What is the solution for y^{''}-5y^'+4y=0,y(0)= 4/3 ,y^'(0)= 1/3 ?

The solution for y^{''}-5y^'+4y=0,y(0)= 4/3 ,y^'(0)= 1/3 is y=(4/3-5/3)e^{4t}+5/3 e^t

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y^{''}-5y^'+4y=0,y(0)= 4/3 ,y^'(0)= 1/3

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = \frac{4}{3}, y^{^{'}} (0) = \frac{1}{3}

y = (\frac{4}{3} - \frac{5}{3}) e^{4 t} + \frac{5}{3} e^{t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = (\frac{4}{3} - \frac{5}{3}) e^{4 t} + \frac{5}{3} e^{t}

y = (\frac{4}{3} - \frac{5}{3}) e^{4 t} + \frac{5}{3} e^{t}

y^{^{''}} = - 64 y

y^{^{''}} - 30 y^{^{'}} + 325 y = 0

- y^{^{''}} + a y^{^{'}} + b y = 0

f^{^{''}} (x) + 4 a^{2} f (x) = 0

y^{^{''}} - 5 y^{^{'}} + 10 y = 0

What is the solution for y^{''}-5y^'+4y=0,y(0)= 4/3 ,y^'(0)= 1/3 ?
The solution for y^{''}-5y^'+4y=0,y(0)= 4/3 ,y^'(0)= 1/3 is y=(4/3-5/3)e^{4t}+5/3 e^t