What is the solution for y^'=e^{2y}sin(3x) ?

The solution for y^'=e^{2y}sin(3x) is y=-(ln(2/3 cos(3x)-c_{1}))/2

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y^'=e^{2y}sin(3x)

y^{^{'}} = e^{2 y} sin (3 x)

y = - \frac{ln ( \frac{2}{3} cos ( 3 x ) - c _{1} )}{2}

Solution steps

y^{'} (x) = e^{2 y} sin (3 x)

Solve separable ODE:

y = - \frac{ln ( \frac{2}{3} cos ( 3 x ) - c _{1} )}{2}

y = - \frac{ln ( \frac{2}{3} cos ( 3 x ) - c _{1} )}{2}

t a y l or sin (x), \frac{π}{4}

y^{^{''}} + 6 y^{^{'}} + 8 y = 4 sin (2 t)

\int \frac{x ^{2} + 1}{( x - 7 ) ( x - 6 ) ^{2}} d x

y^{^{'}} = - 0.5 y + x

\int \frac{y}{2} d y

What is the solution for y^'=e^{2y}sin(3x) ?
The solution for y^'=e^{2y}sin(3x) is y=-(ln(2/3 cos(3x)-c_{1}))/2