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

The solution for y^'=sin(x)y+e^x is y=(integral of e^{x+cos(x)})/(e^{cos(x))}

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y^'=sin(x)y+e^x

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

y = \frac{\int e ^{x + c o s (x)} d x}{e ^{c o s (x)}}

Solution steps

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

Solve linear ODE:

y = \frac{\int e ^{x + c o s (x)} d x}{e ^{c o s (x)}}

y = \frac{\int e ^{x + c o s (x)} d x}{e ^{c o s (x)}}

\frac{\partial}{\partial θ} (\frac{cos ( θ )}{r ( θ )})

x \to 2 lim (\frac{0}{( x - 2 ) ^{2}})

\int 4 e^{6} d x

x \to 3 lim (x - x + 1)

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

What is the solution for y^'=sin(x)y+e^x ?
The solution for y^'=sin(x)y+e^x is y=(integral of e^{x+cos(x)})/(e^{cos(x))}