What is the solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-29/2 ?

The solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-29/2 is y=-3e^{5t}+(e^{5t}t)/2

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y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-29/2

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = - \frac{29}{2}

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

Solution steps

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

Solve linear ODE:

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

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

What is the solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-29/2 ?
The solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-29/2 is y=-3e^{5t}+(e^{5t}t)/2