What is the solution for y^'+5y=2sin(e^{5x}) ?

The solution for y^'+5y=2sin(e^{5x}) is y=-(2cos(e^{5x}))/(5e^{5x)}+(c_{1})/(e^{5x)}

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y^'+5y=2sin(e^{5x})

y^{^{'}} + 5 y = 2 sin (e^{5 x})

y = - \frac{2 cos ( e ^{5 x} )}{5 e ^{5 x}} + \frac{c _{1}}{e ^{5 x}}

Solution steps

y^{'} (x) + 5 y = 2 sin (e^{5 x})

Solve linear ODE:

y = - \frac{2 cos ( e ^{5 x} )}{5 e ^{5 x}} + \frac{c _{1}}{e ^{5 x}}

y = - \frac{2 cos ( e ^{5 x} )}{5 e ^{5 x}} + \frac{c _{1}}{e ^{5 x}}

\int_{2}^{6} t^{3} ln (5 x) d x

\frac{1}{( s ^{2} - 1 )}

\frac{\partial}{\partial x} (7 + x ln (x y - 5))

\frac{d}{d y} (2 y)

\int_{0}^{x} y lo g_{10} (y) d y

What is the solution for y^'+5y=2sin(e^{5x}) ?
The solution for y^'+5y=2sin(e^{5x}) is y=-(2cos(e^{5x}))/(5e^{5x)}+(c_{1})/(e^{5x)}