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

The solution for y^{''}-4y^'+3y=t,y(0)=1,y^'(0)=0 is y=(-1+5/9)e^{3t}+e^t+t/3+4/9

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y^{''}-4y^'+3y=t,y(0)=1,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 3 y = t, y (0) = 1, y^{^{'}} (0) = 0

y = (- 1 + \frac{5}{9}) e^{3 t} + e^{t} + \frac{t}{3} + \frac{4}{9}

Solution steps

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

Solve linear ODE:

y = (- 1 + \frac{5}{9}) e^{3 t} + e^{t} + \frac{t}{3} + \frac{4}{9}

y = (- 1 + \frac{5}{9}) e^{3 t} + e^{t} + \frac{t}{3} + \frac{4}{9}

\int_{0}^{1} \frac{x - 8}{x ^{2} - 5 x + 6} d x

\frac{d}{d x} (\frac{x ^{2}}{ln ( x )})

\frac{d}{d x} (x^{\frac{3}{5}} x^{3} + 2 x - 6)

\frac{d}{d x} (2 x^{2} - 9 x + 10)

\frac{2 s - 44}{7 s ^{2} - 14 s + 70}

What is the solution for y^{''}-4y^'+3y=t,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-4y^'+3y=t,y(0)=1,y^'(0)=0 is y=(-1+5/9)e^{3t}+e^t+t/3+4/9