What is the solution for 9y^{''}-6y^'+y=0,y(0)=3,y^'(0)=4 ?

The solution for 9y^{''}-6y^'+y=0,y(0)=3,y^'(0)=4 is y=3e^{t/3}+3e^{t/3}t

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9y^{''}-6y^'+y=0,y(0)=3,y^'(0)=4

9 y^{^{''}} - 6 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = 4

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + \frac{16}{9} y = 0

u, \frac{d ^{2} u}{d x ^{2}} = u

16 y^{^{''}} + 8 y^{^{'}} + 5 y = 0

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

3 y^{^{''}} + 11 y^{^{'}} - 4 y = 0, y (0) = 4, y^{^{'}} (0) = - \frac{5}{3}

What is the solution for 9y^{''}-6y^'+y=0,y(0)=3,y^'(0)=4 ?
The solution for 9y^{''}-6y^'+y=0,y(0)=3,y^'(0)=4 is y=3e^{t/3}+3e^{t/3}t