# Find the solution of Initial value problem in explicit form

Problem:$$sin(2x)dx+cos(6y)dy = 0$$ Initial value: $$y(\frac{π}{2})=\frac{π}{6}$$

Integrate both sides: $$\frac16sin(6x)= \frac12cos(2x)+c$$ $$sin(6y)= 3cos(2x)+c$$ Use initial condition to find: c=3 $$sin(6y)=3cos(2x)+3$$ Use trig IDs $$sin(6y)=6cos^2(x)$$ Solve for y $$y=\frac16sin^{-1}(6cos^2(x))$$ Is this a valid solution?

• Does it satisfy the differential equation? Does it go through the point you need it to? – qbert Sep 16 '18 at 2:52
• I dont know could you help – Doldrums Sep 16 '18 at 2:57
• Plug the solution you found into the equation to see if it satisfies. Can you do that? – Dylan Sep 16 '18 at 6:34

Given $$\sin2x\ dx+\cos 6y\ dy=0,\ \ y\left(\dfrac{\pi}{2}\right)=\dfrac{\pi}{6}$$
$$\cos6y\dfrac{dy}{dx}=-\sin2x$$ $$\dfrac{d}{dx}\left(\dfrac16\sin6y\right)=\dfrac{d}{dx}(\cos^2x)$$Integrate both sides with respect to $x$ $$\dfrac16\sin6y=\cos^2x+C$$$$\sin6y=6\cos^2x+C$$ We just obtained an implicit formula for the general solution. To determine $C$ set $x=\dfrac{\pi}{2}$ and $y=\dfrac{\pi}{6}$$\sin\pi=6\cos^2\left(\dfrac{\pi}{2}\right)+C$$$$0=0+C$$$$C+0$$ Hence the solution of the initial value problem is given implicitly by $$\sin6y=6\cos^2x$$To solve for y we must exercise a little caution. The answer$y(x)=\dfrac16\arcsin(6\cos^2x)$is wrong because then$y(\pi/2)=0$, not$\pi/6$To match$y(\pi/2)=\pi/6$we must choose$y=\dfrac{\pi}{6}-\dfrac16\arcsin(6\cos^2x)\$.