $$z=\frac{C}{x^2}$$ Now, variation of parameter $$2x^2\frac{dz}{dx}+4xz-1=0\implies 2 C'-1=0\implies C=\frac x2+K_1\implies z=\frac{1}{2 x}+\frac{K_1}{x^2}$$ So, $$\frac{dy}{dx}=\frac{1}{2 x}+\frac{K_1}{x^2}$$ which seems to be easy.

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  • $\begingroup$ I don't understand how you got from your second line to your third line, did you divide by $2x$? $\endgroup$ – Sebastian TG Mar 9 '17 at 9:30
  • $\begingroup$ @SebastianTG. For sure, I did since just concerned by the homogeneous equation. Is my way clear for you ? $\endgroup$ – Claude Leibovici Mar 9 '17 at 9:33
  • $\begingroup$ if you divided your second line by $2x$ shouldn't that leave you with $x\frac{dz}{dx}+2z-\frac{1}{2x}$? $\endgroup$ – Sebastian TG Mar 9 '17 at 9:36
  • $\begingroup$ @SebastianTG. Getting the solution of the homogeneous equation, I went back to my second line to apply the method of variation of parameters. $\endgroup$ – Claude Leibovici Mar 9 '17 at 9:42
  • $\begingroup$ I see what you did, thank you so much this has expanded my understanding! $\endgroup$ – Sebastian TG Mar 9 '17 at 10:01
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Note that we can rewrite the equation: $$2x^2\frac{d^2y}{dx^2}+4x\frac{dx}{dy}-1=\frac{d}{dx}\left(2x^2\frac{dx}{dy}-x\right)=0$$ Ergo, we have the following for some constant $C$: $$2x^2\frac{dy}{dx}-x=C$$ Now we can solve quite easily: \begin{align*} &2x^2\frac{dy}{dx}-x=C\\ &\frac{dy}{dx}=\frac{C}{2x^2}+\frac{1}{2x}\\ &y=\int\left[ \frac{C}{2x^2}+\frac{1}{2x}\right]dx\\ &y=-\frac{C}{2x}+\frac{1}{2}\ln\left(x\right)+B \end{align*} Letting $A=\frac{C}{2}$, we have a final answer: $$y=\frac{1}{2}\ln\left(x\right)-\frac{A}{x}+B$$

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Let $x=e^{s}$, then $$\frac{dy}{ds}=\frac{dy}{dx}\frac{dx}{ds}=\frac{dy}{dx}e^{s}=x\frac{dy}{dx}$$ and $$\frac{d^{2}y}{ds^{2}} =\frac{d}{ds}\left(x\frac{dy}{dx}\right) =\frac{dx}{ds}\frac{dy}{dx}+x\frac{d}{ds}\frac{dy}{dx} =x\frac{dy}{dx}+x^{2}\frac{d^{2}y}{dx^{2}} $$

So $$x\frac{dy}{dx}=\frac{dy}{ds},\quad x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{ds^{2}}-\frac{dy}{ds} $$ The coefficients are now constant and you can use the usual techniques to solve the new equation

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$2x^{2}\frac{d^{2}y}{dx^{2}} + 4x\frac{dy}{dx} -1 = 0$

Using the substitution $x=e^{z}$

$\frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz}$

$\frac{dx}{dz} = e^{z}$

$\frac{dy}{dz} = e^{z}\frac{dy}{dx}$

$\frac{dy}{dx} = e^{-z}\frac{dy}{dz}$

$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(e^{-z}\frac{dy}{dz}) = \frac{dz}{dx}\cdot \frac{d}{dz}(e^{-z}\frac{dy}{dz}) = e^{-z}(e^{-z}\frac{dy}{dz} + e^{-z}\frac{d^{2}y}{dz^{2}}) = e^{-2z}(\frac{dy}{dz} + \frac{d^{2}y}{dz^{2}})$

Then substituting into the original equation:

$2e^{2z}\cdot e^{-2z}(\frac{dy}{dz} + \frac{d^{2}y}{dz^{2}}) + 4e^{z}\cdot e^{-z}\frac{dy}{dx}-1 = 0$

$\Rightarrow 2\frac{d^{2}y}{dz^{2}} + 6\frac{dy}{dz} -1 = 0$

This is now in the form you are familiar with, which you should be able to solve. If you get stuck let me know.

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