What is the solution for 100y^{''}-60y^'+13y+30=0 ?

The solution for 100y^{''}-60y^'+13y+30=0 is y=e^{(3t)/(10)}(c_{1}cos(t/5)+c_{2}sin(t/5))-30/13

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100y^{''}-60y^'+13y+30=0

100 y^{^{''}} - 60 y^{^{'}} + 13 y + 30 = 0

y = e^{\frac{3 t}{10}} (c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5})) - \frac{30}{13}

Solution steps

100 y^{''} (t) - 60 y^{'} (t) + 13 y + 30 = 0

Solve linear ODE:

y = e^{\frac{3 t}{10}} (c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5})) - \frac{30}{13}

y = e^{\frac{3 t}{10}} (c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5})) - \frac{30}{13}

y^{^{''}} - 2 y^{^{'}} - 8 y = 2 e^{4 x} + 3 x e^{2 x} + 4 x

y^{^{''}} + 16 y = 4 sin (2 x)

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 15 y = 6 + 4 t

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

(4 D^{2} + 4 D + 1) y = x e^{- \frac{x}{2}} sin (x)

What is the solution for 100y^{''}-60y^'+13y+30=0 ?
The solution for 100y^{''}-60y^'+13y+30=0 is y=e^{(3t)/(10)}(c_{1}cos(t/5)+c_{2}sin(t/5))-30/13