What is the solution for 3y^{''}-4y^'+y=5 ?

The solution for 3y^{''}-4y^'+y=5 is y=c_{1}e^t+c_{2}e^{t/3}+5

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3y^{''}-4y^'+y=5

3 y^{^{''}} - 4 y^{^{'}} + y = 5

y = c_{1} e^{t} + c_{2} e^{\frac{t}{3}} + 5

Solution steps

3 y^{''} (t) - 4 y^{'} (t) + y = 5

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{\frac{t}{3}} + 5

y = c_{1} e^{t} + c_{2} e^{\frac{t}{3}} + 5

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{- x} (9 x^{2} - 3 x - 4)

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 64 e^{t}, y (0) = 4, y^{^{'}} (0) = 7

x^{^{''}} + x^{^{'}} + 2 x = sin (t)

q^{^{''}} + 10 q^{^{'}} + 10000 q = 20 (t)

What is the solution for 3y^{''}-4y^'+y=5 ?
The solution for 3y^{''}-4y^'+y=5 is y=c_{1}e^t+c_{2}e^{t/3}+5