Answer to Question #125356 in Differential Equations for jse

Question

1. 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).

y(t) = _____

2. Use the method of variation of parameters to find a particular solution of the differential equation

y′′−2y′−15y=384e^−t.

Y(t) = ______

Answer 1

1.The given equation is, $(D^{2}+4)y = t^{2}+6e^{t}$ . The auxiliary equation is, $m^{2} + 4 = 0.$

$\begin{array}{l} m^{2}=-4\\ \Rightarrow m = \pm 2i \end{array}$

The complementary function is,

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

The particular integral is,

$\begin{aligned} y_{p}&=\dfrac{1}{D^{2}+4}\left(t^{2}+6e^{t}\right)\\ &=\dfrac{1}{D^{2}+4} \cdot t^{2}+ \dfrac{1}{D^{2}+4} \cdot 6e^{t}\\ &=\dfrac{1}{4}\left(1+\frac{D^{2}}{4}\right)^{-1}\cdot t^{2} + \frac{6e^{t}}{5}\\ &\qquad \qquad \qquad\qquad (\text{Using}~ D=1~ \text{in $2^{nd}$ term} )\\ &=\dfrac{1}{4}\left(1-\frac{D^{2}}{4}+\cdots\right)\cdot t^{2} + \frac{6e^{t}}{5}\\ &=\dfrac{1}{4}\left(t^{2}-\frac{1}{2}\right) + \frac{6e^{t}}{5}\\ &=\dfrac{t^{2}}{4} + \frac{6e^{t}}{5} -\frac{1}{8} \end{aligned}$

The general solution is,

$y = c_{1}\cos(2t)+c_{2}\sin(2t)+\dfrac{t^{2}}{4}+\dfrac{6e^{t}}{5}-\dfrac{1}{8}$

Then,

$y'=-2c_{1}\sin(2t)+2c_{2}\cos(2t)+\dfrac{t}{2}+\dfrac{6e^{t}}{5}$

Using the given initial conditions, $y(0)=0, y'(0)=5$ we get

$c_{1} = -\dfrac{43}{40}, c_{2} = \dfrac{19}{10}$ .

Hence, the general solution is

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

2.The given equation is, $(D^{2}-2D-15)y = 384e^{-t}$ . The auxiliary equation is,

$m^{2}-2m-15=0$. Solving we get, $m = 5,-3$. The complementary function is,

$y_{c} = c_{1}e^{5t}+c_{2}e^{-3t} = c_{1}f_{1}+c_{2}f_{2}$.

The particular solution is,

$y_{p} = Pf_{1}+Qf_{2},~\text{where} \\~\\ P = -\displaystyle\int \dfrac{f_{2}g(t)}{W(f_{1},f_{2})}dt; \quad Q = \displaystyle\int \dfrac{f_{1}g(t)}{W(f_{1},f_{2})}dt\\~\\ \text{and}~ g(t) = 384e^{-t}, W(f_{1},f_{2}) = -8e^{2t}.$

$\begin{aligned} P &= -\displaystyle\int \dfrac{f_{2}g(t)}{W(f_{1},f_{2})}dt\\ &= -\displaystyle\int \dfrac{e^{-3t}\cdot 384e^{-t}}{-8e^{2t}}dt\\ &= 48\displaystyle\int e^{-6t}dt = -8e^{-6t}\\ \end{aligned}$

$\begin{aligned} Q &= \displaystyle\int \dfrac{f_{1}g(t)}{W(f_{1},f_{2})}dt\\ &= \displaystyle\int \dfrac{e^{5t}\cdot 384e^{-t}}{-8e^{2t}}dt\\ &= -48\displaystyle\int e^{2t}dt = -24e^{2t}\\ \end{aligned}$

Thus,

$y_{p} = -8e^{-6t} \cdot e^{5t} -24e^{2t} \cdot e^{-3t} = -32e^{-t}$

The general solution is,

$y= c_{1}e^{5t}+c_{2}e^{-3t}-32e^{-t}$