Question

solve the initial value problem y''-4y'+4y=64sin2t for y(0)=0 and y'(0)=1

Answer 1

$\text{Solve } \frac{d^{2} y(t)}{d t^{2}}-4 \frac{d y(t)}{d t}+4 y(t)=64 \sin (2 t), \text{ such that } y(0)=0 \text{ and } y^{\prime}(0)=1 :\\[4mm] \text{The general solution will be the sum of the complementary solution and particular solution.}\\ \text{ Find the complementary solution by solving } \frac{d^{2} y(t)}{d t^{2}}-4 \frac{d y(t)}{d t}+4 y(t)=0 :\\ \text{Assume a solution will be proportional to } e^{\lambda t} \text{ for some constant }\lambda.\\[4mm] \text{Substitute } y(t)=e^{\lambda t} \text{ into the differential equation:}\\ \frac{d^{2}}{d t^{2}}\left(e^{x t}\right)-4 \frac{d}{d t}\left(e^{x t}\right)+4 e^{\lambda t}=0\\[4mm] \text{Substitute } \frac{d^{2}}{d t^{2}}\left(e^{\lambda t}\right)=\lambda^{2} e^{\lambda t} \text{ and } \frac{d}{d t}\left(e^{\lambda t}\right)=\lambda e^{\lambda t}\\ \lambda^{2} e^{\lambda t}-4 \lambda e^{\lambda t}+4 e^{\lambda t}=0\\[4mm] \text{Factor out } e^{\lambda t}\\[2mm] \left(\lambda^{2}-4 \lambda+4\right) e^{\lambda t}=0\\ \text{Since } e^{\lambda t} \neq 0 \text{for any finite } \lambda, \text{ the zeros must come from the polynomial:}\\[2mm] \lambda^{2}-4 \lambda+4=0$

$\text{Factor:}\\ (\lambda-2)^{2}=0\\[4mm] \text{Solve for } \lambda :\\[2mm] \lambda=2 \text { or } \lambda=2\\[4mm] \text{The multiplicity of the root } \lambda=2 \text{ is }2 \text{ which gives } y_{1}(t)=c_{1} e^{2 t}, y_{2}(t)=c_{2} e^{2 t} t \text{ as solutions, where } c_{1} \text{ and } c_{2} \text{ are arbitrary constants.}\\[4mm] \text{The general solution is the sum of the above solutions:}\\[2mm] y(t)=y_{1}(t)+y_{2}(t)=c_{1} e^{2 t}+c_{2} e^{2 t} t\\[4mm] \text{Determine the particular solution to } \frac{d^{2} y(t)}{d t^{2}}-4 \frac{d y(t)}{d t}+4 y(t)=64 \sin (2 t) \text{ by the method of undetermined coefficients:}\\[4mm] \text{The particular solution to } \frac{d^{2} y(t)}{d t^{2}}-4 \frac{d y(t)}{d t}+4 y(t)=64 \sin (2 t) \text{ is of the form: } y_{p}(t)=a_{1} \cos (2 t)+a_{2} \sin (2 t)\\[4mm] \text{Solve for the unknown constants } a_{1} \text{ and } a_{2} :\\[4mm] \text{Compute } \frac{d y_{p}(t)}{d t} :\\ \begin{aligned} \frac{d y_{p}(t)}{d t} &=\frac{d}{d t}\left(a_{1} \cos (2 t)+a_{2} \sin (2 t)\right) \\ &=-2 a_{1} \sin (2 t)+2 a_{2} \cos (2 t) \end{aligned}\\$

$\text{Compute } \frac{d^{2} y_{p}(t)}{d t^{2}} :\\[2mm] \frac{d^{2} y_{p}(t)}{d t^{2}} =\frac{d^{2}}{d t^{2}}\left(a_{1} \cos (2 t)+a_{2} \sin (2 t)\right)\\ \qquad \;\; =-4 a_{1} \cos (2 t)-4 a_{2} \sin (2 t)\\[4mm] \text{Substitute the particular solution } y_{p}(t) \text{ into the differential equation:}\\[2mm] \begin{aligned} &\frac{d^{2} y_{p}(t)}{d t^{2}}-4 \frac{d y_{p}(t)}{d t}+4 y_{p}(t)=64 \sin (2 t) \\ &-4 a_{1} \cos (2 t)-4 a_{2} \sin (2 t)- \\ &\quad 4\left(-2 a_{1} \sin (2 t)+2 a_{2} \cos (2 t)\right)+4\left(a_{1} \cos (2 t)+a_{2} \sin (2 t)\right)=64 \sin (2 t) \end{aligned}\\[4mm] \text{ Simplify:}\\[2mm] -8 a_{2} \cos (2 t)+8 a_{1} \sin (2 t)=64 \sin (2 t)\\[4mm] \text{Equate the coefficients of } \cos (2 t) \text{ on both sides of the equation:}\\[2mm] -8 a_{2}=0\\[4mm] \text{Equate the coefficients of } \sin (2 t) \text{ on both sides of the equation:}\\[2mm] 8 a_{1}=64\\[4mm] \text{Solve the system:}\\[2mm] \begin{aligned} &a_{1}=8 \\ &a_{2}=0 \end{aligned}$

$\text{Substitute } a_{1} \text{ and } a_{2} \text{ into} y_{p}(t)=\cos (2 t) a_{1}+\sin (2 t) a_{2} :\\[2mm] y_{p}(t)=8 \cos (2 t)\\[4mm] \text{ The general solution is: }\\[2mm] y(t)=y_{c}(t)+y_{p}(t)=8 \cos (2 t)+c_{1} e^{2 t}+c_{2} e^{2 t} t\\[4mm] \text{ Solve for the unknown constants using the initial conditions: }\\[2mm] \text{ Compute }\frac{d y(t)}{d t} :\\[2mm] \begin{aligned} \frac{d y(t)}{d t} &=\frac{d}{d t}\left(8 \cos (2 t)+c_{1} e^{2 t}+c_{2} e^{2 t} t\right) \\ &=-16 \sin (2 t)+2 c_{1} e^{2 t}+c_{2} e^{2 t}+2 c_{2} e^{2 t} t \end{aligned}\\[4mm] \text{ Substitute } y(0)=0 \text{ into } y(t)=8 \cos (2 t)+e^{2 t} c_{1}+e^{2 t} t c_{2} :\\[2mm] c_{1}+8=0\\[4mm] \text{Substitute } y^{\prime}(0)=1 \text{ into } \frac{d y(t)}{d t}=-16 \sin (2 t)+2 e^{2 t} c_{1}+e^{2 t} c_{2}+2 e^{2 t} t c_{2} :\\[2mm] 2 c_{1}+c_{2}=1\\[4mm] \text{ Solve the system:}\\[2mm] \begin{aligned} &\mathcal{c}_{1}=-8 \\ &\mathcal{c}_{2}=17 \end{aligned} \text{Substitute } c_{1}=-8 \text{ and } c_{2}=17 \text{ into } y(t)=8 \cos (2 t)+e^{2 t} c_{1}+e^{2 t} t c_{2}\\[4mm] \text{ Answer:}\\[4mm] y(t)=8 \cos (2 t)+e^{2 t}(17 t-8)$