Question

1. Find the Wronskian of the functions f(t)= (3e^t)sin t and g(t)= (e^t) cos t. Simplify your answer.

W(f,g)(t)= ________

2. Solve the initial value problem y′′+4y′+4y=0, y(−1)=4, y′(−1)=4.

Answer 1

1).$W(f,g)(t)=\begin{vmatrix} 3e^t\sin(t) & e^t\cos(t) \\ (3e^t\sin(t))' & (e^t\cos(t))' \end{vmatrix}=\begin{vmatrix} 3e^t\sin(t) & e^t\cos(t) \\ 3\sqrt{2}e^t\sin(t+\pi/4) & \sqrt{2}e^t\cos(t+\pi/4) \end{vmatrix}$

Thus,

$$W(f,g)(t)=-3e^{2t}$$

2).Given

$$y′′+4y′+4y=0, y(−1)=4, y′(−1)=4.$$

Now, Auxiliary equation will be

$$p^2+4p+4=0\implies (p+2)^2=0\implies p=-2$$

Thus, general solution will be

$$y=(a+bx)e^{-2x}$$

Now,

$$y'=-2(a+bx)e^{-2x}+be^{-2x}$$

From, initial conditions we get,

$$y(-1)=(a-b)e^2=4\\y'(-1)=-2(a-b)e^2+be^2=4\\ \implies b=12e^{-2},a=16e^{-2}$$

Hence, the solution wil be

$$y=(16+12x)e^{-2(x+1)}$$