What is the solution for y^'+1/2 y=17sin(2t) ?

The solution for y^'+1/2 y=17sin(2t) is y=2(sin(2t)-4cos(2t))+(c_{1})/(e^{t/2)}

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y^'+1/2 y=17sin(2t)

y^{^{'}} + \frac{1}{2} y = 17 sin (2 t)

y = 2 (sin (2 t) - 4 cos (2 t)) + \frac{c _{1}}{e ^{\frac{t}{2}}}

Solution steps

y^{'} (t) + \frac{1}{2} y = 17 sin (2 t)

Solve linear ODE:

y = 2 (sin (2 t) - 4 cos (2 t)) + \frac{c _{1}}{e ^{\frac{t}{2}}}

y = 2 (sin (2 t) - 4 cos (2 t)) + \frac{c _{1}}{e ^{\frac{t}{2}}}

y = 4^{9^{x^{2}}}

\int \frac{x ^{2} + 8 x - 1}{x ^{3} - x} d x

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

\frac{\partial}{\partial x} (x^{9} y^{4} - x^{6} + y^{8})

(- 5, 1), (- 4, - 7)

What is the solution for y^'+1/2 y=17sin(2t) ?
The solution for y^'+1/2 y=17sin(2t) is y=2(sin(2t)-4cos(2t))+(c_{1})/(e^{t/2)}