What is the solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-68/5 ?

The solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-68/5 is y=-3e^{5t}+(7e^{5t}t)/5

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y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-68/5

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = - \frac{68}{5}

y = - 3 e^{5 t} + \frac{7 e ^{5 t} t}{5}

Solution steps

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

Solve linear ODE:

y = - 3 e^{5 t} + \frac{7 e ^{5 t} t}{5}

y = - 3 e^{5 t} + \frac{7 e ^{5 t} t}{5}

y^{^{''}} + y^{^{'}} + y 2 = 0

y^{^{'''}} - 5 y^{^{''}} - 33 y^{^{'}} - 27 y = 0

- \frac{( y ^{^{''}} )}{( y ^{^{'}} )} = 4

49 y^{^{''}} - 70 y^{^{'}} + 25 y = 0

y^{^{''}} + (\frac{9.8}{1.5}) y = 0

What is the solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-68/5 ?
The solution for y^{''}-10y^'+25y=0,y(0)=-3,y^'(0)=-68/5 is y=-3e^{5t}+(7e^{5t}t)/5