Science:Math Exam Resources/Courses/MATH253/December 2012/Question 01 (a)

MATH253 December 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  

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Question 01 (a)
Put your answer in the box provided and show your work. No credit will be given for the answer without the correct accompanying work. 1. Find the value of the constant k such that $\displaystyle u(k,t)=e^{kt}\cos(2x)$ is a solution of the heat equation $\displaystyle u_{t}=u_{xx}$.

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Hint
Find $\displaystyle u_{t}$ and $\displaystyle u_{xx}$.

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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. To determine the constant k such that the equation $\displaystyle u_{t}=u_{xx}$ holds, we take the appropriate partial derivatives of the solution $\displaystyle u(x,t)=e^{kt}\cos(2x)$: ${\begin{aligned}u_{t}(x,t)&=ke^{kt}\cos(2x)\\u_{x}(x,t)&=-2e^{kt}\sin(2x)\\u_{xx}(x,t)&=-4e^{kt}\cos(2x)\\\end{aligned}}$ We then substitute the $\displaystyle u_{t},\,u_{xx}$ terms into the diffusion equation: ${\begin{aligned}u_{t}&=u_{xx}\\ke^{kt}\cos(2x)&=-4e^{kt}\cos(2x)\end{aligned}}$ Cancelling out the $e^{kt}\cos(2x)$ on both sides of the equation gives us the final answer: ${\color {blue}k=-4}.$