What is the solution for y^'+3y=4sin(e^{3x}) ?

The solution for y^'+3y=4sin(e^{3x}) is y=-(4cos(e^{3x}))/(3e^{3x)}+(c_{1})/(e^{3x)}

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y^'+3y=4sin(e^{3x})

y^{^{'}} + 3 y = 4 sin (e^{3 x})

y = - \frac{4 cos ( e ^{3 x} )}{3 e ^{3 x}} + \frac{c _{1}}{e ^{3 x}}

Solution steps

y^{'} (x) + 3 y = 4 sin (e^{3 x})

Solve linear ODE:

y = - \frac{4 cos ( e ^{3 x} )}{3 e ^{3 x}} + \frac{c _{1}}{e ^{3 x}}

y = - \frac{4 cos ( e ^{3 x} )}{3 e ^{3 x}} + \frac{c _{1}}{e ^{3 x}}

\int \frac{13}{25 - 9 x ^{2}} d x

\frac{( 2 x ^{2} + 3 x )}{4 x ^{2}}

\frac{d}{d x} (a x sin (2 x) + b x cos (2 x))

\frac{d x}{d t} = - \frac{50}{π} x^{- \frac{3}{2}}

\int (sin (x) - 2 e^{x} + 3 x) d x

What is the solution for y^'+3y=4sin(e^{3x}) ?
The solution for y^'+3y=4sin(e^{3x}) is y=-(4cos(e^{3x}))/(3e^{3x)}+(c_{1})/(e^{3x)}