Question

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

Answer 1

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 \(y=\frac{51}{8}e^{3t}\ -\ \frac{11}{8}e^{5t}\ -\ \frac{9}{4}e^{-3t}(t\ +\ t^{2})\)