Systems of Differential Equations Nonhomogeneous Systems Math 4B Systems of Differential Equations Nonhomogeneous Systems Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 𝑥 𝑡 + 𝑡 2 𝑡+1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 𝑥 𝑡 + 𝑡 2 𝑡+1 Eigenvalues Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 𝑥 𝑡 + 𝑡 2 𝑡+1 Eigenvalues Eigenvectors Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 𝑥 𝑡 + 𝑡 2 𝑡+1 Fundamental Matrix Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 𝑥 𝑡 + 𝑡 2 𝑡+1 Fundamental Matrix Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 = −1 0 2 2 𝑥 𝑡 + 𝑡 2 𝑡+1 Particular Solution 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 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 1 3 𝑒 𝑡 0 2 𝑒 −2𝑡 3𝑒 −2𝑡 ∙ 𝑡 2 𝑡+1 𝑑𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 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 1 3 𝑒 𝑡 0 2 𝑒 −2𝑡 3𝑒 −2𝑡 ∙ 𝑡 2 𝑡+1 𝑡 2 𝑒 𝑡 (2 𝑡 2 +3𝑡+3) 𝑒 −2𝑡 𝑑𝑡 Need to integrate these two functions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 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 ∙ 1 3 ( 𝑡 2 −2𝑡+2) 𝑒 𝑡 (− 𝑡 2 − 5 2 𝑡− 11 4 ) 𝑒 −2𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 1: Find the general solution to the given system. 𝑥 ′ 𝑡 = −1 0 2 2 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 ∙ 1 3 ( 𝑡 2 −2𝑡+2) 𝑒 𝑡 (− 𝑡 2 − 5 2 𝑡− 11 4 ) 𝑒 −2𝑡 𝑥 𝑝 𝑡 = 𝑡 2 −2𝑡+2 − 𝑡 2 + 1 2 𝑡− 9 4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 = −1 0 2 2 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 2 + 𝑡 2 −2𝑡+2 − 𝑡 2 + 1 2 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 0 1 + 𝑡 2 1 −1 +𝑡 −2 1 2 + 2 − 9 4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡 Example 2: Find the general solution to the given system. 𝑥 ′ 𝑡 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

−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 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡 Eigenvalues −𝑟 1 4 −𝑟 =0→ 𝑟 2 −4=0→𝑟=±2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

−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 = 0 1 4 0 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 1 4 2 →2 𝑣 1 + 𝑣 2 =0→ 𝑣 2 = 1 −2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡 Fundamental Matrix 𝑋 𝑡 = 𝑒 2𝑡 𝑒 −2𝑡 2 𝑒 2𝑡 −2𝑒 −2𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2: Find the general solution to the given system. 𝑥 ′ 𝑡 = 0 1 4 0 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 = 1 2 𝑒 −2𝑡 1 4 𝑒 −2𝑡 1 2 𝑒 2𝑡 − 1 4 𝑒 2𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 = 0 1 4 0 𝑥 𝑡 + sin⁡(3𝑡) 𝑡 Particular Solution 𝑥 𝑝 𝑡 =𝑋(𝑡) 𝑋 −1 (𝑡)∙ 𝑔 𝑡 𝑑𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2: Find the general solution to the given system. 𝑥 ′ 𝑡 = 0 1 4 0 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 1 2 𝑒 −2𝑡 1 4 𝑒 −2𝑡 1 2 𝑒 2𝑡 − 1 4 𝑒 2𝑡 ∙ sin⁡(3𝑡) 𝑡 𝑑𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2: Find the general solution to the given system. 𝑥 ′ 𝑡 = 0 1 4 0 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 1 2 𝑒 −2𝑡 1 4 𝑒 −2𝑡 1 2 𝑒 2𝑡 − 1 4 𝑒 2𝑡 ∙ sin⁡(3𝑡) 𝑡 1 2 𝑒 −2𝑡 sin 3𝑡 + 1 4 𝑡𝑒 −2𝑡 1 2 𝑒 2𝑡 sin 3𝑡 − 1 4 𝑡𝑒 2𝑡 𝑑𝑡 Need to integrate these two functions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 = 0 1 4 0 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 (− 3 26 cos 3𝑡 − 1 13 sin 3𝑡 − 1 8 𝑡− 1 16 ) 𝑒 2𝑡 (− 3 26 cos 3𝑡 + 1 13 sin 3𝑡 + 1 8 𝑡− 1 16 ) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2: Find the general solution to the given system. 𝑥 ′ 𝑡 = 0 1 4 0 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 (− 3 26 cos 3𝑡 − 1 13 sin 3𝑡 − 1 8 𝑡− 1 16 ) 𝑒 2𝑡 (− 3 26 cos 3𝑡 + 1 13 sin 3𝑡 + 1 8 𝑡− 1 16 ) 𝑥 𝑝 𝑡 = − 3 13 cos 3𝑡 − 1 8 − 4 13 sin 3𝑡 − 1 2 𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example 2: Find the general solution to the given system. 𝑥 ′ 𝑡 = 0 1 4 0 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 2 + − 3 13 cos 3𝑡 − 1 8 − 4 13 sin 3𝑡 − 1 2 𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 = 2 1 + 1 0 𝑖 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 2 1 − sin 𝑡 1 0 = 2 cos 𝑡 −sin⁡(𝑡) cos⁡(𝑡) 𝑥 (2) = sin 𝑡 2 1 + cos 𝑡 1 0 = 2 sin 𝑡 +cos(𝑡) sin⁡(𝑡) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 2 1 − sin 𝑡 1 0 = 2 cos 𝑡 −sin⁡(𝑡) cos⁡(𝑡) 𝑥 (2) = sin 𝑡 2 1 + cos 𝑡 1 0 = 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) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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) 2 sin 𝑡 cos 𝑡 + 𝑐𝑜𝑠 2 (𝑡) −2 𝑐𝑜𝑠 2 𝑡 + sin 𝑡 cos⁡(𝑡) 𝑑𝑡 Need to integrate these two functions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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) ∙ 1 2 𝑡+ 1 4 sin 2𝑡 − 1 2 cos⁡(2𝑡) −𝑡− 1 2 sin 2𝑡 − 1 4 cos⁡(2𝑡) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 = − 5 2 𝑡sin 𝑡 − sin 𝑡 cos⁡(2𝑡)− 5 4 sin t sin⁡(2𝑡)− 5 4 cos t cos⁡(2𝑡) − 1 2 𝑡cos 𝑡 + 1 4 cos 𝑡 sin 2𝑡 − 1 2 cos 𝑡 cos 2𝑡 −𝑡sin 𝑡 − 1 2 sin 𝑡 sin 2𝑡 − 1 4 sin 𝑡 cos⁡(2𝑡) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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 + − 5 2 𝑡sin 𝑡 − sin 𝑡 cos⁡(2𝑡)− 5 4 sin t sin⁡(2𝑡)− 5 4 cos t cos⁡(2𝑡) − 1 2 𝑡cos 𝑡 + 1 4 cos 𝑡 sin 2𝑡 − 1 2 cos 𝑡 cos 2𝑡 −𝑡sin 𝑡 − 1 2 sin 𝑡 sin 2𝑡 − 1 4 sin 𝑡 cos⁡(2𝑡) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB