What is the solution for y^{''}-10y^'+25y=t,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-10y^'+25y=t,y(0)=0,y^'(0)=1 is y=-2/125 e^{5t}+(26e^{5t}t)/(25)+t/(25)+2/125

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y^{''}-10y^'+25y=t,y(0)=0,y^'(0)=1

y^{^{''}} - 10 y^{^{'}} + 25 y = t, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{2}{125} e^{5 t} + \frac{26 e ^{5 t} t}{25} + \frac{t}{25} + \frac{2}{125}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = t

Solve linear ODE:

y = - \frac{2}{125} e^{5 t} + \frac{26 e ^{5 t} t}{25} + \frac{t}{25} + \frac{2}{125}

y = - \frac{2}{125} e^{5 t} + \frac{26 e ^{5 t} t}{25} + \frac{t}{25} + \frac{2}{125}

y^{^{'''}} - 49 y^{^{'}} = t e^{- 7 t} + 8 cos (7 t)

y^{^{''}} - y^{^{'}} - 2 y = e^{- t}

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (6 t)

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

y^{^{''}} + y^{^{'}} + 2 = e^{t}

What is the solution for y^{''}-10y^'+25y=t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-10y^'+25y=t,y(0)=0,y^'(0)=1 is y=-2/125 e^{5t}+(26e^{5t}t)/(25)+t/(25)+2/125