Q1E

Question

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$y'' + 4 y = \tan 2 t$

Step-by-Step Solution

Verified

Answer

The general solution is $y (t) = c_{1} \cos 2 t + c_{2} \sin 2 t - \frac{\cos 2 t}{4} ([\ln (\tan 2 t + \sec 2 t)])$ .

1Step 1: Find a particular solution .

The homogenous equation is $r^{2} + 4 = 0$ .

Two independent solutions are $r = \pm 2 i$ .

Then $y_{1} = \cos 2 t, y_{2} = \sin 2 t$

$y_{h} (t) = c_{1} \cos 2 t + c_{2} \sin 2 t$

The particular solution is $y_{p} = v_{1} (t) \cos 2 t + v_{2} (t) \sin 2 t$

2Step 2: Evaluate v 1    and    v 2

Here $y_{p} = v_{1} (t) \cos 2 t + v_{2} (t) \sin 2 t$

And referring to (9) and solve the system then

$\begin{matrix} v_{1}' \cos 2 t + v_{2}' \sin 2 t = 0 \\ - 2 v_{1}' \sin 2 t + 2 v_{2}' \cos 2 t = \frac{f}{a} \\ - 2 v_{1}' \sin 2 t + 2 v_{2}' \cos 2 t = \tan 2 t \end{matrix}$

3Step 3: Find v 1 ' and v 1

$\begin{matrix} v_{1}' = \frac{- f (t) y_{2} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} \\ = \frac{- \tan 2 t . \sin 2 t}{1 [2 \cos^{2} 2 t + 2 \sin^{2} 2 t]} \\ = \frac{- \tan 2 t . \sin 2 t}{2 [\cos^{2} 2 t + \sin^{2} 2 t]} \\ = - \frac{\sin^{2} 2 t}{2 \cos 2 t} \end{matrix}$

Now integrating this.

$\begin{matrix} v_{1} (t) = \int - \frac{\sin^{2} 2 t}{2 \cos 2 t} dt \\ = - \frac{1}{4} \int \frac{\sin^{2} x}{\cos x} dx \\ = \frac{1}{4} \int \frac{\cos^{2} x - 1}{\cos x} dx \\ = \frac{1}{4} [\int cosxdx - \int secx dx] \\ = \frac{1}{4} [sinx - \ln (tanx + secx)] + C \\ = \frac{1}{4} [\sin 2 t - \ln (\tan 2 t + \sec 2 t)] + C \end{matrix}$

4Step 4. Determine v 2 ' and v 2

$\begin{matrix} v_{2}' = \frac{f (t) y_{1} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} \\ = \frac{\tan 2 t . \cos 2 t}{1 [2 \cos^{2} 2 t + 2 \sin^{2} 2 t]} \\ = \frac{- \tan 2 t . \sin 2 t}{2 [\cos^{2} 2 t + \sin^{2} 2 t]} \\ = \frac{\sin 2 t}{2} \end{matrix}$

Now integrate this.

$\begin{matrix} v_{2} (t) = \int \frac{\sin 2 t}{2} dt \\ = - \frac{1}{4} \cos 2 t + C \end{matrix}$

Thus, a particular solution is:

$\begin{matrix} y_{p} = (\frac{1}{4} [\sin 2 t - \ln (\tan 2 t + \sec 2 t)] + C) \cos 2 t + (- \frac{1}{4} \cos 2 t + C) \sin 2 t \\ = - \frac{\cos 2 t}{4} ([\ln (\tan 2 t + \sec 2 t)]) \end{matrix}$

Thus, the general solution is:

$\begin{array}{l} y (t) = y_{h} (t) + y_{p} (t) \\ y (t) = c_{1} \cos 2 t + c_{2} \sin 2 t - \frac{\cos 2 t}{4} ([\ln (\tan 2 t + \sec 2 t)]) \end{array}$

Q48E

Q2E

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Q3E

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