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So for these problems it asks us to use reduction of order to find a second solution to the differential equations. My professor said that there exist other methods to solve these problems. I am working on problem 29 and as you can see the "reduction of order" method will get very messy with this one. I was wondering what the other way of solving this may be and if it exists it is even any easier? Thanks so much.

P.S. I have the solution already and it appears as if there could be an easy way to do it since it is just $y_2=x^{1/4}e^{-2{\sqrt x}}$

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Use Abel's Theorem: if your equation is of the form $y''+p(t)y'+q(t)y=g(t)$, then the Wronskian of your solution set is given by $W=e^{\int -p(t)dt}$. We also know the Wronskian of two solutions is given by $y_1y_2'-y_2y_1'$, so setting those two equal gives a simple first order equation. In this case, rewriting your equation in standard form gives us $$y''+0y'+(0.1875x^{-2}-x^{-1})y=0$$ (note ever symbol is a + and the leading coefficient of $y''$ is a $1$, that's standard form) so we get $W=e^{\int 0dx}=e^C$ We might as well pick the simplest $C$ and go for $C=0$, so we get $W=1$.

Then, we get, choosing $y_1=x^{1/4}e^{2\sqrt x},y_1'=\frac 1 4x^{-\frac 3 4}e^{2\sqrt x}+x^{1/4}e^{2\sqrt x}x^{-1/2}=e^{2\sqrt x}(\frac 1 4x^{-\frac 3 4}+x^{-\frac 1 4})$

So, $1=x^{1/4}e^{2\sqrt x}y_2'-e^{2\sqrt x}(\frac 1 4x^{-\frac 3 4}+x^{-\frac 1 4})y_2$. Divide through by the coefficient of $y_2'$ to put this into standard first order form, and change the - to a +, to get $y_2'+(-\frac 1 4x^{-1}-x^{-\frac 1 2})y_2=x^{-\frac 1 4}e^{-2\sqrt x}.$

So our integrating factor is $\mu=e^{\int-\frac 1 4x^{-1}-x^{-\frac 1 2}dx}=e^{-\frac 1 4\ln x-2x^\frac 1 2}=x^{-\frac 1 4}e^{-2x^{\frac 1 2}}$

You'll note that the integrating factor is exactly $(y_1^{-1})$. This is not a coincidence, it should always work that way. If you don't get that, you made a mistake setting up your problem.

So, setting up your answer, we have $y_2=\frac {\int g(t)\mu (t)dt} {\mu (t)}$, so $y_2=\frac {\int x^{-\frac 1 4}e^{-2x^{\frac 1 2}}x^{-\frac 1 4}e^{-2\sqrt x}dx}{x^{-\frac 1 4}e^{-2x^{\frac 1 2}}}=\frac {\int x^{-\frac 1 2}e^{-4\sqrt x}dx}{x^{-\frac 1 4}e^{-2x^{\frac 1 2}}}$

Now, setting $u=-4\sqrt x$, $du=-2x^{-\frac 1 2}dx,-\frac 1 2 du=x^{-\frac 1 2}dx$we have the integral in the top converts to $\int -\frac 1 2 e^u du=-\frac 1 2 e^u +C=-\frac 1 2 e^{-4\sqrt x} +C$

Plugging that back into the top, we get $y_2=\frac {-\frac 1 2 e^{-4\sqrt x} +C}{x^{-\frac 1 4}e^{-2x^{\frac 1 2}}}=-\frac 1 2e^{-2\sqrt x}x^{-\frac 1 4}+Cx^{\frac 1 4}e^{2\sqrt x}$

Now, the second part with the C is just your original $y_1$ solution times an arbitrary constant. The $-\frac 1 2$ came up in this calculation because we CHOSE an arbitrary $C$ when we calculated the wronskian, any other constant multiple will also work, so we might as well take the constant multiple to be 1, and that gets to the second solution you have.

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