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$$ty'' − y' + (4t^3)y = 0, \quad t > 0;\quad y_1(t) = \sin(t^2)$$

The problem states:

"If y1 is a known nonvanishing solution of y" + p(t)y' + q(t)y = 0, show that a second solution $y_2$ satisfies

$$\left(\frac{y_2}{y_1}\right)' = \frac{W(y_1, y_2)}{y_1^2}$$

where $W(y_1, y_2)$ is the Wronskian of $y_1$ and $y_2$.

2 Then use Abel’s formula to determine $y_2$."

Abel's formula as given in the book: $$W(y_1, y_2)(t) = Ce^{-\int p(t) \, dt}$$

I am having a very hard time with 2. I don't understand why it is useful or how to utilize it. I've tried doing reduction of order but just ended up with an unsolvable integral ($\int \cot(x^2) \, dx$).

This problem has given me multiple headaches and has wasted many sheets of paper. Could I get a shove in the right direction or some outline of how to do such a problem?

Any and all help is greatly appreciated!!

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  • $\begingroup$ Have you studied method of variation of parameters? $\endgroup$ Commented Feb 21, 2016 at 21:30
  • $\begingroup$ Looking in the book right now. "Variation of Parameters" is the subject of next week's section. This problem comes from the section titled "Repeated Roots; Reduction of Order". $\endgroup$ Commented Feb 21, 2016 at 21:37
  • $\begingroup$ So you know how to use reduction of order techniques! $\endgroup$ Commented Feb 21, 2016 at 21:44
  • $\begingroup$ Step 1: Compute wronskian, here $W(t)=t$. Step 2. Apply Ansatz $z'=W/y_1$ to compute $z=y_2/y_1$, here $$z(t)=\int^t\frac{s\,ds}{\sin(s^2)}=\int^{t^2}\frac{du}{2\sin u}=\int^{t^2}\frac{d(\cos u)}{-2\sin^2 u}$$ hence $$-2z(t)=\int^{\cos(t^2)}\frac{dx}{1-x^2}=\frac12\int^{\cos(t^2)}\left(\frac1{1-x}+\frac1{1+x}\right)dx=\cdots$$ $\endgroup$ Commented Feb 23, 2016 at 12:17

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So you know how to use reduction of order techniques! Just assume $y_2=uy_1$ and plug in the ode and try to find $u$

added Here is anther way using the definition of Wronskian

$$W=y_1y'_2-y'y_2 \implies \frac{W}{y_{1}^2} = \frac{y'_2}{y_{1}} - \frac{y_2}{y_{1}^2} = \left( \frac{y_2}{y_{1}} \right)' $$

Which implies by integrating both sides of the last eq. with respect to x

$$ \int \frac{W}{y_{1}^2} = \frac{y_2}{y_{1}} . $$

The last eq. gives you $ y_2$

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  • $\begingroup$ I tried doing that, ending up with $tv"sin(t^2) + 4tv'cos(t^2) - v'sin(t^2) = 0$ after simplifying what I could. From there I substituted w = v', arriving at $(1/w)dw = (1/t)dt - 4cot(t^2)dt$ After that, I didn't know how to find w by integration because I can't integrate $cot(t^2)$... $\endgroup$ Commented Feb 21, 2016 at 22:19
  • $\begingroup$ See here. Note that in your question you are supposed to apply the method to the equation$y''+py'+q=0$ and then use the result to solve your specific problem! $\endgroup$ Commented Feb 21, 2016 at 22:58
  • $\begingroup$ See the other way that you can solve your problem instead of reduction of order technique! $\endgroup$ Commented Feb 21, 2016 at 23:25
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    $\begingroup$ Thank you very much, I will be trying to figure this equation out using your non- reduction of order technique. I think that is exactly what they want. :) $\endgroup$ Commented Feb 21, 2016 at 23:47
  • $\begingroup$ I have tried using that equation with $y1 = sin(t^2)$ and $y2 = v(t)sin(t^2)$ which gives me a Wronskian of $v'sin^2(t^2)$ , but all I end up with is $/int v' = v$ Is there something I am missing? $\endgroup$ Commented Feb 21, 2016 at 23:59

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