What is the solution for xy^'=4x^4cos(2x)+3y ?

The solution for xy^'=4x^4cos(2x)+3y is y=2sin(2x)x^3+c_{1}x^3

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xy^'=4x^4cos(2x)+3y

x y^{^{'}} = 4 x^{4} cos (2 x) + 3 y

y = 2 sin (2 x) x^{3} + c_{1} x^{3}

Solution steps

x y^{'} (x) = 4 x^{4} cos (2 x) + 3 y

Solve linear ODE:

y = 2 sin (2 x) x^{3} + c_{1} x^{3}

y = 2 sin (2 x) x^{3} + c_{1} x^{3}

x \to 0 lim (x sin (\frac{π}{x}))

y = 5 x - x^{2}, y = 2 x, (3, 6)

\frac{\partial}{\partial x} (4 y - 3 x^{2})

f (x) = x^{3} - 3 x + 2 at (2.4)

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

What is the solution for xy^'=4x^4cos(2x)+3y ?
The solution for xy^'=4x^4cos(2x)+3y is y=2sin(2x)x^3+c_{1}x^3