# Reduction of order where $y_1=(\beta \tan^2x+1)$ with $\beta \in R$

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?

• 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? – Vid Feb 20 '18 at 22:10

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 $\beta=3$.

• @Vid, wait a minute, you were right! You need to find a particular value for $\beta$ that satisfies the equation – Yuriy S Feb 20 '18 at 22:16
• @Vid, see the edit – Yuriy S Feb 20 '18 at 22:17
• 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 – Vid Feb 20 '18 at 22:19
• @Vid, well, you've already done all the necessary work. Now you are left with some simplifications, and solving a first order separable ODE – Yuriy S Feb 20 '18 at 22:23
• 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! – Vid Feb 20 '18 at 22:27