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

MATH101 April 2005

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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).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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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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. 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 $\displaystyle {}x(t)=-2\cos(3t)+4\sin(3t)-3e^{-t}+5e^{-3t}$

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