What is the solution for 2y^{''}+y=9e^{2t} ?

The solution for 2y^{''}+y=9e^{2t} is y=c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2)))+e^{2t}

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2y^{''}+y=9e^{2t}

2 y^{^{''}} + y = 9 e^{2 t}

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + e^{2 t}

Solution steps

2 y^{''} (t) + y = 9 e^{2 t}

Solve linear ODE:

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + e^{2 t}

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + e^{2 t}

y^{^{''}} + 6 y = cos (7 t)

(d^{2} + 144) (d^{2} - 144) 2 (d + 12) 4 x = 144

\frac{d ^{2} x}{d t ^{2}} + 36 x = 5 sin (4 t), x (0) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = x^{2}

x y^{^{'''}} - x y^{^{'}} = x^{4}

What is the solution for 2y^{''}+y=9e^{2t} ?
The solution for 2y^{''}+y=9e^{2t} is y=c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2)))+e^{2t}