Science:Math Exam Resources/Courses/MATH215/December 2011/Question 04 (b)

MATH215 December 2011

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Question 04 (b)
Suppose the motion of a spring-mass system is described by the following differential equation $2u''+2\gamma u'+18u=0,\quad u(0)=1,u'(0)=-{\frac {11}{2}}$ (b) Solve the following equation for u(t) in explicit form. $u''+6u'+25u=15\cos(5t),\quad u(0)=1,u'(0)=-{\frac {11}{2}}$

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Hint
What are the particular and homogeneous solutions?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Notice that the ODE $\displaystyle {}u''+6u'+25u=15\cos(5t)$ has a nonzero right-hand side. We will begin by solving the homogeneous equation $\displaystyle {}u''+6u'+25u=0.$ By making the ansatz $u=\exp(rt)$ we arrive at $\displaystyle {}r^{2}+6r+25=0$ with roots $r=-3\pm 4i$. Complex roots of the form $a+ib$ lead to solutions of the form ${\begin{aligned}&e^{at}\sin(bt)\\&e^{at}\cos(bt).\end{aligned}}$ Therefore the homogeneous solution is $\displaystyle {}u_{h}=C_{1}e^{-3t}\sin(4t)+C_{2}e^{-3t}\cos(4t).$ The right-hand side is sinusoidal with argument $5t$. We need to think of which functions (and their derivatives) that could lead to this possible right-hand side. The only such possible functions are ${\begin{aligned}\cos(5t)\\\sin(5t)\end{aligned}}$ and so, we will find a particular solution of the form $\displaystyle {}u_{p}=A\sin(5t)+B\cos(5t).$ Taking this particular solution and plugging it in to the ODE, we have: ${\begin{aligned}(-25A\sin(5t)-25B\cos(5t))+6(5A\cos(5t)-5B\sin(5t))+25(A\sin(5t)+B\cos(5t))&=15\cos(5t)\\-30B\sin(5t)+30A\cos(5t)&=15\cos(5t)\end{aligned}}$ In comparing the coefficients of $\sin(5t)$ and $\cos(5t)$, we find that ${\begin{aligned}B&=0\\A&=1/2.\end{aligned}}$ The general solution to the ODE will be $u=u_{h}+u_{p}$, giving $u={\frac {1}{2}}\sin(5t)+C_{1}e^{-3t}\sin(4t)+C_{2}e^{-3t}\cos(4t).$ The initial conditions tell us ${\begin{aligned}u(0)&=1\\u'(0)&=-{\frac {11}{2}}\end{aligned}}$ From this, we have, $\displaystyle {}u(0)=C_{2}=1$ and $u'(0)=\left.\left({\frac {5}{2}}\cos(5t)-3C_{1}e^{-3t}\sin(4t)+4C_{1}e^{-3t}\cos(4t)-3C_{2}e^{-3t}\cos(4t)-4C_{2}e^{-3t}\sin(4t)\right)\right|_{t=0}={\frac {5}{2}}+4C_{1}-3={\frac {-11}{2}}.$ Therefore we get that $\displaystyle {}C_{1}=-5/4.$ The solution is thus $u(t)={\frac {1}{2}}\sin(5t)-{\frac {5}{4}}e^{-3t}\sin(4t)+e^{-3t}\cos(4t).$