Answer to Question #125757 in Differential Equations for jse

Question

Find the solution of the initial value problem y′′+4y=t^2+6e^t, y(0)=0, y′(0)=5.
Enter an exact answer.
Enclose arguments of functions in parentheses. For example, sin(2x).

Answer 1

The given equation can be written as, $\left(D^{2}+4\right) y=t^{2}+6e^{t}$. The auxiliary equation is, $m^{2}+4=0$. Solving, we get $m=\pm 2i$. The complementary function is,

$C.F=c_{1} \cos (2 t)+c_{2} \sin (2 t).$ The particular integral is given by,

$\begin{aligned} P.I &=\frac{1}{D^{2}+4}\left(t^{2}+6 e^{t}\right) \\ &=\frac{1}{D^{2}+4} \cdot t^{2}+\frac{1}{D^{2}+4} \cdot 6 e^{t} \\ &=\frac{1}{4}\left(1+\frac{D^{2}}{4}\right)^{-1} \cdot t^{2}+\frac{6 e^{t}}{5} \\ &=\frac{1}{4}\left(1-\frac{D^{2}}{4}+\cdots\right) \cdot t^{2}+\frac{6 e^{t}}{5} \\ &=\frac{1}{4}\left(t^{2}-\frac{1}{2}\right)+\frac{6 e^{t}}{5} \\ &=\frac{t^{2}}{4}+\frac{6 e^{t}}{5}-\frac{1}{8} \end{aligned}$

The general solution is, $\displaystyle y=c_{1} \cos (2 t)+c_{2} \sin (2 t)+\frac{t^{2}}{4}+\frac{6 e^{t}}{5}-\frac{1}{8}$.

Hence, $\displaystyle y'=-2 c_{1} \sin (2 t)+2 c_{2} \cos (2 t)+\frac{t}{2}+\frac{6 e^{t}}{5}$ .

Given the initial conditions, $y(0)=0, y^{\prime}(0)=5$. Using the initial conditions we get,

$\begin{aligned} \displaystyle y(0) & = c_{1} +\frac{6 }{5}-\frac{1}{8}\\ 0&=c_{1} +\frac{6 }{5}-\frac{1}{8}\\ c_{1}&=-\frac{43}{40} \end{aligned}$

$\begin{aligned} \displaystyle y'(0) & = 2c_{2} +\frac{6 }{5}\\ 5&=2c_{2} +\frac{6}{5}\\ 2c_{2}&= 5-\frac{6}{5}\\ c_{2}&=\frac{19}{10} \end{aligned}$

Hence, the general solution is,

$\begin{aligned} y&=\frac{19}{10} \sin (2 t)-\frac{43}{40} \cos (2 t)+\frac{t^{2}}{4}+\frac{6 e^{t}}{5}-\frac{1}{8}\\ y&=\frac{1}{40}\bigg(76 \sin (2 t)-43 \cos (2 t)+10 t^{2}+48 e^{t}-5\bigg) \end{aligned}$