What is the solution for y^'=e^{3y}sin(4x) ?

The solution for y^'=e^{3y}sin(4x) is y=-(ln(3/4 cos(4x)-c_{1}))/3

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y^'=e^{3y}sin(4x)

y^{^{'}} = e^{3 y} sin (4 x)

y = - \frac{ln ( \frac{3}{4} cos ( 4 x ) - c _{1} )}{3}

Solution steps

y^{'} (x) = e^{3 y} sin (4 x)

Solve separable ODE:

y = - \frac{ln ( \frac{3}{4} cos ( 4 x ) - c _{1} )}{3}

y = - \frac{ln ( \frac{3}{4} cos ( 4 x ) - c _{1} )}{3}

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f (x) = x^{3} - 5 x^{2}

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What is the solution for y^'=e^{3y}sin(4x) ?
The solution for y^'=e^{3y}sin(4x) is y=-(ln(3/4 cos(4x)-c_{1}))/3