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

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

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

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

y = (- 1 - \frac{1}{3}) e^{4 t} + \frac{1}{3} e^{t}

Solution steps

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

Solve linear ODE:

y = (- 1 - \frac{1}{3}) e^{4 t} + \frac{1}{3} e^{t}

y = (- 1 - \frac{1}{3}) e^{4 t} + \frac{1}{3} e^{t}

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y (1) = 5

0.05 y^{^{''}} + 2 y^{^{'}} + 100 y = 0

x^{^{''}} + \frac{0}{\frac{1}{32}} x^{^{'}} + \frac{13}{\frac{1}{32}} x = 0

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (π) = 5, y (0) = 6

(D^{3} - 4 D^{2} + 3 D) y = 0

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