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Substition of these equations in (6) yiedts \[ \begin{array}{l} A e^{t}+B e^{3 t}+C e^{-2 t}+E+A e^{t}+3 B e^{3 t}-2 C e^{-2 t}=4 e^{t}-8 c_{1} e^{2 t}-3 c_{2} e^{-2 t}-5 \\ 2 A e^{t}+4 B e^{3 t}-C e^{-2 t}+E=4 e^{t}-8 c_{1} e^{3 t}-3 C_{2} e^{-2 t}-5 \end{array} \] Comparing cefficients we find that - \( e^{t}: 2 A=4 \Rightarrow A=2 \) - \( e^{3 t}: 4 B=-8 c_{1} \Rightarrow B=-2 c_{1} \) - \( e^{-2 t}:-C=-3 c_{1} \Rightarrow C=3 c_{2} \) + Const. \( E=-5 \) \[ \Rightarrow y_{P, I}(t)=2 e^{t}-2 c_{1} e^{3 t}+3 c_{2} e^{-2 t}-5 \] Therefora, \[ \begin{aligned} y(t) & =y_{C, F}(t)+y_{P, I}(t) \\ & =C_{3} e^{-t}+2 e^{t}-2 C_{1} e^{3 t}+3 C_{2} e^{-2 t}-5 \end{aligned} \] Tho solution therefore contains is arbitrary constants \( C_{1}, C_{2} \) and \( C_{3} \) wheress the determinant \[ \operatorname{dat}\left|\begin{array}{ll} D+5 & D+1 \\ D+11 & 2 D+1 \end{array}\right|=D^{2}-D-6 \] is of the seand order. Substitution of the solution \( (x(t), y(t)) \) inte (2) yields \( C_{3}=0 \). Consequently a general solution of the systan is given by \[ \begin{array}{l} x(t)=c_{1} e^{3 t}+c_{2} e^{-2 t}-\frac{1}{2} e^{t}+1 \\ y(t)=2 e^{t}-2 c_{1} e^{3 t}+3 c_{2} e^{2 t}-5 \end{array} \] with anly 2 arbitrary constants \( C_{1} \) and \( C_{2} \). 2.1. Solve the system: \( -2 \dot{x}+\ddot{y}+y=t^{2} \) \[ \dot{x}-2 x+2 \dot{y}-y=3 t^{2}+3, \] Powered by CamScai by using the elimination methed (qoerator method). Eliminate a first. Edution \( D \)-gerator form: \( -2 D[x]+D^{2}[y]+[y]=-t^{2} \) \[ \begin{aligned} & \Rightarrow-2 D[x]+\left(D^{2}+1\right)[y]=-t^{2} \ldots(1 \\ \text { and } & D[x]-2[x]+2 D[y]-[y]=3 t^{2}+3 \\ \Rightarrow & (D-2)[x]+(2 D-1)[y]=8 t^{2}+3 \ldots \end{aligned} \] Eliminating \( x \) first we find: \[ \begin{aligned} \cdot(D-2)[(1)] & :(D-2)(-2 D[x])+(D-2)\left(D^{2}+1\right)[y] \\ \Rightarrow & =\left(D-2 D^{2}+4 D\right)[x]+\left(D^{2}\right) \\ \cdot-2 D[(2)] & :-2 D(D-2)[x]-2 D(2 D-1)[y]=-2 D\left(3 t^{2}+3\right) \\ \Rightarrow & \left(-2 D^{2}+4 D\right)[x]+\left(-4 D^{2}+2 D\right)[y]=-12 t \end{aligned} \] Naw, 

          Substition of these equations in (6) yiedts
\[
\begin{array}{l}
A e^{t}+B e^{3 t}+C e^{-2 t}+E+A e^{t}+3 B e^{3 t}-2 C e^{-2 t}=4 e^{t}-8 c_{1} e^{2 t}-3 c_{2} e^{-2 t}-5 \\
2 A e^{t}+4 B e^{3 t}-C e^{-2 t}+E=4 e^{t}-8 c_{1} e^{3 t}-3 C_{2} e^{-2 t}-5
\end{array}
\]

Comparing cefficients we find that
- \( e^{t}: 2 A=4 \Rightarrow A=2 \)
- \( e^{3 t}: 4 B=-8 c_{1} \Rightarrow B=-2 c_{1} \)
- \( e^{-2 t}:-C=-3 c_{1} \Rightarrow C=3 c_{2} \)
+ Const. \( E=-5 \)
\[
\Rightarrow y_{P, I}(t)=2 e^{t}-2 c_{1} e^{3 t}+3 c_{2} e^{-2 t}-5
\]

