1. Systems of Differential Equations Nonhomogeneous Systems
Math 4B
Systems of Differential Equations
Nonhomogeneous Systems
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2. 𝑥 ′ 𝑡 =𝐴 𝑥 + 𝑔 (𝑡) 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
A system of 1st order linear differential equations will have the form
𝑥 ′ 𝑡 =𝐴 𝑥 + 𝑔 (𝑡)
We have already seen to how to solve the homogeneous case g(t)=0.
Now we will find a particular solution to match the nonhomogeneous part.
The full solution will be the sum of the homogeneous and particular parts.
Since the system is 1st-order, we can use either variation of parameters or an integrating factor to find the particular solution. In some cases undetermined coefficients might also be appropriate, but it is usually cumbersome.
After some work, we will find the solution to be:
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
Here X(t) is the fundamental matrix obtained from the solution of the homogeneous case.
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3. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
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4. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Eigenvalues
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5. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Eigenvalues
Eigenvectors
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6. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Fundamental Matrix
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7. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Fundamental Matrix
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8. 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 𝑥 ′ 𝑡 = −1 0 2 2 𝑥 𝑡 + 𝑡 2 𝑡+1 Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
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9. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 3 𝑒 −𝑡 0 −2 𝑒 −𝑡 𝑒 2𝑡 𝑒 𝑡 0 2 𝑒 −2𝑡 3𝑒 −2𝑡 ∙ 𝑡 2 𝑡+1 𝑑𝑡
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10. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 3 𝑒 −𝑡 0 −2 𝑒 −𝑡 𝑒 2𝑡 𝑒 𝑡 0 2 𝑒 −2𝑡 3𝑒 −2𝑡 ∙ 𝑡 2 𝑡 𝑡 2 𝑒 𝑡 (2 𝑡 2 +3𝑡+3) 𝑒 −2𝑡 𝑑𝑡
Need to integrate these two functions
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11. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 3 𝑒 −𝑡 0 −2 𝑒 −𝑡 𝑒 2𝑡 ∙ ( 𝑡 2 −2𝑡+2) 𝑒 𝑡 (− 𝑡 2 − 𝑡− ) 𝑒 −2𝑡
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12. Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 3 𝑒 −𝑡 0 −2 𝑒 −𝑡 𝑒 2𝑡 ∙ ( 𝑡 2 −2𝑡+2) 𝑒 𝑡 (− 𝑡 2 − 𝑡− ) 𝑒 −2𝑡
𝑥 𝑝 𝑡 = 𝑡 2 −2𝑡+2 − 𝑡 𝑡− 9 4
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13. 𝑥 𝑡 = 3 𝑒 −𝑡 0 −2 𝑒 −𝑡 𝑒 2𝑡 ∙ 𝑐 1 𝑐 2 + 𝑡 2 −2𝑡+2 − 𝑡 2 + 1 2 𝑡− 9 4
Example 1:
Find the general solution to the given system.
𝑥 ′ 𝑡 = − 𝑥 𝑡 + 𝑡 2 𝑡+1
General Solution
Add the homogeneous and particular solutions
𝑥 𝑡 = 3 𝑒 −𝑡 0 −2 𝑒 −𝑡 𝑒 2𝑡 ∙ 𝑐 1 𝑐 𝑡 2 −2𝑡+2 − 𝑡 𝑡− 9 4
Solution can also be multiplied out and written like this:
𝑥 𝑡 = 𝑐 1 𝑒 −𝑡 3 −2 + 𝑐 2 𝑒 2𝑡 𝑡 −1 +𝑡 − − 9 4
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14. 𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡 Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
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15. −𝑟 1 4 −𝑟 =0→ 𝑟 2 −4=0→𝑟=±2 𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Eigenvalues
−𝑟 1 4 −𝑟 =0→ 𝑟 2 −4=0→𝑟=±2
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16. −𝑟 1 4 −𝑟 =0→ 𝑟 2 −4=0→𝑟=±2 𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Eigenvalues
−𝑟 1 4 −𝑟 =0→ 𝑟 2 −4=0→𝑟=±2
Eigenvectors
𝑟 1 =2→ −2 1 4 −2 →−2 𝑣 1 + 𝑣 2 =0→ 𝑣 1 = 1 2
𝑟 2 =−2→ →2 𝑣 1 + 𝑣 2 =0→ 𝑣 2 = 1 −2
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17. 𝑋 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡 𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Fundamental Matrix
𝑋 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡
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18. Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Fundamental Matrix
𝑋 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡
𝑋 −1 (𝑡)= 1 −4 −2𝑒 −2𝑡 −𝑒 −2𝑡 −2 𝑒 2𝑡 𝑒 2𝑡 = 𝑒 −2𝑡 𝑒 −2𝑡 𝑒 2𝑡 − 𝑒 2𝑡
