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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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