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Term Test 2 / Re: Problem 1 (main sitting)

« on: November 19, 2019, 07:38:12 PM »

(Almost perfect typing appreciated). V.I.

a)
Solve the homogeneous equation: $y''+4y = 0$
$$r^2+4=0$$
$$r=\pm2i$$
fundamental set of solutions:
$$y_1 (t)=\cos(2t)$$
$$y_2 (t)=\sin(2t)$$
homogeneous solution:
$$y_h (t)=c_1 \cos(2t) + c_2 \sin(2t)$$
Solve for a particular solution using variation of parameters:
Find the Wronskian:
$$W=\begin{vmatrix}
    \cos(2t) & \sin(2t) \\
    -2\sin(2t) & 2\cos(2t)\\
    \end{vmatrix}$$
$$=2\cos^2 (2t) + 2\sin^2 (2t) = 2$$
Solve for $\mu_1 (t)$:
$$\mu_1 (t)=-\int \frac {y_2 (t) g(t)}{W} dt$$
$$=\int \frac {-\sin(2t)}{2\cos^2 t} dt$$

use identity $2\cos^2 t = \cos(2t) +1$

$$=\int \frac {-\sin(2t)}{\cos(2t)+1} dt$$

let $u=\cos(2t)+1$ then $\frac{1}{2}du=-\sin(2t) dt$

$$=\frac{1}{2} \int \frac {1}{u} du$$
$$=\frac{1}{2} \ln(u)$$
$$=\frac{1}{2} \ln(\cos(2t)+1)$$
Solve for $\mu_2 (t)$:
$$\mu_2 (t)=\int \frac {y_1 (t) g(t)}{W} dt$$
$$=\int \frac {\cos(2t)}{2\cos^2 t} dt$$

use identity $\cos(2t)=2\cos^2 (t)-1$

$$=\int \frac {2\cos^2 t-1}{2\cos^2 t} dt$$
$$=\int 1 - \frac {1}{2\cos^2 t} dt$$
$$=\int 1 - \frac {1}{2}\sec^2 t \, dt$$
$$=t - \frac{1}{2}\tan t$$
particular solution:
$$y_p(t)=\frac{1}{2}\cos(2t)\ln [\cos(2t)+1] + t\sin(2t) - \frac{1}{2}\sin(2t)\tan t$$
Therefore, general solution to ODE is:
$$y(t) = y_h(t) + y_p(t)$$
$$y(t) = c_1\cos(2t) + c_2\sin(2t) + \frac{1}{2}\cos(2t)\ln [\cos(2t)+1] + t\sin(2t) - \frac{1}{2}\sin(2t)\tan t$$

b)
Take the derivative of the general solution:
$$y'(t)=-2c_1\sin(2t)+2c_2\cos(2t)-\sin(2t)\ln[\cos(2t)+1] - \frac{(\cos(2t))(\sin(2t))}{\cos(2t)+1} + \sin(2t) + 2t\cos(2t) - \cos(2t)\tan t - \frac{1}{2}\sin(2t)\sec^2t$$
Use the initial conditions to solve for constants $c_1$ and $c_2$:
$$y(0)=0=c_1+\frac{\ln2}{2}$$
$$c_1=-\frac{\ln2}{2}$$
$$y'(0)=0=2c_2$$
$$c_2=0$$
Therefore the solution that satisfies the initial conditions is:
$$y(t)=-\frac{\ln2}{2}\cos(2t)+\frac{1}{2}\cos(2t)\ln[\cos(2t)+1]+t\sin(2t)-\frac{1}{2}\sin(2t)\tan t$$