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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int + (1 + x)^{3} d x

\frac{\partial}{\partial x} (x e^{x} sin (y))

\int \frac{x ^{4}}{1 - x ^{2}} d x

(x^{3} - 12 x) (3 x^{2} + 2 x)

\frac{d}{d x} ((2^{x} \cdot x^{- 2})^{3 x - 1})

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