What is the solution for y^{''}-6y^'+34y=25e^{3t},y(0)=1 ?

The solution for y^{''}-6y^'+34y=25e^{3t},y(0)=1 is y=c_{2}e^{3t}sin(5t)+e^{3t}

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y^{''}-6y^'+34y=25e^{3t},y(0)=1

y^{^{''}} - 6 y^{^{'}} + 34 y = 25 e^{3 t}, y (0) = 1

y = c_{2} e^{3 t} sin (5 t) + e^{3 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 34 y = 25 e^{3 t}

Solve linear ODE:

y = c_{2} e^{3 t} sin (5 t) + e^{3 t}

y = c_{2} e^{3 t} sin (5 t) + e^{3 t}

(D^{2} - 9 D + 18) y = e^{e^{- 3 x}}

x^{^{''}} + x = 6 sin (2 t), x (0) = 3, x^{^{'}} (0) = 1

2 y^{^{''}} + y^{^{'}} + 3 y = 33 cos (3 t) - 22 sin (3 t)

x^{^{''}} + 3 x^{^{'}} + 2 x = 4 t^{2}, x^{0} = 0, x^{^{'}} 0 = 0

d (x) = \frac{x ^{2} - 12 x + 20}{3 x}

What is the solution for y^{''}-6y^'+34y=25e^{3t},y(0)=1 ?
The solution for y^{''}-6y^'+34y=25e^{3t},y(0)=1 is y=c_{2}e^{3t}sin(5t)+e^{3t}