Therefora,
\[
\begin{aligned}
y(t) & =y_{C, F}(t)+y_{P, I}(t) \\
& =C_{3} e^{-t}+2 e^{t}-2 C_{1} e^{3 t}+3 C_{2} e^{-2 t}-5
\end{aligned}
\]

Tho solution therefore contains is arbitrary constants \( C_{1}, C_{2} \) and \( C_{3} \) wheress the determinant
\[
\operatorname{dat}\left|\begin{array}{ll}
D+5 & D+1 \\
D+11 & 2 D+1
\end{array}\right|=D^{2}-D-6
\]
is of the seand order. Substitution of the solution \( (x(t), y(t)) \) inte (2) yields \( C_{3}=0 \). Consequently a general solution of the systan is given by
\[
\begin{array}{l}
x(t)=c_{1} e^{3 t}+c_{2} e^{-2 t}-\frac{1}{2} e^{t}+1 \\
y(t)=2 e^{t}-2 c_{1} e^{3 t}+3 c_{2} e^{2 t}-5
\end{array}
\]
with anly 2 arbitrary constants \( C_{1} \) and \( C_{2} \).
2.1. Solve the system: \( -2 \dot{x}+\ddot{y}+y=t^{2} \)
\[
\dot{x}-2 x+2 \dot{y}-y=3 t^{2}+3,
\]

Powered by
CamScai

by using the elimination methed (qoerator method). Eliminate a first.

Edution \( D \)-gerator form: \( -2 D[x]+D^{2}[y]+[y]=-t^{2} \)
\[
\begin{aligned}
& \Rightarrow-2 D[x]+\left(D^{2}+1\right)[y]=-t^{2} \ldots(1 \\
\text { and } & D[x]-2[x]+2 D[y]-[y]=3 t^{2}+3 \\
\Rightarrow & (D-2)[x]+(2 D-1)[y]=8 t^{2}+3 \ldots
\end{aligned}
\]

Eliminating \( x \) first we find:
\[
\begin{aligned}
\cdot(D-2)[(1)] & :(D-2)(-2 D[x])+(D-2)\left(D^{2}+1\right)[y] \\
\Rightarrow & =\left(D-2 D^{2}+4 D\right)[x]+\left(D^{2}\right) \\
\cdot-2 D[(2)] & :-2 D(D-2)[x]-2 D(2 D-1)[y]=-2 D\left(3 t^{2}+3\right) \\
\Rightarrow & \left(-2 D^{2}+4 D\right)[x]+\left(-4 D^{2}+2 D\right)[y]=-12 t
\end{aligned}
\]

Naw,
Show more…
Substition of these equations in (6) yiedts A e^t+B e^3 t+C e^-2 t+E+A e^t+3 B e^3 t-2 C e^-2 t=4 e^t-8 c1 e^2 t-3 c2 e^-2 t-5 2 A e^t+4 B e^3 t-C e^-2 t+E=4 e^t-8 c1 e^3 t-3 C2 e^-2 t-5 Comparing cefficients we find that - e^t: 2 A=4 β‡’ A=2 - e^3 t: 4 B=-8 c1β‡’ B=-2 c1 - e^-2 t:-C=-3 c1β‡’ C=3 c2 + Const. E=-5 β‡’ yP, I(t)=2 e^t-2 c1 e^3 t+3 c2 e^-2 t-5 Therefora, y(t) =yC, F(t)+yP, I(t) =C3 e^-t+2 e^t-2 C1 e^3 t+3 C2 e^-2 t-5 Tho solution therefore contains is arbitrary constants C1, C2 and C3 wheress the determinant dat| D+5 D+1 D+11 2 D+1 |=D^2-D-6 is of the seand order. Substitution of the solution (x(t), y(t)) inte (2) yields C3=0. Consequently a general solution of the systan is given by x(t)=c1 e^3 t+c2 e^-2 t-(1)/(2) e^t+1 y(t)=2 e^t-2 c1 e^3 t+3 c2 e^2 t-5 with anly 2 arbitrary constants C1 and C2. 2.1. Solve the system: -2 αΊ‹+ΓΏ+y=t^2 αΊ‹-2 x+2 ẏ-y=3 t^2+3, Powered by CamScai by using the elimination methed (qoerator method). Eliminate a first. Edution D-gerator form: -2 D[x]+D^2[y]+[y]=-t^2 β‡’-2 D[x]+(D^2+1)[y]=-t^2…(1 and D[x]-2[x]+2 D[y]-[y]=3 t^2+3 β‡’ (D-2)[x]+(2 D-1)[y]=8 t^2+3 … Eliminating x first we find: Β·(D-2)[(1)] :(D-2)(-2 D[x])+(D-2)(D^2+1)[y] β‡’ =(D-2 D^2+4 D)[x]+(D^2) Β·-2 D[(2)] :-2 D(D-2)[x]-2 D(2 D-1)[y]=-2 D(3 t^2+3) β‡’ (-2 D^2+4 D)[x]+(-4 D^2+2 D)[y]=-12 t Naw,

