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

MATH215 December 2011

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

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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 $\displaystyle t^{2}y''-4ty'+6y=0$

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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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $y=f(t)t^{2}+g(t)t^{3}=C_{1}t^{3}+C_{2}t^{2}+{\frac {3}{2}}t.$