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

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

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y^'+y=5sin(2x)

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (sin (n x) e^{- n^{2} t})

x \to \infty lim (1 - e^{x})

x \to - 1 lim (x^{2} + 4)

\int \frac{1}{5 + 2 x} d x

\int x - 9 cos (x) d x

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