Question

Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

Answer 1

`y=(-3\pm√(4sin2x+1))/(2)`

`x={\pi}{4}`

Explanation:

We are given that

`y'=(2cos2x)/(3+2y)`

y(0)=-1

`(dy)/(dx)=(2cos2x)/(3+2y)`

`(3+2y)dy=2cos2x dx`

Taking integration on both sides then we get

`\int (3+2y)dy=2\int cos 2xdx`

`3y+y^2=sin2x+C`

Using formula

`\int x^n=(x^(n+1))/(n+1)+C`

`\int cosx dx=sinx`

Substitute x=0 and y=-1

`-3+1=sin0+C`

`-2=C`

`sin0=0`

Substitute the value of C

`y^2+3y=sin2x-2`

`y^2+3y-sin 2x+2=0`

`y=(-3\pm√((3)^2-4(1)(-sin2x+2)))/(2)`

By using quadratic formula

`x=(-b\pm√(b^2-4ac))/(2a)`

`y=(-3\pm√(9+4sin2x-8))/(2)=(-3\pm√(4sin2x+1))/(2)`

Hence, the solution
`y=(-3\pm√(4sin2x+1))/(2)`

When the solution is maximum then y'=0

`(2cos2x)/(3+2y)=0`

`2cos2x=0`

`cos2x=0`

`cos2x=cos(\pi)/(2)`

`cos(\pi)/(2)=0`

`2x=(\pi)/(2)`

`x=(\pi)/(4)`