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vredina [299]
3 years ago
5

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

Mathematics
1 answer:
zloy xaker [14]3 years ago
3 0

Answer:

y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

x={\pi}{4}

Step-by-step explanation:

We are given that

y'=\frac{2cos2x}{3+2y}

y(0)=-1

\frac{dy}{dx}=\frac{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=\frac{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=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}

By using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}

Hence, the solution y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

When the solution is maximum then y'=0

\frac{2cos2x}{3+2y}=0

2cos2x=0

cos2x=0

cos2x=cos\frac{\pi}{2}

cos\frac{\pi}{2}=0

2x=\frac{\pi}{2}

x=\frac{\pi}{4}

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