What is the solution for y^'+2y=4sin(e^{2x}) ?

The solution for y^'+2y=4sin(e^{2x}) is y=-(2cos(e^{2x}))/(e^{2x)}+(c_{1})/(e^{2x)}

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y^'+2y=4sin(e^{2x})

y^{^{'}} + 2 y = 4 sin (e^{2 x})

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

Solution steps

y^{'} (x) + 2 y = 4 sin (e^{2 x})

Solve linear ODE:

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

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

y^{^{'}} = \frac{( e ^{- x} - e ^{x} )}{( 3 + 4 y )}, y (0) = 1

(0, - 9), (- 1, - 4)

\frac{d y}{d x} = cos^{2} (x)

\frac{d}{d x} (2 x^{2} sin (x) tan (x))

\int \frac{8000 ln ( x + 40 )}{( x + 40 ) ^{2}} d x

What is the solution for y^'+2y=4sin(e^{2x}) ?
The solution for y^'+2y=4sin(e^{2x}) is y=-(2cos(e^{2x}))/(e^{2x)}+(c_{1})/(e^{2x)}