What is the solution for y^{''}-5y^'+4y=0,y(0)=-1/2 ,y^'(0)=-17/4 ?

The solution for y^{''}-5y^'+4y=0,y(0)=-1/2 ,y^'(0)=-17/4 is y=(-1/2-3/4)e^{4t}+3/4 e^t

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y^{''}-5y^'+4y=0,y(0)=-1/2 ,y^'(0)=-17/4

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{17}{4}

y = (- \frac{1}{2} - \frac{3}{4}) e^{4 t} + \frac{3}{4} e^{t}

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{2} - \frac{3}{4}) e^{4 t} + \frac{3}{4} e^{t}

y = (- \frac{1}{2} - \frac{3}{4}) e^{4 t} + \frac{3}{4} e^{t}

\frac{d}{d x} (x^{\frac{3}{2}} + 4)

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

\frac{d}{d x} ((1 + e^{- x})^{x})

\int 3 x^{3} e^{3 x} d x

t an g e n t f (x) = x^{2} + x, 4, at x = 2

What is the solution for y^{''}-5y^'+4y=0,y(0)=-1/2 ,y^'(0)=-17/4 ?
The solution for y^{''}-5y^'+4y=0,y(0)=-1/2 ,y^'(0)=-17/4 is y=(-1/2-3/4)e^{4t}+3/4 e^t