1. Differential Equations
Second-Order Linear DEs
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2. The general solution will always have the form:
A Second-Order Linear Differential Equation can always be put into the form:
The general solution will always have the form:
yh is the solution to the corresponding homogeneous equation, where g(t)=0
yp is a particular solution to the original DE.
There should be two independent solutions to the homogeneous equation, and yh will be a linear combination of these two.
In fact, the two independent solutions form a basis for a 2-dimensional solution space (an actual vector space like we saw last quarter).
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3. Here are a few examples of linear, second order DEs with constant coefficients.
These equations are all homogeneous. We will solve some inhomogeneous equations later. Also, we will add some initial conditions to these problems.
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5. For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION.
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6. For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION.
Solve for the roots of this equation, then each root will correspond to a solution of the DE.
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7. For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION.
Solve for the roots of this equation, then each root will correspond to a solution of the DE.
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8. The general solution is a linear combination of these:
For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION.
Solve for the roots of this equation, then each root will correspond to a solution of the DE.
The general solution is a linear combination of these:
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9. Now add some initial conditions.
We will need to determine the constants in our general solution that match up with the given conditions.
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10. Now add some initial conditions.
We will need to determine the constants in our general solution that match up with the given conditions.
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11. Now add some initial conditions.
We will need to determine the constants in our general solution that match up with the given conditions.
Here is our final solution.
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13. Start with the characteristic equation:
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14. Start with the characteristic equation:
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15. Start with the characteristic equation:
This time there is only one repeated root, so it seems like there is only one solution:
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16. Start with the characteristic equation:
This time there is only one repeated root, so it seems like there is only one solution:
This is one of the solutions, but we need to find another independent solution. The trick is to multiply this solution by t:
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17. Start with the characteristic equation:
This time there is only one repeated root, so it seems like there is only one solution:
This is one of the solutions, but we need to find another independent solution. The trick is to multiply this solution by t:
Now the general solution is a linear combination of these:
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18. Now add some initial conditions.
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19. Now add some initial conditions.
Calculate the derivative, then plug in to find the constants.
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20. Now add some initial conditions.
Calculate the derivative, then plug in to find the constants.
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21. Now add some initial conditions.
Calculate the derivative, then plug in to find the constants.
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23. This time the roots are complex numbers.
The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula.
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24. This time the roots are complex numbers.
The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula.
This is Euler’s formula.
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25. This time the roots are complex numbers.
The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula.
This is Euler’s formula.
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26. This time the roots are complex numbers.
The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula.
This is Euler’s formula.
cos(-t)=cos(t) and sin(-t)=-sin(t)
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27. This time the roots are complex numbers.
The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula.
This is Euler’s formula.
cos(-t)=cos(t) and sin(-t)=-sin(t)
Any linear combination of these solutions will also be a solution to the DE, specifically:
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28. Plug in the given values to find the solution that matches up.
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29. Plug in the given values to find the solution that matches up.
The solution is simply an oscillation because the roots of the characteristic equation were completely imaginary.
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33. These roots have both a real and imaginary part.
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34. These roots have both a real and imaginary part.
The corresponding general solution is:
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35. Now add some initial values
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36. Now add some initial values
The general solution is:
Compute the first derivative, then plug in the given values:
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37. Now add some initial values
The solution is
Using some trigonometry this can be re-written as a single cosine function.
Here is the general formula:
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38. Now add some initial values
The solution is
Using some trigonometry this can be re-written as a single cosine function.
The plot of the solution shows the function oscillating between the exponential “envelope” and eventually approaching 0.
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