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17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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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

\(\displaystyle{ \mathcal{L}\{y'\} = sY(s) - y(0) }\)

\(\displaystyle{ \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) }\)

where \(\mathcal{L}\{y(t)\} = Y(s)\)

\(\displaystyle{ \mathcal{L}\{ f^{(n)}(t) \} = }\) \(\displaystyle{ s^nF(s) - s^{n-1}f(0) - }\) \(\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }\)

The idea is that we are given a differential equation which we first transform using the Laplace Transform, we solve it in the s-domain and then perform an inverse Laplace transform back to the t-domain to get our answer. Here is a very quick video giving these steps in a flow-chart.

PatrickJMT - Laplace Transform flow-chart [1min-33secs]

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.

Dr Chris Tisdell - solve differential equations using Laplace Transforms [28min-50secs]

video by Dr Chris Tisdell

This video expands on the one above taking you step-by-step 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.

Dr Chris Tisdell - big picture [12min-45secs]

video by Dr Chris Tisdell

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, solve these initial value problems using Laplace transforms. Give your answers in exact, completely factored form.
Notes
1. In the following problems, \(u(t)\) is the unit step function (Heaviside function).
2. Some of these problems may require convolution to solve.

Basic Problems

\( y' + y = u(t-1) \); \( y(0) = 0 \)

Problem Statement

\( y' + y = u(t-1) \); \( y(0) = 0 \)

Solution

646 solution video

video by Dr Chris Tisdell

close solution

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

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

648 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

Intermediate Problems

\(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) 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

647 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

649 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

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

2242 solution video

video by Krista King Math

Final Answer

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

close solution

\( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)

Problem Statement

\( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)

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

\( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)

Solution

2259 solution video

video by Michel vanBiezen

Final Answer

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

close solution

\( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)

Problem Statement

\( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)

Final Answer

\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \)

Problem Statement

\( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)

Solution

2261 solution video

video by Michel vanBiezen

Final Answer

\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \)

close solution
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