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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + 49 y = 0, y (π) = 0, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} + a y = 0

\frac{d ^{2} y}{d x ^{2}} + 5 y = 0

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 0, y (0) = 3, y^{^{'}} (0) = 1

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