MAT244--2019F > Term Test 2

Problem 1 (main sitting)

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ZYR:
For the third solution, I believe there are some typo in Question a). For the last two equations, $y_2$ is missing, it should be $\frac{1}{2} \sin2t( 2t - tant)$, or simplify $\sin2t(t - \frac{1}{2}tant)$.

Vy Nguyen:
(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:
$$y(t)=-\frac{\ln2}{2}\cos(2t)+\frac{1}{2}\cos(2t)\ln[\cos(2t)+1]+t\sin(2t)-\frac{1}{2}\sin(2t)\tan t$$

Mingdi Xie:
This is my solutions for problem1.