What is the solution for y^{''}-4y^'+3y=1,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-4y^'+3y=1,y(0)=1,y^'(0)=0 is y=(-1+2/3)e^{3t}+e^t+1/3

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y^{''}-4y^'+3y=1,y(0)=1,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 3 y = 1, y (0) = 1, y^{^{'}} (0) = 0

y = (- 1 + \frac{2}{3}) e^{3 t} + e^{t} + \frac{1}{3}

Solution steps

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

Solve linear ODE:

y = (- 1 + \frac{2}{3}) e^{3 t} + e^{t} + \frac{1}{3}

y = (- 1 + \frac{2}{3}) e^{3 t} + e^{t} + \frac{1}{3}

(D^{2} + 1) y = sin (x)

y^{^{''}} + y = 16 cos (3 x)

y^{^{''}} + 7 y^{^{'}} + 12 y = e^{t}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = e^{- x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{(- x)} cos (2 x)

What is the solution for y^{''}-4y^'+3y=1,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-4y^'+3y=1,y(0)=1,y^'(0)=0 is y=(-1+2/3)e^{3t}+e^t+1/3