What is the solution for (x-y^4+y^2sin(x))dx=(4xy^3+2ycos(x))dy ?

The solution for (x-y^4+y^2sin(x))dx=(4xy^3+2ycos(x))dy is

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(x-y^4+y^2sin(x))dx=(4xy^3+2ycos(x))dy

(x - y^{4} + y^{2} sin (x)) d x = (4 x y^{3} + 2 y cos (x)) d y

y = - \frac{cos ( x ) + 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - - \frac{cos ( x ) + 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - \frac{cos ( x ) - 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - - \frac{cos ( x ) - 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}

Solution steps

(x - y^{4} + y^{2} sin (x)) d x = (4 x y^{3} + 2 y cos (x)) d y

Solve exact ODE:

y = - \frac{cos ( x ) + 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - - \frac{cos ( x ) + 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - \frac{cos ( x ) - 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - - \frac{cos ( x ) - 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}

y = - \frac{cos ( x ) + 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - - \frac{cos ( x ) + 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - \frac{cos ( x ) - 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}, y = - - \frac{cos ( x ) - 2 x ( x ^{2} - c _{1} ) + cos ^{2} ( x )}{2 x}

y^{^{''}} + 4 y = 7 sin (2 x)

5 x y (\frac{d y}{d x}) = 2

\frac{3}{t ^{4}}

\frac{\partial}{\partial u} (uv - 1)

\frac{2 s}{( s - 1 ) ( s ^{2} + 25 )}

What is the solution for (x-y^4+y^2sin(x))dx=(4xy^3+2ycos(x))dy ?
The solution for (x-y^4+y^2sin(x))dx=(4xy^3+2ycos(x))dy is