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}}

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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)