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This is the question:

For which $A,\varphi$ does the function $A\sin(x+\varphi)$ solve the differential equation $(y')^2+y^2=1$?

I'm new to differential equations and I have no idea what this is supposed to mean. How do I use the function $A\sin(x+\varphi)$ to solve the differential equation? I can not find any example in my textbook that looks anything like the problem I have to solve here. Thanks.

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We replace $y$ by $A\sin(x+\varphi)$ in the differential equation and find the appropriate values of $A$ and $\varphi$.

Since $y=A\sin(x+\varphi)$ then $y'=A\cos(x+\varphi)$ and then $$(y')^2+y^2=1\iff A^2\cos^2(x+\varphi)+A^2\sin^2(x+\varphi)=A^2=1\iff A=\pm1$$ hence $$y=\pm\sin(x+\varphi),\ \forall \varphi\in\mathbb R$$ is a solution for the differential equation.

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  • $\begingroup$ Thank you! I didn't realize it was that simple. $\endgroup$ – user91971 Aug 28 '13 at 0:26
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Here is a start

$$ y(x)= A\sin(x+\phi) \implies y'(x)= A\cos(x+\phi)$$

Substitute back in the ode and you will find the answer.

Note:

$$ \sin^2(t)+\cos^2(t)=1 $$

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