2
$\begingroup$

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

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!!

$\endgroup$
  • $\begingroup$ Have you studied method of variation of parameters? $\endgroup$ – Mhenni Benghorbal Feb 21 '16 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". $\endgroup$ – A.Roy Feb 21 '16 at 21:37
  • $\begingroup$ So you know how to use reduction of order techniques! $\endgroup$ – Mhenni Benghorbal Feb 21 '16 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 $$-2z(t)=\int^{\cos(t^2)}\frac{dx}{1-x^2}=\frac12\int^{\cos(t^2)}\left(\frac1{1-x}+\frac1{1+x}\right)dx=\cdots$$ $\endgroup$ – Did Feb 23 '16 at 12:17
1
$\begingroup$

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$

$\endgroup$
  • $\begingroup$ 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 $cot(t^2)$... $\endgroup$ – A.Roy Feb 21 '16 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! $\endgroup$ – Mhenni Benghorbal Feb 21 '16 at 22:58
  • $\begingroup$ See the other way that you can solve your problem instead of reduction of order technique! $\endgroup$ – Mhenni Benghorbal Feb 21 '16 at 23:25
  • $\begingroup$ 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. :) $\endgroup$ – A.Roy Feb 21 '16 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? $\endgroup$ – A.Roy Feb 21 '16 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.