3
$\begingroup$

I need to apply the reduction of order into differential equation $$\cos^2xy''-6y=0$$where the first solution is of the form $$y_1=(\beta \tan^2x+1)$$ with $\beta \in R$.

I started solving equation with reduction of order, however I don't know how should I use the knowledge of "$\beta \in R$."

$$y_1=(\beta \tan^2x+1)$$ $$y_2=u\cdot y_1=u\cdot(\beta \tan^2x+1)$$ $$y_2'=(u\cdot y_1)'= \dot u(\beta \tan^2x+1)+u(2\beta \cdot \tan x\cdot \sec^2x)$$

$$y_2''= (\dot u(\beta \tan^2x+1)+u(2\beta \cdot \tan x\cdot \sec^2x))'= \ddot u (\beta \tan^2x+1)+ \dot u(4\beta \tan x \cdot \sec^2x) +u(2\beta \sec^4x+4\beta \tan^2x \sec^2x) $$

After inserting $y_2, y_2',y_2''$ into $\cos^2xy''-6y=0$ I got the form where any of the element of the equation want to reduce. Therefore, I would like to ask should I assume do with $\beta$, can I assume that $\beta$ equals for instance 1?

$\endgroup$
  • $\begingroup$ The form of the equation after substitution look like this: $$\ddot u (\beta sin^2x+cos^2x) + \dot u(4 \beta tanx) + u(2\beta sec^2x + 4 \beta tan^2x) - 6 (\beta tan^2x +1) = 0$$ Always when I solve the reduction of order problem one of the elements $\ddot u, \dot u$ or $u$ is reduced. In this case I can't do it. any tips? $\endgroup$ – Vid Feb 20 '18 at 22:10
2
$\begingroup$

What you are doing is too complicated. You know $y_1$ is a solution, you just use that knowledge:

$$y_2=u y_1$$

$$y_2'=u'y_1+u y_1'$$

$$y_2''=u''y_1+2u' y_1'+u y_1''$$

Now substitute in the equation:

$$\cos^2 x~(u''y_1+2u' y_1'+u y_1'')-6 u y_1=0$$

Using the fact that:

$$\cos^2 x~y_1''-6y_1=0$$

We have:

$$\cos^2 x~(u''y_1+2u' y_1')=0$$

Now you only need to compute $y_1'$ (already done) and take $u'=v$ as the new function.

Edit

However, $y_1$ is not a solution for any $\beta$, the OP was right.

To find $\beta$ we need to substitute just $y_1$ in the equation and find for which value of $\beta$ it is correct.

In other words, you need to simplify:

$$\cos^2 x ( \beta \tan^2 x+1)''-6 ( \beta \tan^2 x+1)=0$$

I haven't done this by hand, but Wolfram Alpha gives $\beta=3$.

$\endgroup$
  • 1
    $\begingroup$ @Vid, wait a minute, you were right! You need to find a particular value for $\beta$ that satisfies the equation $\endgroup$ – Yuriy S Feb 20 '18 at 22:16
  • $\begingroup$ @Vid, see the edit $\endgroup$ – Yuriy S Feb 20 '18 at 22:17
  • $\begingroup$ Ok, so now I will calculate first and second derivative for $y_1$ to get the value of beta. It can't be short problem because it costed 33 out of 100 point in my exam :D $\endgroup$ – Vid Feb 20 '18 at 22:19
  • $\begingroup$ @Vid, well, you've already done all the necessary work. Now you are left with some simplifications, and solving a first order separable ODE $\endgroup$ – Yuriy S Feb 20 '18 at 22:23
  • $\begingroup$ thank you @Yuriy because it wasn't obvious for me that I should start from inserting $y_1$ and derivatives of $y_1$ into equation! $\endgroup$ – Vid Feb 20 '18 at 22:27

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.