Term Test 2 Question 1 Solve Integral

[Qingyang Wei]

For question 1 a) of Term Test 2, i.e. Solve equation $y'' + 4y = 2\tan (t)$, how do you solve the integral to arrive at the particular solution of the equation? I see the solution to the question being posted in the Term test section of the forum but I am not sure how the particular solution $$y_p(t)=-\cos 2t(t-\sin (t)\cos (t))+\sin 2t(\log (\cos (t)) - 1/2 \cos 2t)$$ is calculated, specifically, how the integral is solved to arrive at this answer.

[Tzu-Ching Yen]

There is this method call variation of parameter that give particular solution as linear combination of general solution $y_i$ where coefficients are determined by integrals.

[Shlok Somani]

In these type of question, I just find the wronskian and W₁ and W₂ as this equation was second order and then find their respective U's, So for U₁ integrate the division of W₁ and W and for the U₂ integrate the division of W₂ and W. Then you multiply respective U's to your Y's in this case y₁ and y₂ then add it to your general solution. For a better understanding of this process, I would recommend seeing the example 1 of section 4.4 in the textbook. I think that will clarify how the integral was solved in this question.

[Qingyang Wei]

So I used variation of parameter and concluded that the particular solution follows:
$$y_p = 2\cos (2t) \int{\tan (t) \sin (2t) dt} + 2 \sin (2t) \int{\tan(t) \cos(2t) dt}$$
I am not sure how to solve the two integrals $\int{\tan (t) \sin (2t) dt}$ and $\int{\tan(t) \cos(2t) dt}$ since it involves the $2t$ term inside the sine and cosine function. Can anyone help me solve this?

[Monika Dydynski]

So I used variation of parameter and concluded that the particular solution follows:
$$y_p = 2\cos (2t) \int{\tan (t) \sin (2t) dt} + 2 \sin (2t) \int{\tan(t) \cos(2t) dt}$$
I am not sure how to solve the two integrals $\int{\tan (t) \sin (2t) dt}$ and $\int{\tan(t) \cos(2t) dt}$ since it involves the $2t$ term inside the sine and cosine function. Can anyone help me solve this?

First, recall the double-angle trig identity, $\sin(2t)=2\sin(t)\cos(t)$. Thus, the integrand $\tan(t)\sin(2t)=\tan(t)(2\sin(t)\cos(t))=2\sin^2(t)$.


The half-angle trig identity allows us to rewrite $2\sin^2(t)$ as $1-\cos(2t)$.

Now, we have
$$\int{1}dt-\int{\cos(2t)}dt$$

Let $u=2t \Rightarrow du=2dt$
$$\int{1}dt-\int{\cos(2t)}dt=t-\frac{1}{2}\int{\cos(u)}du=t-\frac{1}{2}\sin(u)=t-\frac{1}{2}\sin(2t)=t-\sin(t)\cos(t)$$

The process is similar for $\int{\tan(t)\cos(2t)}dt$. Hope that helps!