Answer:
![y=\frac{-3\pm\sqrt{4sin2x+1}}{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B-3%5Cpm%5Csqrt%7B4sin2x%2B1%7D%7D%7B2%7D)
![x={\pi}{4}](https://tex.z-dn.net/?f=x%3D%7B%5Cpi%7D%7B4%7D)
Step-by-step explanation:
We are given that
![y'=\frac{2cos2x}{3+2y}](https://tex.z-dn.net/?f=y%27%3D%5Cfrac%7B2cos2x%7D%7B3%2B2y%7D)
y(0)=-1
![\frac{dy}{dx}=\frac{2cos2x}{3+2y}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B2cos2x%7D%7B3%2B2y%7D)
![(3+2y)dy=2cos2x dx](https://tex.z-dn.net/?f=%283%2B2y%29dy%3D2cos2x%20dx)
Taking integration on both sides then we get
![\int (3+2y)dy=2\int cos 2xdx](https://tex.z-dn.net/?f=%5Cint%20%283%2B2y%29dy%3D2%5Cint%20cos%202xdx)
![3y+y^2=sin2x+C](https://tex.z-dn.net/?f=3y%2By%5E2%3Dsin2x%2BC)
Using formula
![\int x^n=\frac{x^{n+1}}{n+1}+C](https://tex.z-dn.net/?f=%5Cint%20x%5En%3D%5Cfrac%7Bx%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%2BC)
![\int cosx dx=sinx](https://tex.z-dn.net/?f=%5Cint%20cosx%20dx%3Dsinx)
Substitute x=0 and y=-1
![-3+1=sin0+C](https://tex.z-dn.net/?f=-3%2B1%3Dsin0%2BC)
![-2=C](https://tex.z-dn.net/?f=-2%3DC)
![sin0=0](https://tex.z-dn.net/?f=sin0%3D0)
Substitute the value of C
![y^2+3y=sin2x-2](https://tex.z-dn.net/?f=y%5E2%2B3y%3Dsin2x-2)
![y^2+3y-sin 2x+2=0](https://tex.z-dn.net/?f=y%5E2%2B3y-sin%202x%2B2%3D0)
![y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B-3%5Cpm%5Csqrt%7B%283%29%5E2-4%281%29%28-sin2x%2B2%29%7D%7D%7B2%7D)
By using quadratic formula
![x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B-3%5Cpm%5Csqrt%7B9%2B4sin2x-8%7D%7D%7B2%7D%3D%5Cfrac%7B-3%5Cpm%5Csqrt%7B4sin2x%2B1%7D%7D%7B2%7D)
Hence, the solution ![y=\frac{-3\pm\sqrt{4sin2x+1}}{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B-3%5Cpm%5Csqrt%7B4sin2x%2B1%7D%7D%7B2%7D)
When the solution is maximum then y'=0
![\frac{2cos2x}{3+2y}=0](https://tex.z-dn.net/?f=%5Cfrac%7B2cos2x%7D%7B3%2B2y%7D%3D0)
![2cos2x=0](https://tex.z-dn.net/?f=2cos2x%3D0)
![cos2x=0](https://tex.z-dn.net/?f=cos2x%3D0)
![cos2x=cos\frac{\pi}{2}](https://tex.z-dn.net/?f=cos2x%3Dcos%5Cfrac%7B%5Cpi%7D%7B2%7D)
![cos\frac{\pi}{2}=0](https://tex.z-dn.net/?f=cos%5Cfrac%7B%5Cpi%7D%7B2%7D%3D0)
![2x=\frac{\pi}{2}](https://tex.z-dn.net/?f=2x%3D%5Cfrac%7B%5Cpi%7D%7B2%7D)
![x=\frac{\pi}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Cpi%7D%7B4%7D)
The answer to
This is rational
Answer:
2x+2y
Step-by-step explanation:
Answer: This is because you want to find a quad root of it and the nunber would be 4/3 to do that
Step-by-step explanation: I am very busy