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

The solution for y^{''}-5y^'+4y=0,y(0)=5,y^'(0)=0 is y=(5-20/3)e^{4t}+20/3 e^t

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

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

y = (5 - \frac{20}{3}) e^{4 t} + \frac{20}{3} e^{t}

Solution steps

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

Solve linear ODE:

y = (5 - \frac{20}{3}) e^{4 t} + \frac{20}{3} e^{t}

y = (5 - \frac{20}{3}) e^{4 t} + \frac{20}{3} e^{t}

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y^{^{'}} (- 1) = 9, y (- 1) = 3

y^{^{'''}} - 2 y^{^{''}} - 9 y^{^{'}} + 4 y = 0

y^{^{''}} + a y = 0, y (0) = 0, y (1) = 0

(D^{3} + D^{2} + D + 1) y = 0

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (1) = 1, y^{^{'}} (1) = 0

What is the solution for y^{''}-5y^'+4y=0,y(0)=5,y^'(0)=0 ?
The solution for y^{''}-5y^'+4y=0,y(0)=5,y^'(0)=0 is y=(5-20/3)e^{4t}+20/3 e^t