1 problem found
Given that \[ \frac{\mathrm{d}x}{\mathrm{d}t}=4(x-y)\qquad\mbox{ and }\qquad\frac{\mathrm{d}y}{\mathrm{d}t}=x-12(\mathrm{e}^{2t}+\mathrm{e}^{-2t}), \] obtain a differential equation for \(x\) which does not contain \(y\). Hence, or otherwise, find \(x\) and \(y\) in terms of \(t\) given that \(x=y=0\) when \(t=0\).
Solution: \begin{align*} && \frac{\d x}{\d t} &= 4(x-y) \\ && \frac{\d y}{\d t} &= x - 12(e^{2t}+e^{-2t}) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} &= 4 \frac{\d x}{\d t}-4\frac{\d y}{\d t} \\ &&&= 4 \frac{\d x}{\d t}-4 \left ( x - 12(e^{2t}+e^{-2t}) \right) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} - 4 \frac{\d x}{\d t}+4x &= 48 (e^{2t}+e^{-2t}) \end{align*} This differential equation has characteristic polynomial \(\lambda^2 - 4\lambda + 4 = (\lambda-2)^2\). Therefore we should expect a general solution of \((At+B)e^{2t}\). For particular integrals we should try \(ke^{-2t}\) and \(Ct^2 e^{2t}\). For the former, we have: \begin{align*} && 48 &= 4k+8k+k \\ \Rightarrow && k &= \frac{48}{13} \end{align*} For the latter we have: \begin{align*} &&4Ct^2e^{2t} -4C(2te^{2t}+2t^2e^{2t})+2C((1+2t)e^{2t}+2t^2e^{2t}) &= 48e^{2t} \\ \Rightarrow && 2C &= 48 \\ \Rightarrow && C &= 24 \end{align*} Therefore the solution should be: \begin{align*} x = (At+B)e^{2t} + \frac{48}{13}e^{-2t} + 24t^2 e^{2t} \\ x(0) = B + \frac{48}{13} \\ x'(0) = 2B+A-\frac{96}{13} \\ x =\frac{48}{13}((4t-1)e^{2t}+e^{-2t})+24t^2e^{2t} \\ y = x - \frac{1}{4} \frac{\d x}{\d t} \end{align*}