What is the solution for y^'=(cos^2(x))/(y^2),y(pi)=-1 ?

The solution for y^'=(cos^2(x))/(y^2),y(pi)=-1 is y=\sqrt[3]{(3(2x+sin(2x))-4-6pi)/4}

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y^'=(cos^2(x))/(y^2),y(pi)=-1

y^{^{'}} = \frac{cos ^{2} ( x )}{y ^{2}}, y (π) = - 1

y = 3 \frac{3 ( 2 x + sin ( 2 x ) ) - 4 - 6 π}{4}

Solution steps

y^{'} (x) = \frac{cos ^{2} ( x )}{y ^{2}}

Solve separable ODE:

y = 3 \frac{3 ( 2 x + sin ( 2 x ) ) - 4 - 6 π}{4}

y = 3 \frac{3 ( 2 x + sin ( 2 x ) ) - 4 - 6 π}{4}

y \frac{d y}{d x} = 4 x (y^{2} + 7)^{9}

\frac{d y}{d x} = y - 10

3 x + x y = \frac{d y}{d x}

5 y \frac{d y}{d x} = 9 x^{2}

\frac{d y}{d x} y = 3 x + 7

What is the solution for y^'=(cos^2(x))/(y^2),y(pi)=-1 ?
The solution for y^'=(cos^2(x))/(y^2),y(pi)=-1 is y=\sqrt[3]{(3(2x+sin(2x))-4-6pi)/4}