Answer:
[tex]y=\frac{-3\pm\sqrt{4sin2x+1}}{2}[/tex]
[tex]x={\pi}{4}[/tex]
Step-by-step explanation:
We are given that
[tex]y'=\frac{2cos2x}{3+2y}[/tex]
y(0)=-1
[tex]\frac{dy}{dx}=\frac{2cos2x}{3+2y}[/tex]
[tex](3+2y)dy=2cos2x dx[/tex]
Taking integration on both sides then we get
[tex]\int (3+2y)dy=2\int cos 2xdx[/tex]
[tex]3y+y^2=sin2x+C[/tex]
Using formula
[tex]\int x^n=\frac{x^{n+1}}{n+1}+C[/tex]
[tex]\int cosx dx=sinx[/tex]
Substitute x=0 and y=-1
[tex]-3+1=sin0+C[/tex]
[tex]-2=C[/tex]
[tex]sin0=0[/tex]
Substitute the value of C
[tex]y^2+3y=sin2x-2[/tex]
[tex]y^2+3y-sin 2x+2=0[/tex]
[tex]y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}[/tex]
By using quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}[/tex]
Hence, the solution [tex]y=\frac{-3\pm\sqrt{4sin2x+1}}{2}[/tex]
When the solution is maximum then y'=0
[tex]\frac{2cos2x}{3+2y}=0[/tex]
[tex]2cos2x=0[/tex]
[tex]cos2x=0[/tex]
[tex]cos2x=cos\frac{\pi}{2}[/tex]
[tex]cos\frac{\pi}{2}=0[/tex]
[tex]2x=\frac{\pi}{2}[/tex]
[tex]x=\frac{\pi}{4}[/tex]
