Solve left(d^{2}frac{y}{dt^{2}}right) + 7left(frac{dy}{dt}right) + 10y=4te^{-3}t with y(0)=0, y'(0)= -1

Question
Differential equations
Solve
$$\left(d^{2}\frac{y}{dt^{2}}\right)\ +\ 7\left(\frac{dy}{dt}\right)\ +\ 10y=4te^{-3}t$$ with
$$y(0)=0,\ y'(0)=\ -1$$

2021-02-16
Given:
$$\frac{d^{2}y}{dt^{2}}\ +\ 7\frac{dy}{dt}\ +\ 10y=4te^{-3}t$$
with $$y(0) = 0,\ y'(0) =\ -1$$
The auxiliary equation is given by $$m^{2}\ +\ 7m\ +\ 10 = 0$$
solving this we get $$m =\ -5,\ -2$$
Hence the complimentary function is $$y_{c} = C_{1}e^{2t}\ +\ C_{2}e^{-5}t$$
Now The P.I. of given differential equation is
$$y_{p}=\frac{1}{D^{2}\ +\ 7D\ +\ 10}4tx^{-3}t=\frac{1}{(d\ +\ 5)(D\ +\ 2)}4te^{-3}t$$
$$=e^{3}t\frac{1}{(D\ -\ 3\ +\ 5)(D\ -\ 3\ +\ 2)}4t$$
$$=e^{3}t\frac{1}{(D\ +\ 2)(D\ -\ 1)}4t$$
$$=e^{3}t\frac{1}{-2(1\ +\ \frac{D}{2})(1\ -\ D)}4t$$
$$=\ -\frac{e^{-3}t}{2}\left(1\ +\ \frac{D}{2}\right)^{1}(1\ -\ D)^{-1}4t$$
$$=\ -\frac{e^{-3}t}{2}\left(1\ -\ \frac{D}{6}\ +\ \cdots\right)(1\ +\ D\ +\ D^{2})4t$$
$$=\ -\frac{e^{-3}t}{2}\left(1\ +\ \left(1\ -\ \frac{1}{2}\right)D\ +\ \cdots\right)4t$$
$$=\ -\frac{e^{-3}t}{2}(4t\ +\ 2)$$
$$=\ - e^{-3}t (2t\ +\ 1)$$
Hence the general solution is $$y = y_{c}\ +\ y_{p} = C_{1}v^{-2}t\ +\ C_}2}e^{5}t\ -\ e^{-3}t (2t\ +\ 1) Now \(y(0) = 0\ \Rightarrow\ C_{1}\ +\ e_{2}\ -\ 1 = 0\ \Rightarrow\ C_{1} = C_{2} = 1\ \cdots$$(1)
Also $$y' =\ -2C_{1}e^{-2}t\ -\ 5C_{2}e^{-5}t\ +\ 3e^{-3}t (2t\ +\ 1)\ -\ 2e^{-3}t$$
Hence $$y' (0) =\ -1\Rightarrow\ -2C_{1}\ -\ 5C_{2}\ -\ 2\ +\ 3 =\ -1\ \Rightarrow\ 2C_{1}\ +\ 5C_{2} = 2\ \cdots$$ (2)
solving (1) and (2) we get $$C_{1} = 1,\ C_{2} = 0$$
Hence the solution is $$y = e^{-2}t\ -\ e^{-3}t (2t\ +\ 1)$$

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