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Differential Equations
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Find a particular solution to y′′+8y′+16y=−2e^−4t.
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i t is Undetermined Coefficients kind problem
1/ homogenous part give you \(y_c = C_1 e^{-4t}+ C_2t e^{-4t}\) 2/ therefore, partial solution must have the form of \(y_p=At^2e^{-4t}\) 3/ take first and second derivative of \(y_p\), then replace them into the original function to solve for A, you will have A = -1 replace to \(y_p\), you have \(y_p = -t^2e^{-4t}\) 4/ combine homogeneous and partial solution, your answer should be \[y = C_1e^{-4t}+C_2te^{-4t}-t^2e^{-4t}\)
thank you very much! that helped a lot
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