Added by Gregg M.

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A First Course in Differential Equations with Modeling Applications
A First Course in Differential Equations with Modeling Applications
Dennis G. Zill 11th Edition
Chapter 8
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Substition of these equations in (6) yiedts \[ \begin{array}{l} A e^{t}+B e^{3 t}+C e^{-2 t}+E+A e^{t}+3 B e^{3 t}-2 C e^{-2 t}=4 e^{t}-8 c_{1} e^{2 t}-3 c_{2} e^{-2 t}-5 \\ 2 A e^{t}+4 B e^{3 t}-C e^{-2 t}+E=4 e^{t}-8 c_{1} e^{3 t}-3 C_{2} e^{-2 t}-5 \end{array} \] Comparing cefficients we find that - \( e^{t}: 2 A=4 \Rightarrow A=2 \) - \( e^{3 t}: 4 B=-8 c_{1} \Rightarrow B=-2 c_{1} \) - \( e^{-2 t}:-C=-3 c_{1} \Rightarrow C=3 c_{2} \) + Const. \( E=-5 \) \[ \Rightarrow y_{P, I}(t)=2 e^{t}-2 c_{1} e^{3 t}+3 c_{2} e^{-2 t}-5 \] Therefora, \[ \begin{aligned} y(t) & =y_{C, F}(t)+y_{P, I}(t) \\ & =C_{3} e^{-t}+2 e^{t}-2 C_{1} e^{3 t}+3 C_{2} e^{-2 t}-5 \end{aligned} \] Tho solution therefore contains is arbitrary constants \( C_{1}, C_{2} \) and \( C_{3} \) wheress the determinant \[ \operatorname{dat}\left|\begin{array}{ll} D+5 & D+1 \\ D+11 & 2 D+1 \end{array}\right|=D^{2}-D-6 \] is of the seand order. Substitution of the solution \( (x(t), y(t)) \) inte (2) yields \( C_{3}=0 \). Consequently a general solution of the systan is given by \[ \begin{array}{l} x(t)=c_{1} e^{3 t}+c_{2} e^{-2 t}-\frac{1}{2} e^{t}+1 \\ y(t)=2 e^{t}-2 c_{1} e^{3 t}+3 c_{2} e^{2 t}-5 \end{array} \] with anly 2 arbitrary constants \( C_{1} \) and \( C_{2} \). 2.1. Solve the system: \( -2 \dot{x}+\ddot{y}+y=t^{2} \) \[ \dot{x}-2 x+2 \dot{y}-y=3 t^{2}+3, \] Powered by CamScai by using the elimination methed (qoerator method). Eliminate a first. Edution \( D \)-gerator form: \( -2 D[x]+D^{2}[y]+[y]=-t^{2} \) \[ \begin{aligned} & \Rightarrow-2 D[x]+\left(D^{2}+1\right)[y]=-t^{2} \ldots(1 \\ \text { and } & D[x]-2[x]+2 D[y]-[y]=3 t^{2}+3 \\ \Rightarrow & (D-2)[x]+(2 D-1)[y]=8 t^{2}+3 \ldots \end{aligned} \] Eliminating \( x \) first we find: \[ \begin{aligned} \cdot(D-2)[(1)] & :(D-2)(-2 D[x])+(D-2)\left(D^{2}+1\right)[y] \\ \Rightarrow & =\left(D-2 D^{2}+4 D\right)[x]+\left(D^{2}\right) \\ \cdot-2 D[(2)] & :-2 D(D-2)[x]-2 D(2 D-1)[y]=-2 D\left(3 t^{2}+3\right) \\ \Rightarrow & \left(-2 D^{2}+4 D\right)[x]+\left(-4 D^{2}+2 D\right)[y]=-12 t \end{aligned} \] Naw,
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