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\begin{align*} y” + 2y' + 2y = 0 \end{align*} Find that \(r = \{-1 - i, -1 + i\}\). Then \begin{align*} y(t) = C_1 e^{-t} \cos (t) + C_2 e^{-t} \sin (t) \end{align*}

\begin{align*} y” - 2y' + 5y = 0, \quad y(\pi/2) = 0, \quad y'(\pi / 2) = 2 \end{align*} Find that \(r = \{1+2i,1-2i\}\). Then \begin{align*} y(t) = C_1 e^t \cos (2t) + C_2 e^t \sin (2t) \end{align*} Let us apply the initial conditions \begin{align*} \begin{cases} y(\pi/2)=0\\ y'(\pi/2)=2 \end{cases} \implies \begin{cases} C_1 = 0\\ C_2 = -\frac{1}{e^{\pi/2}} \end{cases} \end{align*} Finally, \begin{align*} y(t) = -\frac{1}{e^{\pi/2}} e^t \sin (2t) \end{align*}

\begin{align*} 5u” + 2u' + 7u = 0, \quad u(0) = 2, \quad u'(0) = 1 \end{align*}

Find that \(r = \{-\frac{1}{5} - i\frac{\sqrt{34}}{5}, -\frac{1}{5} + i\frac{\sqrt{34}}{5}\}\). Then \begin{align*} \begin{cases} u(0) = 2\\ u'(0) = 1 \end{cases} \implies \begin{cases} C_1 = 2\\ C_2 = \frac{7}{\sqrt{34}} \end{cases} \end{align*} Finally, \begin{align*} u(t) = 2 e^{-t/5} \cos\left(\frac{\sqrt{34}t}{5}\right) + \frac{7}{\sqrt{34}} e^{-t/5} \sin\left(\frac{\sqrt{34}t}{5}\right) \end{align*}

\begin{align*} T \approx 14.512 \end{align*}

\begin{aligned} W\left(e^{\lambda t} \cos \mu t, e^{\lambda t} \sin \mu t\right) & =\left|\begin{array}{cc}e^{\lambda t} \cos \mu t & e^{\lambda t} \sin \mu t \\ \frac{d}{d t}\left(e^{\lambda t} \cos \mu t\right) & \frac{d}{d t}\left(e^{\lambda t} \sin \mu t\right)\end{array}\right| \\ &=\left|\begin{array}{cc}e^{\lambda t} \cos \mu t & e^{\lambda t} \sin \mu t \\ \lambda e^{\lambda t} \cos \mu t-\mu e^{\lambda t} \sin \mu t \quad \lambda e^{\lambda t} \sin \mu t+\mu e^{\lambda t} \cos \mu t\end{array}\right| \\ &=e^{\lambda t} \cos \mu t\left(\lambda e^{\lambda t} \sin \mu t+\mu e^{\lambda t} \cos \mu t\right)-e^{\lambda t} \sin \mu t\left(\lambda e^{\lambda t} \cos \mu t-\mu e^{\lambda t} \sin \mu t\right) \\ &=\lambda e^{2 \lambda t} \cos \mu t \sin \mu t+\mu e^{2 \lambda t} \cos ^{2} \mu t\lambda e^{2 \lambda t} \cos \mu t \sin \mu t+\mu e^{2 \lambda t} \sin ^{2} \mu t \\ &=\mu e^{2 \lambda t}\left(\cos ^{2} \mu t+\sin ^{2} \mu t\right) \\ &=\mu e^{2 \lambda t} \end{aligned}

\begin{align*} y” - 2y' + y = 0 \end{align*} Then \begin{align*} r^2 - 2r + 1 = 0 \end{align*} So \(r = 1\), finally \begin{align*} y(t) = C_1 e^t + C_2 t e^t \end{align*}

\begin{align*} 2y” + 2y' + y = 0 \end{align*} Then \begin{align*} 2r^2 + 2r + 1 = 0 \end{align*} So \(r = \{-1/2 - 1/2i, -1/2 + 1/2i\}\). Finally, \begin{align*} y(t) = C_1 e^{-t/2} \cos \left(\frac{1}{2}t\right) + C_2 e^{-t/2} \sin \left(\frac{1}{2}t\right) \end{align*}

\begin{align*} y” + 4y' + 4y = 0, \quad y(-1)=2, \quad y'(-1) = 1 \end{align*} Then \(r=\{-2\}\). So \begin{align*} y(t) = C_1 e^{-2t} + C_2 t e^{-2t} \end{align*} Taking the derivative \begin{align*} y'(t) = C_1 e^{-2t} - 2 C_1t e^{-2t} - 2 C_2 e^{-2t} \end{align*} Applying the initial conditions \begin{align*} \begin{cases} y(-1) = -C_1e^2 + C_2 e^2 = 2\\ y'(-1) = C_1 e^2 + 2 C_1 e^2 - 2 C_2 e^2 = 1 \end{cases} \implies \begin{cases} C_1 = \frac{5}{e^2}\\ C_2 = \frac{7}{e^2} \end{cases} \end{align*} Therefore \begin{align*} y(t) = \frac{5}{e^2}te^{-2t} + \frac{7}{e^2}e^{-2t} \end{align*} As \(t \to \infty\), \(y \to 0\).

\begin{align*} t^2 y” - 4 t y' + 6 y = 0, \quad t > 0; \quad y_1(t) = t^2 \end{align*} Let us apply the method of reduction Let the general solution be of the form \begin{align*} y(t) = v(t) y_1(t) = t^2v(t) \end{align*} Let us find the derivatives \begin{align*} y_2'(t) &= t^2v'(t) + 2tv(t)\\ y_2”(t) &= t^2v”(t) + 2tv'(t) + 2tv'(t) + 2v(t) \end{align*} Let us substitute the above into our original ODE \begin{align*} t^2 (t^2v”(t) + 2tv'(t) + 2tv'(t) + 2v(t)) -4t (t^2v'(t) + 2tv(t)) + 6(t^2v(t)) = 0 \end{align*} Simplifying yields \begin{align*} t^4 v”(t) + 3t^3v'(t) + t^2v(t) - 4t^3v'(t) -8t^2v(t) + 6t^2v(t) = 0\\ t^4v”(t) = 0 \end{align*} Then \begin{align*} v”(t) = 0 \end{align*} Integrate both sides \begin{align*} v'(t) = C_1\\ v(t) = C_1 t + C_2 \end{align*} Finally, the general solution is \begin{align*} y(t) = (C_1t + C_2)t^2 = C_1t^3 + C_2t^2 = C_1 y_1(t) + C_2 y_2(t) \end{align*} and the second solution is \(y_2(t) = t^3\).