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Section 7.6 Summary & Exercises
Summary of the Key Ideas.
The method of undetermined coefficients is used to solve non-homogeneous linear differential equations.
The general solution to a non-homogeneous equation is the sum of the general solution to the corresponding homogeneous equation and a particular solution.
The method involves guessing the form of the particular solution based on the form of the non-homogeneous term and solving for the coefficients.
The method is applicable when the non-homogeneous term can be expressed as a linear combination of known functions.
Exercises Exercises
1.
Solve the IVP.
\begin{equation}
2z'' + z = 9e^{2t}, \quad z(0) = 3, \quad z'(0) = -1 \tag{43}
\end{equation}
2. \(z_h\text{,}\) of the general solution of (43) .Answer .
\begin{equation*}
z_h = c_1\cos\left(\frac{1}{\sqrt{2}}t\right) + c_2\sin\left(\frac{1}{\sqrt{2}}t\right)
\end{equation*}
3. \(z_p\text{,}\) by looking at the non-homogenous term of equation (43) .Answer . \(\ds z_p = Ae^{2t} \)
4. \(z_p \) into (43) to find the coefficient, \(A\text{.}\) Then give \(z_p\text{.}\) Answer .
\begin{equation*}
z_p = e^{2t}
\end{equation*}
5. \(z = z_c + z_p. \) Answer .
\begin{equation*}
z = c_1\cos\left(\frac{1}{\sqrt{2}}t\right) + c_2\sin\left(\frac{1}{\sqrt{2}}t\right) + e^{2t}
\end{equation*}
6. Answer .
\begin{equation*}
z(t) = -3\sqrt{2}\cos\left(\frac{1}{\sqrt{2}}t\right) + 2\sin\left(\frac{1}{\sqrt{2}}t\right) + e^{2t}
\end{equation*}
Find the Initial Form of \(y_p\text{.}\) .
\(y_p\text{.}\) DO NOT determine the values of the coefficients \(A, \ds B, \) etc.
7. \(\ds y'' - 3y' - 17y = x^2 + \cos x \) Answer . \(\ds y_p = \Big[ Ax^2 + Bx + C \Big] + \Big[ D\cos x + E\sin x \Big] \)
8. \(\ds y'' + y' - 12y = 4e^{3t} + \cos(2t) \) Answer . \(\ds y_p = \Big[ Ate^{3t} \Big] + \Big[ B\cos(2t) + C\sin(2t) \Big] \)
9. \(\ds y'' - 4y = 2t - e^{2t}\sin(3t) \) Answer . \(\ds y_p = \Big[ At + B \Big] + \Big[ Ce^{2t}\cos(3t) + De^{2t}\sin(3t) \Big] \)
10. \(\ds x^3 \sin(4x) + x^2 \sin(4x) + 7y'' - y' + 2y = 0 \) 11. \(\ds w'' - 4w' + 13w = e^{3x} + x^2e^{-x} \) Answer . \(\ds w_p = \Big[ Ae^{3x} \Big] + \Big[ (Bx^2 + Cx + D)e^{-x} \Big] \)
12. \(\ds z'' - 4z' + 5z = te^{2t} + 2e^{2t}\sin t \) Answer . \(\ds z_p = \Big[ (At + B)e^{2t} \Big] + \Big[ Cet^{2t}\cos t + Dte^{2t}\sin t \Big] \)
13. \(\ds y'' - 2y' + y = x^2 + e^x \) Answer . \(\ds y_p = \Big[ Ax^2+ Bx + C\Big] + \Big[ Dx^2 e^x \Big] \)
14. \(\ds w'' - 4w' + 13 = e^{3x} + x^2e^{-x} \) Answer . \(\ds w_p = \Big[ Ae^{3x} \Big] + \Big[ (Bx^2 + Cx + D)e^{-x} \Big] \)
15. \(\ds 3z'' - 4z' - 12z = 17 + 2\cos(2\theta) \) Answer . \(\ds z_p = \Big[ A \Big] + \Big[ B\cos(2\theta) + C\sin(2\theta) \Big] \)
Find the general solution for each differential equation.
16. \(\ds 2x' + x = 3t^2 \) Answer . \(\ds x = c_1e^{-\frac{1}{2}t} + 3t^2 -12 t + 24 \)
17. \(\ds y'' - 4y' - 5y = t + 2e^{-t} \) Answer . \(\ds y = c_1e^{5t} + c_2e^{-t} - \frac{1}{5}t + \frac{4}{25} - \frac{1}{3}te^{-t} \)
18. \(\ds z'' -650\sin(6x) +34z = 6z' \) Answer . \(\ds z = c_1e^{3x}\cos(5x) + c_2e^{3x}\sin(5x) + 18\cos(6x) - \sin(6x) \)
19. \(\ds y'' - 4y' + 4y = 8\cos x + 12 e^{2x} \) Answer . \(\ds y = c_1e^{2x} + c_2xe^{2x} + \frac{8}{9}\cos x - \frac{32}{27}\sin x + 6x^2 e^{2x} \)
20. \(\ds y'' - y' - 6y = 2t + 3e^{3t} - e^{-2t} \) Answer . \(\ds y = c_1e^{3t} + c_2e^{-2t} -\frac{1}{3}t + \frac{1}{18} + \frac{3}{5}te^{3t} + \frac{1}{5}te^{-2t} \)
Exercise Group. 21.
\begin{equation*}
2x' + x = 3t^2, \hspace{0.5cm} x(0) = 15 \text{.}
\end{equation*}
Answer . \(\ds x = -9e^{-\frac{1}{2}t} + 3t^2 -12 t + 24 \)
Exercise Group. 22. Exercise 7.6.21 is actually a first-order linear differential equation, so we can use another method of solution to solve it (and we should get the same answer!). Solve the same IVP using an integrating factor and compareyour answers.Answer . \(\ds x = 3t^2 - 12t + 24 - 9e^{-\frac{1}{2}t} \)
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