# Solve dfrac{d^{2}y}{dt^{2}} - 8dfrac{dy}{dt} + 15y=9te^{3t} with y(0)=5, y'(0)=10

Question
Differential equations
Solve $$d\frac{d^{2}y}{dt^{2}}\ -\ 8d\frac{dy}{dt}\ +\ 15y=9te^{3t}\ with\ y(0)=5,\ y'(0)=10$$

2020-12-07
Given:$$\frac{d^{2}y}{dt^{2}}\ -\ 8\frac{dy}{dt}\ +\ 15y=9te^{3t}$$
with $$y(0) = 5,\ y'(0) = 10$$
The auxiliary equation is given by $$m^{2}\ -\ 8m\ +\ 15 = 0$$
solving this we get $$m = 5,\ 3$$
Hence the complimentary function is $$y_{c} = C_{1}e^{-3t}\ +\ C_{2}e^{-5t}$$
Now The P.I. of given differential equation is
$$y_{p}=\frac{1}{D^{2}\ -\ 8D\ +\ 15}9tx^{3}t=\frac{1}{(D\ -\ 3)(D\ -\ 5)}9te^{3}t$$
$$=e^{3}t\frac{1}{(D\ +\ 3\ -\ 3)(D\ +\ 3\ -\ 3)}9t$$
$$=e^{3}t\frac{1}{D(D\ -\ 2)}9t$$
$$=e^{3}t\frac{1}{-2D\left(1\ -\ \frac{D}{2}\right)}9t$$
$$=\ -\frac{e^{3t}}{2}\frac{1}{D}\left(1\ -\ \frac{D}{2}\right)^{-1}(1\ -\ D)^{-1}9t$$
$$=\ -\frac{e^{3t}}{2}\frac{1}{D}\left(1\ +\ \frac{D}{2}\ +\ \cdots\right)9t$$
$$=\ -\frac{e^{-3t}}{2}\frac{1}{D}\left(9t\ +\ \frac{9}{2}\right)$$
$$=\ -\frac{e^{-3t}}{2}\frac{1}{D}\left(\frac{9}{2}t^{2} +\ \frac{9}{2}t\right)$$
$$=\ -\frac{9}{4}e^{-3}t(t\ +\ t^{2})$$
Hence the general solution is $$y=y_{c}\ +\ y{p}=C_{1}v^{-3t}\ +\ C_{2}e^{5t}\ -\ \frac{9}{2}e^{-3}t(t\ +\ t^{2})$$
Now $$y(0)=0\ \Rightarrow\ C_{1}\ +\ e_{2}=5\cdots$$(1)
Also $$y'=3C_{1}e^{-2t}\ +\ 5C_{2}e^{5t}\ -\ \frac{9}{4}[3e^{-3t}(1\ +\ 2t)=3e^{3t}(t\ +\ t^{2})]$$
Hence $$y'(0)=10\ \Rightarrow\ 3C_{1}\ +\ 5C_{2}\ -\ \frac{9}{4}=10\ \Rightarrow\ 12C_{1}\ +\ 20C_{2}=49\ \cdots$$(2) solving (1) and (2) we get
$$C_{1} = \frac{51}{8},\ C_{2} = \frac{-11}{8}$$
Hence the solution is $$y=\frac{51}{8}e^{3t}\ -\ \frac{11}{8}e^{5t}\ -\ \frac{9}{4}e^{-3t}(t\ +\ t^{2})$$

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