What is the solution for y^{''}-6y^'+9y=0,y(0)=2,y^'(0)= 25/3 ?

The solution for y^{''}-6y^'+9y=0,y(0)=2,y^'(0)= 25/3 is y=2e^{3t}+(7e^{3t}t)/3

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y^{''}-6y^'+9y=0,y(0)=2,y^'(0)= 25/3

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = \frac{25}{3}

y = 2 e^{3 t} + \frac{7 e ^{3 t} t}{3}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = 2 e^{3 t} + \frac{7 e ^{3 t} t}{3}

y = 2 e^{3 t} + \frac{7 e ^{3 t} t}{3}

What is the solution for y^{''}-6y^'+9y=0,y(0)=2,y^'(0)= 25/3 ?
The solution for y^{''}-6y^'+9y=0,y(0)=2,y^'(0)= 25/3 is y=2e^{3t}+(7e^{3t}t)/3