1. Differential Equations
Second-Order Linear DEs
Variation of Parameters
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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. Variation of Parameters
To solve a non-homogeneous equation, find a solution to the corresponding homogeneous equation first, then find a particular solution.
The method of VARIATION of PARAMETERS will modify the homogeneous solution to find the particular solution. Below is the method applied to a general equation, and later will be a few specific examples.
Above is the general linear 2nd order D.E., below is the associated homogeneous equation
Find a fundamental solution set for this homogeneous equation (we know how to do this if p(t) and q(t) are constants). Call these solutions y1(t) and y2(t).
So the solution to the homogeneous equation is:
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4. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
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5. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
We need to find u1 and u2 that make this the solution to the non-homogeneous DE.
Find the derivative (using the product rule):
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6. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
We need to find u1 and u2 that make this the solution to the non-homogeneous DE.
Find the derivative (using the product rule):
Next we make the simplifying assumption that the terms involving the derivatives of u1 and u2 cancel out. This is a crucial, non-obvious step, but it simplifies the problem.
So at this point we have:
This is our simplifying assumption
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7. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
We need to find u1 and u2 that make this the solution to the non-homogeneous DE.
Find the derivative (using the product rule):
Next we make the simplifying assumption. This is a crucial, non-obvious step, but it works because the u1 and u2 functions were originally constants, so terms involving their derivatives ought to cancel out. At this point we have:
This is our simplifying assumption
Taking the derivative again, we obtain y”. Next we will substitute into the original DE.
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8. Variation of Parameters
Here is the original DE:
Our proposed solution and its derivatives are below:
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9. Variation of Parameters
Here is the original DE:
Our proposed solution and its derivatives are below:
Substituting these in yields the following condition:
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10. Variation of Parameters
Here is the original DE:
Our proposed solution and its derivatives are below:
Substituting these in yields the following condition:
Next we rearrange, then notice that most of the left side cancels out.
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11. Variation of Parameters
Here is the original DE:
Our proposed solution and its derivatives are below:
Substituting these in yields the following condition:
Next we rearrange, then notice that most of the left side cancels out.
We are left with this condition on u1 and u2
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12. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
At this point we have found 2 conditions on our unknowns u1 and u2:
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For Campus Learning Assistance Services at UCSB

13. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
At this point we have found 2 conditions on our unknowns u1 and u2:
This is a system of 2 equations in 2 unknowns – u1’ and u2’. We can set it up as a matrix and find the solution using our wicked linear algebra skills:
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For Campus Learning Assistance Services at UCSB

14. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
At this point we have found 2 conditions on our unknowns u1 and u2:
This is a system of 2 equations in 2 unknowns – u1’ and u2’. We can set it up as a matrix and find the solution using our wicked linear algebra skills:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

15. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
At this point we have found 2 conditions on our unknowns u1 and u2:
This is a system of 2 equations in 2 unknowns – u1’ and u2’. We can set it up as a matrix and find the solution using our wicked linear algebra skills:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

16. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
At this point we have found 2 conditions on our unknowns u1 and u2:
This is a system of 2 equations in 2 unknowns – u1’ and u2’. We can set it up as a matrix and find the solution using our wicked linear algebra skills:
All we need to do now is integrate to find u1 and u2.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

17. Variation of Parameters
To find the particular solution we will replace the constants c1 and c2 with unknown functions u1(t) and u2(t) – we are “varying the parameters” by letting them be functions.
We have the following formulas for u1 and u2:
We can solve these integrals, then put u1 and u2 back into our expression for yp to get the particular solution.
GOOD NEWS: We can skip the derivation and jump straight to the formulas for u1 and u2 if we so desire.
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18. Example 1: Find the general solution to the following D.E.
First find the homogeneous solution:
Now that we have yh, we can use it to find yp by variation of parameters.
The unknown functions u1 and u2 are found from the following formulas:
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19. Example 1: Find the general solution to the following D.E.
We will need the Wronskian of y1 and y2:
Now we can integrate to find u1 and u2. Both integrals are done by parts.
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20. Example 1: Find the general solution to the following D.E.
Putting these functions into our expression for yp, we get:
Finally we can write down the general solution by adding yh and yp.
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21. Example 2: Find the general solution to the following D.E.
First find the homogeneous solution:
Find the particular solution by assuming it has the form:
The unknown functions u1 and u2 are found by integrating:
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22. Example 2: Find the general solution to the following D.E.
We need to find the Wronskian of y1 and y2:
Now we can find u1 and u2 :
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23. Example 2: Find the general solution to the following D.E.
Now we can write down the particular solution:
When combined with the yh, we get
Note that part of the particular solution simply repeated terms in the homogeneous solution. These extraneous terms were discarded (or, if you like, they were combined into the c1et and c2e2t terms.
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24. Don’t forget to put this one in standard form by dividing by t first.
Example 3: Find the general solution to the following D.E., given the solutions, y1 and y2, of the corresponding homogeneous equation.
Don’t forget to put this one in standard form by dividing by t first.
This is g(t)
Find the particular solution by assuming it has the form:
The unknown functions u1 and u2 are found by integrating:
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25. We need to find the Wronskian of y1 and y2:
Example 3: Find the general solution to the following D.E., given the solutions, y1 and y2, of the corresponding homogeneous equation.
We need to find the Wronskian of y1 and y2:
Now we can find u1 and u2 :
Integrate by parts
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26. Now we can write down the particular solution:
Example 3: Find the general solution to the following D.E., given the solutions, y1 and y2, of the corresponding homogeneous equation.
Now we can write down the particular solution:
The general solution will be the sum of yh and yp.
This piece can be combined with the c2 term so we can ignore it.
The last part of the particular solution repeats one of the homogeneous solutions, so we can ignore it.
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27. Example 4: Find the general solution to the following D.E.
This one does not have constant coefficients, but it should look familiar. We solved something like this before, and you should recognize it as a Cauchy-Euler equation.
The homogeneous solution will be some power of t. Substitute into the equation to find yh.
This gives us one solution to the homogeneous equation: y1=t2.
We need to use “reduction of order” to find the second solution.
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28. Example 4: Find the general solution to the following D.E.
“Reduction of Order” modifies our existing solution by multiplying by an unknown function of t. This should yield a 2nd independent solution to our DE.
Find derivatives and plug in to find a condition for v(t).
This simplifies to:
We need to solve this equation for v(t).
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29. Example 4: Find the general solution to the following D.E.
One way to solve this is to recognize that we have a nice product rule here:
Integrating yields
Integrating again gives us
So our 2nd homogeneous solution is
To recap, we have found the homogeneous solutions to be:
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30. Example 4: Find the general solution to the following D.E.
We need the Wronskian of y1 and y2
Before we can use variation of parameters we need to divide by t2 to put our DE in standard form.
This is g(t)
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31. Example 4: Find the general solution to the following D.E.
Next we can find the particular solution by variation of parameters
The general solution is thus:
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