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19. 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
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20. Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡 𝑒 −2𝑡 𝑒 −2𝑡 𝑒 2𝑡 − 𝑒 2𝑡 ∙ sin⁡(3𝑡) 𝑡 𝑑𝑡
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21. Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡 𝑒 −2𝑡 𝑒 −2𝑡 𝑒 2𝑡 − 𝑒 2𝑡 ∙ sin⁡(3𝑡) 𝑡 𝑒 −2𝑡 sin 3𝑡 𝑡𝑒 −2𝑡 𝑒 2𝑡 sin 3𝑡 − 𝑡𝑒 2𝑡 𝑑𝑡
Need to integrate these two functions
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22. 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡 ∙ 𝑒 −2𝑡 (− cos 3𝑡 − sin 3𝑡 − 𝑡− ) 𝑒 2𝑡 (− cos 3𝑡 sin 3𝑡 𝑡− )
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23. Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡 ∙ 𝑒 −2𝑡 (− cos 3𝑡 − sin 3𝑡 − 𝑡− ) 𝑒 2𝑡 (− cos 3𝑡 sin 3𝑡 𝑡− )
𝑥 𝑝 𝑡 = − cos 3𝑡 − − sin 3𝑡 − 𝑡
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24. Example 2:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 𝑥 𝑡 + sin⁡(3𝑡) 𝑡
General Solution
Add the homogeneous and particular solutions
𝑥 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡 ∙ 𝑐 1 𝑐 − cos 3𝑡 − − sin 3𝑡 − 𝑡
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25. 𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡) Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
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26. 2−𝑟 −5 1 −2−𝑟 =0→ 𝑟 2 +1=0→𝑟=±𝑖 𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Eigenvalues
2−𝑟 −5 1 −2−𝑟 =0→ 𝑟 2 +1=0→𝑟=±𝑖
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27. 2−𝑟 −5 1 −2−𝑟 =0→ 𝑟 2 +1=0→𝑟=±𝑖 𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Eigenvalues
2−𝑟 −5 1 −2−𝑟 =0→ 𝑟 2 +1=0→𝑟=±𝑖
Eigenvectors
𝑟=𝑖→ 2−𝑖 −5 1 −2−𝑖 → 𝑣 1 + (−2−𝑖)𝑣 2 =0→ 𝑣 = 𝑖
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28. 𝑥 (1) = cos 𝑡 2 1 − sin 𝑡 1 0 = 2 cos 𝑡 −sin⁡(𝑡) cos⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Two independent real solutions are:
𝑥 (1) = cos 𝑡 − sin 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) cos⁡(𝑡)
𝑥 (2) = sin 𝑡 cos 𝑡 = 2 sin 𝑡 +cos(𝑡) sin⁡(𝑡)
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29. 𝑥 (1) = cos 𝑡 2 1 − sin 𝑡 1 0 = 2 cos 𝑡 −sin⁡(𝑡) cos⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Two independent real solutions are:
𝑥 (1) = cos 𝑡 − sin 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) cos⁡(𝑡)
𝑥 (2) = sin 𝑡 cos 𝑡 = 2 sin 𝑡 +cos(𝑡) sin⁡(𝑡)
Fundamental Matrix
𝑋 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) sin⁡(𝑡)
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30. 𝑋 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) sin⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
𝑋 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) sin⁡(𝑡)
𝑋 −1 𝑡 = −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) −2 cos 𝑡 +sin⁡(𝑡)
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31. 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) sin⁡(𝑡) −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) −2 cos 𝑡 +sin⁡(𝑡) ∙ 0 cos⁡(𝑡) sin 𝑡 cos 𝑡 + 𝑐𝑜𝑠 2 (𝑡) −2 𝑐𝑜𝑠 2 𝑡 + sin 𝑡 cos⁡(𝑡) 𝑑𝑡
Need to integrate these two functions
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32. 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) sin⁡(𝑡) ∙ 𝑡 sin 2𝑡 − cos⁡(2𝑡) −𝑡− sin 2𝑡 − cos⁡(2𝑡)
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33. 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
Particular Solution
𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡
𝑥 𝑝 𝑡 = − 𝑡sin 𝑡 − sin 𝑡 cos⁡(2𝑡)− sin t sin⁡(2𝑡)− cos t cos⁡(2𝑡) − 𝑡cos 𝑡 cos 𝑡 sin 2𝑡 − cos 𝑡 cos 2𝑡 −𝑡sin 𝑡 − sin 𝑡 sin 2𝑡 − sin 𝑡 cos⁡(2𝑡)
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34. 𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡) Example 3:
Find the general solution to the given system.
𝑥 ′ 𝑡 = 2 −5 1 −2 𝑥 𝑡 + 0 cos⁡(𝑡)
General Solution
Add the homogeneous and particular solutions
𝑥 𝑡 = 2 cos 𝑡 −sin⁡(𝑡) 2 sin 𝑡 +cos(𝑡) cos⁡(𝑡) sin⁡(𝑡) ∙ 𝑐 1 𝑐 2 +
− 𝑡sin 𝑡 − sin 𝑡 cos⁡(2𝑡)− sin t sin⁡(2𝑡)− cos t cos⁡(2𝑡) − 𝑡cos 𝑡 cos 𝑡 sin 2𝑡 − cos 𝑡 cos 2𝑡 −𝑡sin 𝑡 − sin 𝑡 sin 2𝑡 − sin 𝑡 cos⁡(2𝑡)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB