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

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MATH215 December 2011

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• April 2013 • December 2011 • December 2013 •

Question 04 (c)
Suppose the motion of a spring-mass system is described by the following differential equation $2u''+2\gamma u'+18u=0,\quad u(0)=1,\quad u'(0)=-{\frac {11}{2}}$ (c) Find the general solution of $t^{2}y''-4ty'+6y=3t,\quad (t>0)$ provided that y₁(t) = t² and y₂(t) = t³ are two linearly independent solutions of the homogeneous equation $\displaystyle t^{2}y''-4ty'+6y=0$

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
Try something along the lines of $y=f(t)t^{2}+g(t)t^{3}$ . This is known as variation of parameters.

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Solution
We are given the solution to the homogeneous equation. To find the solution with the nonzero right-hand side, we will try variation of parameters. Given $t^{2}$ and $t^{3}$ satisfy the homogeneous ODE, we will make an ansatz $\displaystyle {}y=f(t)t^{2}+g(t)t^{3}$ where f(t) and g(t) are unknown functions. If we differentiate this we get $\displaystyle {}y'=f'(t)t^{2}+2f(t)t+g'(t)t^{3}+3g(t)t^{2}$ and to avoid having more than a single derivative upon f and g in computing $y''$ , we will set $\displaystyle {}f'(t)t^{2}+g'(t)t^{3}=0$ so that $\displaystyle {}y'=2f(t)t+3g(t)t^{2}.$ Notice that by making the assumption, the original first derivative remains unchanged. It is important that we make this step, otherwise when we try to solve the ODE we will have one equation for two unknowns. This assumption adds extra information that allows us to solve the problem. Differentiating again, we get $\displaystyle {}y''=2f'(t)t+2f(t)+3g'(t)t^{2}+6g(t)t.$ Now we take our derivatives and substitute them into the original equation: $\displaystyle {}t^{2}(2f'(t)t+2f(t)+3g'(t)t^{2}+6g(t)t)-4t(2f(t)t+3g(t)t^{2})+6(f(t)t^{2}+g(t)t^{3})=3t.$ After expanding and simplifying, we get $\displaystyle {}2t^{3}f'(t)+3t^{4}g'(t)=3t.$ Now we have two first-order ODEs for f and g: ${\begin{aligned}t^{2}f'+t^{3}g'&=0\\2t^{3}f'+3t^{4}g'&=3t.\end{aligned}}$ The first equation implies $\displaystyle {}f'=-tg'$ which can be substituted into the second equation: ${\begin{aligned}\displaystyle {}-2t^{4}g'+3t^{4}g'&=3t\\t^{4}g'&=3t\\g'&={\frac {3}{t^{3}}}\\g&={\frac {-3}{2t^{2}}}+C_{1}.\end{aligned}}$ Also, from above, ${\begin{aligned}f'&=-tg'={\frac {-3}{t^{2}}}\\f&={\frac {3}{t}}+C_{2}.\end{aligned}}$ Having solve for f(t) and g(t) we get that the general solution to our problem is $y=f(t)t^{2}+g(t)t^{3}=C_{1}t^{3}+C_{2}t^{2}+{\frac {3}{2}}t.$