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

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

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

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

y = (1 - \frac{5}{4}) e^{4 t} + \frac{5}{4} e^{t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{5}{4}) e^{4 t} + \frac{5}{4} e^{t}

y = (1 - \frac{5}{4}) e^{4 t} + \frac{5}{4} e^{t}

n = 4 \sum \infty 3 - 0. 5^{n}

y^{^{'}} = \frac{x}{1 + 2 y}

\int \frac{20}{( x - 4 ) ( x ^{2} + 4 )} d x

sin (x), \frac{3 π}{2}

\frac{d}{d x} (100 x - x^{2})

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