What is the solution for y^'+y+4sin(x)=0 ?

The solution for y^'+y+4sin(x)=0 is y=-2(sin(x)-cos(x))+(c_{1})/(e^x)

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y^'+y+4sin(x)=0

y^{^{'}} + y + 4 sin (x) = 0

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

Solution steps

y^{'} (x) + y + 4 sin (x) = 0

Solve linear ODE:

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

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

\int \frac{31}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\int tan (6 x + 1) d x

\int \frac{8 x ^{2} - 7 x + 12}{x ^{3} + 4 x} d x

\frac{d}{d x} ((- 2 x + 3 x^{- 2}) ln (x))

\frac{d}{d x} (\frac{x ^{7}}{x ^{4}})

What is the solution for y^'+y+4sin(x)=0 ?
The solution for y^'+y+4sin(x)=0 is y=-2(sin(x)-cos(x))+(c_{1})/(e^x)