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Term Test 2 Question 1 Solve Integral

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**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**:

--- Quote from: Qingyang Wei on December 05, 2018, 08:49:38 PM ---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?

--- End quote ---

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!

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