Science:Math Exam Resources/Courses/MATH101/April 2005/Question 04

MATH101 April 2005

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[hide] Question 04
Full-Solution Problem. Justify your answer and show all your work. Simplification of your answer is not required. A mass attached to a spring satisfies the differential equation $\displaystyle x''+4x'+3x=60\cos(3t)$ , where x = x(t) is the position of the mass at time t. Find x(t), given that x(0) = x'(0) = 0 (i.e. solve the initial-value problem).

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[show] Hint
Remember that the solution to the initial value problem will be the sum of a particular solution and a homogeneous solution.

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[show] Solution
Since this is a non-homogeneous problem, we must find a particular solution and a homogeneous solution. Let $x_{p}$ and $x_{h}$ be the particular and homogeneous solutions, resp. Particular solution We will first find the particular solution $x_{p}$ by assuming a oscillatory solution of the same frequency as the right hand side of the differential equation: $\,x_{p}(t)=A\cos(3t)+B\sin(3t)$ where $A,B$ are constants to be determined. ${\begin{aligned}x''_{p}+4x'_{p}+3x_{p}&=60\cos(3t)\\-9A\cos(3t)-9B\sin(3t)&\\-12A\sin(3t)+12B\cos(3t)&\\+3A\cos(3t)+3B\sin(3t)&=60\cos(3t)\end{aligned}}$ Separating the parts of the equation that are multiplied by $\sin(3t)$ and $\cos(3t)$ , we find that we must solve the following equations for $A,B$ : $\left.{\begin{array}{c}-9A+12B+3A=60\\-9B-12A+3B=0\end{array}}\right\}\quad \rightarrow \quad A=-2,\,\,B=4.$ Thus the particular solution is: $\displaystyle {}x_{p}(t)=-2\cos(3t)+4\sin(3t)$ Homogeneous solution Now will find the solution to the homogeneous equation, $x_{h}(t)$ , which satisfies: $\displaystyle {}x''_{h}+4x'_{h}+3x_{h}=0$ As is standard in homogeneous constant coefficient problems, we substitute $x_{h}(t)=e^{mt}$ into the above equation and get $m^{2}e^{mt}+4me^{mt}+3e^{mt}=e^{mt}\left(m^{2}+4m+3\right)=0$ Factoring and solving for $m$ gives $(m+3)(m+1)=0\quad \rightarrow \quad m=-1,-3$ Thus, the homogeneous solution is given by $\left.\right.x_{h}(t)=c_{1}e^{-t}+c_{2}e^{-3t}$ where the constants $c_{1}$ , $c_{2}$ will be chosen to satisfy the initial condition. Initial conditions Putting both the particular and homogeneous solution together we obtain the full solution: ${\begin{aligned}x(t)&=x_{p}(t)+x_{h}(t)\\&=-2\cos(3t)+4\sin(3t)+c_{1}e^{-t}+c_{2}e^{-3t}\end{aligned}}$ Now, apply the initial conditions: $x(0)=x'(0)=0$ . $\left.{\begin{array}{c}x(0)=-2+c_{1}+c_{2}=0\\x'(0)=12-c_{1}-3c_{2}=0\end{array}}\right\}\quad \rightarrow \quad c_{1}=-3,\,\,c_{2}=5$ Final answer Thus, the solution to the initial value problem is $\displaystyle {}x(t)=-2\cos(3t)+4\sin(3t)-3e^{-t}+5e^{-3t}$