What is the solution for y^{''}-10y^'+25y=0,y(0)=3,y^'(0)= 31/2 ?

The solution for y^{''}-10y^'+25y=0,y(0)=3,y^'(0)= 31/2 is y=3e^{5t}+(e^{5t}t)/2

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y^{''}-10y^'+25y=0,y(0)=3,y^'(0)= 31/2

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{31}{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}

4 \frac{d y}{d x} - 3 \frac{d ^{2} y}{d x ^{2}} = 0

y^{^{'''}} + 5 y^{^{''}} - 11 y^{^{'}} - 39 y = 0

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y^{^{''''}} + 5 y^{^{''}} - 36 y = 0

y^{^{''}} + 16 y = 0, y (\frac{π}{8}) = - 2, y^{^{'}} (\frac{π}{8}) = 16

What is the solution for y^{''}-10y^'+25y=0,y(0)=3,y^'(0)= 31/2 ?
The solution for y^{''}-10y^'+25y=0,y(0)=3,y^'(0)= 31/2 is y=3e^{5t}+(e^{5t}t)/2