What is the solution for y^{''}-4y^'+3y=1 ?

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

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y^{''}-4y^'+3y=1

y^{^{''}} - 4 y^{^{'}} + 3 y = 1

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 3 y = 1

Solve linear ODE:

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

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

y^{^{''}} + 9 y = 2 cos (3 x) + 2 sin (3 x)

y^{^{''}} - 32 y = 16 cos (x)

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 91

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

x^{^{''}} + 16 x = cos (4 t), x (0) = 0, x^{^{'}} (0) = 1

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