A. Use Laplace transform to solve the following initial value
problems.
- \(y''-6y'+5y=0; \:\: y(0)=2, \: y'(0)=5\) solution
- \(y^{(4)}-y=0; \:\: y(0)=1,\: y'(0)=0,\: y''(0)=1,
y'''(0)=0\) solution
- \( y^{(4)}-4y'''+6y''-4y'+y=0, \quad y(0)=0, \:
y'(0)=1, \: y''(0)=0, \: y'''(0)=1\) solution
- \(y''-2y'+2y=e^{-t}; \:\: y(0)=0, \: y'(0)=1\) solution
- \(y'+5y=f(t); \quad y(0)=0, \:
f(t)=\begin{cases} 1, \: 0 \leq t <1, \\ 0, \: t \geq
1 \end{cases} \) solution
- \(y''+4y=3\sin t; \quad y(0)=1,\: y'(0)=-1 \) solution
- \(y''+2y'-15y=6\delta(t-1); \quad y(0)=0,\: y'(0)=2 \) solution
- \(y''+y=\delta(t-\pi) +1; \quad y(0)=0, \: y'(0)=0\) solution
- \( y''-2y'-3y=2\delta(t-1)-\delta(t-3);\quad y(0)=2, \: y'(0)=2\) solution
- \(y''+4y=\sin (t) -u(t-2\pi)\sin(t-2\pi), \quad y(0)=0, \: y'(0)=0 \) solution
- \(y''+4y=g(t); \quad y(0)=0, \: y'(0)=0, \:
g(t)=\begin{cases} 0, \: 0\leq t <5,\\
\frac{1}{5} (t-5), \: 5 \leq t < 10,
\\ 1, \: t \geq 10 \end{cases} \) solution
- \(y''+4y = \cos (\alpha t); \quad y(0)=0, \: y'(0)=1\) solution
B. Solve the Volterra integral equation by the Laplace
transform: \(\displaystyle{y(t)-\int_0^t (t-v)y(v)\, dv = 1.} \) solution
C. Solve the integro-differential equation by the Laplace transform:
\(\displaystyle{y'(t)-\frac{1}{2}\int_0^t (t-v)^2y(v)\, dv = -t, \quad
y(0)=1.}\) solution