What is the solution for y^{''}-5y^'+4y=0,y(0)=2,y^'(0)=-3 ?

The solution for y^{''}-5y^'+4y=0,y(0)=2,y^'(0)=-3 is y=(2-11/3)e^{4t}+11/3 e^t

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y^{''}-5y^'+4y=0,y(0)=2,y^'(0)=-3

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

y = (2 - \frac{11}{3}) e^{4 t} + \frac{11}{3} e^{t}

Solution steps

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

Solve linear ODE:

y = (2 - \frac{11}{3}) e^{4 t} + \frac{11}{3} e^{t}

y = (2 - \frac{11}{3}) e^{4 t} + \frac{11}{3} e^{t}

x \to n - lim (f (x))

\int_{0}^{π} 5 sin (x) d x

x \to 1 lim (\frac{x ^{2} - 4}{x - 2})

4 (5 - 9 x)^{5}

\int (\frac{6}{x} + \frac{x}{6}) d x

What is the solution for y^{''}-5y^'+4y=0,y(0)=2,y^'(0)=-3 ?
The solution for y^{''}-5y^'+4y=0,y(0)=2,y^'(0)=-3 is y=(2-11/3)e^{4t}+11/3 e^t