# STEP 3 1987 Q6: Simultanious Differential Equations

I have been doing some of the old STEP papers because they seem to be more challenging and I stumbled across a problem I am not too sure about as I am not very experienced with second-order differential equations. The question is as follows, solve the simultaneous differential equations $$\frac{dy}{dt}+2x-5y=0$$ and $$\frac{dx}{dt} + x -2y=2\cos t$$. I solved for $$x$$ in the first equation afterwards I differentiated it and plugged it back into the second equation to get $$\frac{d^2y}{dx^2}+y=4\cos t - 2\sin t$$. I can get a solution for the complementary function, however, when I try to get a solution for the particular integral I can't seem to get one. I first tried substituting $$y=p\cos t +q\sin t$$ which did not work (as there were similar terms in the complementary solution). Then I tried $$y=tp\cos t +tq\sin t$$ which resulted in the following equation $$(-2p + q)\sin t+(2q+p)\cos t -t(p\cos t- q\sin t)=4\cos t-2\sin t$$. From here I don't know how to proceed. I thought of matching coefficients but this means that $$p\cos t - q\sin t=0$$ and that can't be true for all $$t$$ and constant values of $$p$$ and $$q$$. Any help would be greatly appreciated! Thanks in advance.

• You should double-check your calculations with $y=t(p\cos{t}+q\sin{t})$. When computing $y’’+y$: the term with two derivatives of $t$ vanish; the term with no derivative of $t$ is compensated by $y$, so only $2(-p\sin{t}+q\cos{t})$ remains to be equated to the RHS. – Mindlack Apr 4 at 13:37
• Thank you very much, I just realised my mistake. Could you post it as an answer so I can mark it as answered whilst giving you some virtual points? – Maths Wizzard Apr 4 at 13:48

You should double-check your calculations with $$y=t(p\cos{t}+q\sin{t})$$. When computing $$y’’+y$$: the term with two derivatives of $$t$$ vanishes; the term with no derivative of $$t$$ is compensated by $$y$$, so only $$2(-p\sin{t}+q\cos{t})$$ remains to be equated to the RHS.