User:Egm6321.f12.team4.harris/Homework56 R*6.6 Part 2.8.1

Problem R*6.6 2.8 – Nonhomogeneous solution
sec33-5

Statement
Solve the nonhomogeneous L2-ODE-CC
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$$ a_2y''+a_1y'+a_0y=f(t) $$     (6.6.?) with an excitation of $$ f(t)=e^{-t^2} $$ (Gaussian distribution). For Part 2.8.1, consider the characteristic equation $$ (r+1)(r-2)=0 $$. For Part 2.8.2, consider the characteristic equation $$ (r-4)^2=0 $$.
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Part 2.8.1
From the characteristic equation $$ (r+1)(r-2)=0 $$, one can find that $$ \alpha = -1 $$ and $$ \beta =2 $$. From Part 2.6 above, one can see that the equation for the particular solution is


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$$ y_p(t)=e^{-\beta t} \int ^s (\frac{e^{(\beta - \alpha)s}}{\bar {a_1}} \int ^t (e^{\alpha t} f(t) dt))ds $$     (6.6.?)
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Substituting in for $$ f(t), \alpha, \beta $$, one arrives at the particular solution
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$$ y_p(t)=e^{-2t} \int ^s (\frac{e^{3s}}{\bar {a_1}} \int ^t (e^{-t} e^{-t^2} dt))ds=e^{-2t} \int ^s (\frac{e^{3s}}{\bar {a_1}} \int ^t (e^{-t(1+t)} dt))ds $$     (6.6.?)
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Part 2.8.1
From the characteristic equation $$ (r-4)^2=0 $$, one can find that $$ \alpha = \beta =4 $$. From Part 2.8.1 equation 6.6.? above, one can see that the equation for the particular solution is
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$$ y_p(t)=e^{-4t} \int ^s (\frac{e^{0}}{\bar {a_1}} \int ^t (e^{4t} e^{-t^2} dt))ds=\frac{e^{-4t}}{\bar {a_1}} \int ^s \int ^t (e^{t(4-t)} dt)ds $$     (6.6.?)
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