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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.

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    $\begingroup$ 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. $\endgroup$ – Mindlack Apr 4 at 13:37
  • $\begingroup$ 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? $\endgroup$ – Maths Wizzard Apr 4 at 13:48
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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.

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