What is the solution for (1+cos^2(x))y^'+2ycos(x)sin(x)=0 ?

The solution for (1+cos^2(x))y^'+2ycos(x)sin(x)=0 is y=e^{c_{1}}+e^{c_{1}}cos^2(x)

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(1+cos^2(x))y^'+2ycos(x)sin(x)=0

(1 + cos^{2} (x)) y^{^{'}} + 2 y cos (x) sin (x) = 0

y = e^{c_{1}} + e^{c_{1}} cos^{2} (x)

Solution steps

(1 + cos^{2} (x)) y^{'} (x) + 2 y cos (x) sin (x) = 0

Solve separable ODE:

y = e^{c_{1}} + e^{c_{1}} cos^{2} (x)

y = e^{c_{1}} + e^{c_{1}} cos^{2} (x)

y^{^{'}} = 8 x y^{2}

y^{^{'}} = 8 x y^{3}

\frac{d y}{d x} = 2 y

\frac{d y}{d x} = \frac{x sin ( x )}{y ^{2}}, y (0) = - 9

\frac{d y}{d x} 50 x^{\frac{1}{3}} y^{\frac{2}{3}} = 600

What is the solution for (1+cos^2(x))y^'+2ycos(x)sin(x)=0 ?
The solution for (1+cos^2(x))y^'+2ycos(x)sin(x)=0 is y=e^{c_{1}}+e^{c_{1}}cos^2(x)