What is the solution for y^'+4y=5sin(e^{4x}) ?

The solution for y^'+4y=5sin(e^{4x}) is y=-(5cos(e^{4x}))/(4e^{4x)}+(c_{1})/(e^{4x)}

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y^'+4y=5sin(e^{4x})

y^{^{'}} + 4 y = 5 sin (e^{4 x})

y = - \frac{5 cos ( e ^{4 x} )}{4 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

Solution steps

y^{'} (x) + 4 y = 5 sin (e^{4 x})

Solve linear ODE:

y = - \frac{5 cos ( e ^{4 x} )}{4 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

y = - \frac{5 cos ( e ^{4 x} )}{4 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

\int_{1}^{e} 7 t^{2} ln (t) d t

x^{2} + x - 3

\frac{\partial}{\partial y} (3 x^{2} y^{4} - 3)

f (x) = 4 x^{3}

\frac{d}{d x} (ln (1 + x y))

What is the solution for y^'+4y=5sin(e^{4x}) ?
The solution for y^'+4y=5sin(e^{4x}) is y=-(5cos(e^{4x}))/(4e^{4x)}+(c_{1})/(e^{4x)}