What is the solution for y^'+y=5cos(2t),y(0)=3 ?

The solution for y^'+y=5cos(2t),y(0)=3 is y=2sin(2t)+cos(2t)+2/(e^t)

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y^'+y=5cos(2t),y(0)=3

y^{^{'}} + y = 5 cos (2 t), y (0) = 3

y = 2 sin (2 t) + cos (2 t) + \frac{2}{e ^{t}}

Solution steps

y^{'} (t) + y = 5 cos (2 t)

Solve linear ODE:

y = 2 sin (2 t) + cos (2 t) + \frac{2}{e ^{t}}

y = 2 sin (2 t) + cos (2 t) + \frac{2}{e ^{t}}

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

\int \frac{1}{1 - x ^{2} - 4 x} d x

0.4241 (y - 25) + 70 x = - 10845.81 \cdot \frac{d y}{d x}

\int \frac{3 x ^{3} - 2 x ^{2} + x - 5}{x ^{2}} d x

f (x) = \frac{e ^{x}}{x ^{2}}, at x = 1

What is the solution for y^'+y=5cos(2t),y(0)=3 ?
The solution for y^'+y=5cos(2t),y(0)=3 is y=2sin(2t)+cos(2t)+2/(e^t)