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

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

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

y^{^{'}} = \frac{sin ( x ) e ^{x + 2 y}}{y}

y = - \frac{W ( \frac{2 e ^{x} ( s i n ( x ) - c o s ( x ) ) + c _{1}}{e} )}{2} - \frac{1}{2}

Solution steps

y^{'} (x) = \frac{sin ( x ) e ^{x + 2 y}}{y}

Solve separable ODE:

y = - \frac{W ( \frac{2 e ^{x} ( s i n ( x ) - c o s ( x ) ) + c _{1}}{e} )}{2} - \frac{1}{2}

y = - \frac{W ( \frac{2 e ^{x} ( s i n ( x ) - c o s ( x ) ) + c _{1}}{e} )}{2} - \frac{1}{2}

\frac{d N}{d t} = \frac{3}{2} N

\frac{d y}{d x} - \frac{y}{x} = - \frac{y ^{2}}{x}

x \frac{d y}{d x} = e^{6 y + x}

\frac{d t}{d x} = \frac{ln ( x )}{x ^{2} t}

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}

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