2
$\begingroup$

I have to solve this ODE using the variation of parameters method:

$$4y''+y=\frac{2}{\cos \left( \frac{x}{2} \right)}$$

Solving the homogeneous problem yields

$$y_h(x)= c_1 \cos \left( \frac{x}{2} \right)+ c_2 \sin \left( \frac{x}{2} \right)$$

Now, to solve the variation of parameters problem, you have to solve

$$ \begin{bmatrix} \cos \left( \frac{x}{2} \right) & \sin \left( \frac{x}{2} \right) \\ -\frac{1}{2} \sin \left( \frac{x}{2} \right) & \frac{1}{2} \cos \left( \frac{x}{2} \right) \end{bmatrix} \begin{bmatrix} c_1' \\ c_2' \end{bmatrix}= \begin{bmatrix} 0 \\ 2\sec \left( \frac{x}{2} \right) \end{bmatrix}$$

Solving the first equation $ c_1'\cos \left( \frac{x}{2} \right)+ c_2'\sin \left( \frac{x}{2} \right)=0$. This gives $c_1'=-c_2'\tan \left( \frac{x}{2} \right)$

Solving the second equation gives $ \frac{c_2'}{2} \cos \left( \frac{x}{2} \right)- \frac{c_1'}{2} \sin \left( \frac{x}{2} \right)= \frac{2}{\cos \left( \frac{x}{2} \right)}$.

Substituting gives $ \frac{c_2'}{2} \cos \left( \frac{x}{2} \right)+ \frac{c_2'}{2} \frac{\sin^2 \left( \frac{x}{2} \right)} {\cos \left( \frac{x}{2} \right)}= \frac{2}{\cos \left( \frac{x}{2} \right)}$.

Solving for $c_2'$ gives $\frac{c_2'}{2}=2$.

Finally, $c_2'=4$ and $c_2=4x$. This means that $ c_1'=-4\tan \left( \frac{x}{2} \right)$ and $ c_1=8\ln \left( \cos \left( \frac{x}{2} \right) \right)$.

However, according to Wolfram, $c_1=2\ln \left( \cos\left( \frac{x}{2} \right)\right)$ and $c_2=x$.

Indeed, I tried solving with my values for $c_1$ and $c_2$ and it doesn't work. What did I do wrong?

$\endgroup$
1
$\begingroup$

Try with the equation writen in this form. The method needs the coefficient to be 1 for the highest derivative (standard form).

$$y''+\frac{y}{4}=\frac{1}{2\cos \left( \frac{x}{2} \right)}$$

With it, we get $c_2'/2=1/2$ and $c_1'=-\tan \left( \frac{x}{2} \right)$. With them, the expected solution follows.

$\endgroup$

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.