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

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

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

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

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

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}

y^{^{'}} = - 2 - y

x cos (x) = (2 y + e^{3 y}) y^{^{'}} (x), y (0) = 0

\frac{d y}{d x} + 9 x y = 4 x, y (0) = 5

\frac{d y}{d x} y = (sin^{2} (π - x))

\frac{d P}{d t} = 0.0017 P (1 - \frac{P}{6200})

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