What is the solution for y^'=y^2(x)cos(x) ?

The solution for y^'=y^2(x)cos(x) is y=-1/(xsin(x)+cos(x)+c_{1)}

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y^'=y^2(x)cos(x)

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

y = - \frac{1}{x sin ( x ) + cos ( x ) + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{x sin ( x ) + cos ( x ) + c _{1}}

y = - \frac{1}{x sin ( x ) + cos ( x ) + c _{1}}

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

(\frac{d y}{d x}) - y x = 0

\frac{d y}{d x} = 70 y x^{6}, y = 6

\frac{d v}{d t} = v \cdot t^{3} - 1.5 \cdot v, v (0) = 1

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

What is the solution for y^'=y^2(x)cos(x) ?
The solution for y^'=y^2(x)cos(x) is y=-1/(xsin(x)+cos(x)+c_{1)}