Find the general solution of the differential equation whose…

Find the general solution of the differential equation whose solution is given by $cy=\sin(cx)$.

Step 1: Eliminate $c$

Differentiating $cy=\sin(cx)$ gives $y'=\cos(cx)$, and squaring both relations with $\sin^2+\cos^2=1$ gives $(y')^2+c^2y^2=1$. Differentiating this again and simplifying (using $c^2=(1-(y')^2)/y^2$) eliminates $c$ entirely:

\[\boxed{\,yy''=(y')^2-1.}\]

Step 2: Reduce the order

With $p=y'$ and $y''=p\,\dfrac{dp}{dy}$, the ODE becomes $yp\,\dfrac{dp}{dy}=p^2-1$, which separates as $\dfrac{p\,dp}{p^2-1}=\dfrac{dy}{y}$. Integrating gives $p^2-1=\pm Ky^2$, i.e. $(y')^2=1-c^2y^2$ (taking $c^2=K$).

Step 3: Solve the first-order equation

Separating again, $\dfrac{dy}{\sqrt{1-c^2y^2}}=\pm dx$, so $\dfrac1c\arcsin(cy)=\pm x+C_2$. Taking the sine of both sides gives $cy=\sin(cx+\phi)$; setting $\phi=0$ recovers the original form.