2006 Exercises

Exercice. Find a solution of $x''+3x'+2x=te^{-t}$.
Solution. We use the method of variation of parameters.
Suppose there is a solution of the form $x(t) =u(t)e^{-t}$ for some function $u(t)$. $$x'=(u'-u)e^{-t}$$ $$x''=(u''-2u'+u)e^{-t}$$ and the equation becomes $$\begin{aligned} x''+3x'+x&=[(u''-2u'+u)+3(u'-u)+2u]e^{-t}\\ &=(u''+u')e^{-t}\\ &=te^{-t} \end{aligned}$$ The exponential is never zero, so can cancel $e^{-t}$ from both sides and obtain $$u''+u'=t$$ Lowest order derivative is 1, so we should bump up all degrees by 1. So try $u=At^2+Bt+C$.
$$u'=2At+B$$ $$u''=2A$$ $$u''+u'=2At+(2A+B)=t$$ $$A=1/2\text{ and }B=-1$$ so $$u=\frac{1}{2}t^2-t$$ $$x=(\frac{t^2}{2}-t)e^{-t}$$

Exercice.
(a) Find a solution of $x''+x=5te^{2t}$.
(b) Suppose that $y(t)$ is a solution of the same equation, $x''+x=5te^{2t}$, such that $y(0) = 1$ and $y'(0)=2$. (This is probably not the solution you found in (a).) Use $y(t)$ and other functions to write down a solution $x(t)$ such that $x(0) = 3$ and $x'(0) = 5$.
Solution.
(a) Variation of parameters: $$x=ue^{2t}$$ $$x'=(u'+2u)e^{2t}$$ $$x''=2(u'+2u)e^{2t}+(u''+2u')e^{2t}=(u''+4u'+2u)e^{2t}$$ so $$x''+x=(u''+4u'+5u)e^{2t}$$ and $u$ must satisfy $$u''+4u'+5u=5t$$ Undetermined coefficients: $$u_p=at+b$$ $$u_p'=a$$ $$u_p''=0$$ so $$4a+5(at+b)=5t$$ $$a=1,b=-4/5$$ $$u_p=t-4/5$$ $$x_p=(t-\frac{4}{5})e^{2t}$$ (b) The homogeneous equation has general solution $a\cos t+b\sin t$, so the general solution of $x''+x=5te^{2t}$ is $x = y + a\cos t+b\sin t$. $$x(0)=3=y(0)+a=1+a\rArr a=2$$ $$x'(0)=5=y'(0)+b=2+b\rArr b=3$$ so $$x=y+2\cos t+3\sin t$$