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 \(y = e^{-2}t\ -\ e^{-3}t (2t\ +\ 1)\)

\(\frac{d^{2}y}{dt^{2}}\ +\ 7\frac{dy}{dt}\ +\ 10y=4te^{-3}t\)

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 \(y = e^{-2}t\ -\ e^{-3}t (2t\ +\ 1)\)