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

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

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

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

y = - cos (e^{x}) + \frac{sin ( e ^{x} )}{e ^{x}} + \frac{c _{1}}{e ^{x}}

Solution steps

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

Solve linear ODE:

y = - cos (e^{x}) + \frac{sin ( e ^{x} )}{e ^{x}} + \frac{c _{1}}{e ^{x}}

y = - cos (e^{x}) + \frac{sin ( e ^{x} )}{e ^{x}} + \frac{c _{1}}{e ^{x}}

y^{^{'}} - (y - 1)^{2} = 0

\int X d X

x \to 0 lim ((1 + \frac{1}{x})^{5 x})

\int_{6}^{8} \frac{14}{( x - 6 ) ^{3}} d x

x \to a lim ((\frac{sin ( x )}{sin ( a )})^{\frac{1}{x + a}})

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