If you haven't seen Laplace Transforms before, the Laplace Transform section of 17calculus has a complete discussion of what they are and how to work with them.
Using Laplace Transforms to solve differential equation initial value problems is a great way to streamline solutions and, for forcing functions that are discontinuous, they are about the only way to solve them. The equations we use are
Laplace Transforms of Derivatives 

\(\displaystyle{ \mathcal{L}\{y'(t)\} = sY(s)  y(0) }\) 
\(\displaystyle{ \mathcal{L}\{y''(t)\} = s^2Y(s)  sy(0)  y'(0) }\) 
where \(\mathcal{L}\{y(t)\} = Y(s)\) 
\(\displaystyle{ \mathcal{L}\{ f^{(n)}(t) \} = }\) \(\displaystyle{ s^nF(s)  s^{n1}f(0)  }\) \(\displaystyle{ s^{n2}f'(0)  . . .  f^{(n1)}(0) }\) 
The idea is that we are given a differential equation which we first transform using the Laplace Transform, we solve it in the sdomain and then perform an inverse Laplace transform back to the tdomain to get our answer. Here is a very quick video giving these steps in a flowchart.
video by PatrickJMT 

This video clip explains in detail how to solve differential equations using Laplace Transforms including a quick calculation of the Laplace transform of the first derivative function above.
video by Dr Chris Tisdell 

This video expands on the one above taking you stepbystep through this process and, at the end of the video, gives you a big picture of how to do this. This is a great video.
video by Dr Chris Tisdell 

Practice
Unless otherwise instructed, solve these initial value problems using Laplace transforms. Give your answers in exact, completely factored form.
Notes
1. In these problems, \(u(t)\) is the unit step function (Heaviside function).
2. Some of these problems may require convolution to solve.
Basic
\( y' + y = u(t1) \); \( y(0) = 0 \)
Problem Statement 

\( y' + y = u(t1) \); \( y(0) = 0 \)
Solution 

video by Dr Chris Tisdell 

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\( y^{(4)}y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)
Problem Statement 

\( y^{(4)}y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)
Final Answer 

\(\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }\)
Problem Statement 

\( y^{(4)}y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\(\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }\) 
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Intermediate
\(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)
Problem Statement 

\(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)
Final Answer 

\(\displaystyle{ y(t) = \frac{1}{4} \left[ t \frac{1}{2} \sin(2t) \right]  \frac{u(t\pi/2)}{4}\left[(t\pi/2)  \frac{1}{2} \sin[2(t\pi/2)] \right] }\)
Problem Statement 

\(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\(\displaystyle{ y(t) = \frac{1}{4} \left[ t \frac{1}{2} \sin(2t) \right]  \frac{u(t\pi/2)}{4}\left[(t\pi/2)  \frac{1}{2} \sin[2(t\pi/2)] \right] }\) 
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\( y''  3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t1)  u(t2) \)
Problem Statement 

\( y''  3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t1)  u(t2) \)
Final Answer 

\(\displaystyle{ y(t) = e^t + 2e^{2t} + u(t1)\left[ \frac{1}{2}  e^{t1} + \frac{1}{2}e^{2(t1)} \right]  u(t2)\left[ \frac{1}{2}  e^{t2} + \frac{1}{2}e^{2(t2)} \right] }\)
Problem Statement 

\( y''  3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t1)  u(t2) \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\(\displaystyle{ y(t) = e^t + 2e^{2t} + u(t1)\left[ \frac{1}{2}  e^{t1} + \frac{1}{2}e^{2(t1)} \right]  u(t2)\left[ \frac{1}{2}  e^{t2} + \frac{1}{2}e^{2(t2)} \right] }\) 
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\( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=7 \)
Problem Statement 

\( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=7 \)
Final Answer 

\( y(t) = 3\cos(t/2)  14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t\tau)~d\tau} \)
Problem Statement 

\( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=7 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(t) = 3\cos(t/2)  14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t\tau)~d\tau} \) 
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\( y'' + \omega_0^2 y = F_0 \sin(\omega t); yâ€™(0)=0, y(0)=0 \)
Problem Statement 

\( y'' + \omega_0^2 y = F_0 \sin(\omega t); yâ€™(0)=0, y(0)=0 \)
Final Answer 

\(\displaystyle{ y(t) = \frac{F_0}{\omega_0^2\omega^2} \left[ \sin(\omega t)  \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }\)
Problem Statement 

\( y'' + \omega_0^2 y = F_0 \sin(\omega t); yâ€™(0)=0, y(0)=0 \)
Solution 

video by Michel vanBiezen 

Final Answer 

\(\displaystyle{ y(t) = \frac{F_0}{\omega_0^2\omega^2} \left[ \sin(\omega t)  \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }\) 
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\( yâ€™ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{3t}; y(0)=0 \)
Problem Statement 

\( yâ€™ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{3t}; y(0)=0 \)
Final Answer 

\( y(t) = e^{3t}(3+4e^te^{2t}) \)
Problem Statement 

\( yâ€™ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{3t}; y(0)=0 \)
Solution 

video by Michel vanBiezen 

Final Answer 

\( y(t) = e^{3t}(3+4e^te^{2t}) \) 
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Practice Instructions
Unless otherwise instructed, solve these initial value problems using Laplace transforms. Give your answers in exact, completely factored form.
Notes
1. In these problems, \(u(t)\) is the unit step function (Heaviside function).
2. Some of these problems may require convolution to solve.