IVP using system of equations: Undetermined Coefficients

Question

The given equations and values are $$x^{"}=-5x+2y\space(1)$$ $$y^{"}=2x-8y \space(2)$$ $$x(0)=10 \space,x^{'}(0)=13$$ $$y(0)=5 \space ,y^{'}(0)=-16$$ Rewriting $(1)$ and $(2)$ $$(D^2+5)x-2y=0 \space (1)$$
$$(D^2+8)y-2x=0\space (2)$$ Finding the roots is easy, operate $(1)$ by $(D^2+8)$ and multiplying $(2)$ by $2$ and adding the two equations yields: $$(D^4+13D^2+36)x=0$$ $$m^4+13m^2+36=0$$ $$(m^2+4)(m^2+9)=0$$ $$m_{1,2}=\pm2i$$ $$m_{3,4}=\pm3i$$ Then the Aux. equation is: $$x(t)=C_1\cos(2t)+C_2\sin(2t)+C_3\cos(3t)+C_4\sin(3t)$$ Now, because we are given initial values in terms of both $x$ and $y$ we can assume that we need $y(t)$, which is: $$y(t)=C_5\cos(2t)+C_6\sin(2t)+C_7\cos(3t)+C_8\sin(3t)$$ From the given we can tell we need to take the 2nd derivative of $x(t)$: $$x^{'}(t)=-4C_1\cos(2t)-4C_2\sin(2t)-9C_3\cos(3t)-9C_4\sin(3t)$$ Now from here we use $(1)$ to get rid of the excess constants: $$x^{"}+5x-2y=0$$ What I got was: $$\frac{1}{2}C_{1,2}=C_{5,6}$$ $$-2C_{3,4}=C_{7,8}$$ But when I plugged in for my IV's I got: $$y(0)=5: 5=\frac{1}{2}C_1-2C_3$$ and thats where I stopped because I think I messed something serious up. Am I wrong and my math is correct?

Answer 1

Your calculation is correct: $$10=C_1+C_3$$ $$13=2C_2+3C_4$$ Now we need two more equations from intital condition on $y$: $$x''+5x=2y$$ $$x''(0)=-5x(0)+2y(0)$$ $$x''(0)=-50+10=-40$$ $$x(t)=C_1\cos(2t)+C_2\sin(2t)+C_3\cos(3t)+C_4\sin(3t)$$ $$x''(0)=-4C_1-9C_3=-40$$ And a fourth equation: $$x'''(0)=-5x'(0)+2y'(0)$$ $$x'''(0)=-5*13+2*(-16)=-97$$ $$x'''(0)=-8C_2-27C_4=-97$$ You have four equations. You can solve the system for the $C_i$ and deduce $y(t)$ too. $$ \begin{align} 8C_2+27C_4=&97 \\ 4C_1+9C_3=&40 \\ C_1+C_3=&10 \\ 2C_2+3C_4=&13 \end{align} $$ For $y(t)$ you have: $$2y(t)=x''+5x$$ $$2y(t)=C_1\cos(2t)+C_2\sin(2t)-4C_3\cos(3t)-4C_4\sin(3t)$$ I find for the coefficients: $$(C_1,C_2,C_3,C_4)=(10,2,0,3)$$ Therefore we have : $$\boxed {y(t)=5\cos(2t)+\sin(2t)-6\sin(3t) \\ x(t)=10\cos(2t)+2\sin(2t)+3\sin(3t)} $$