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

The solution for y^{''}-5y^'+4y=0,y(0)=3,y^'(0)=2 is y=(3-10/3)e^{4t}+10/3 e^t

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

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

y = (3 - \frac{10}{3}) e^{4 t} + \frac{10}{3} e^{t}

Solution steps

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

Solve linear ODE:

y = (3 - \frac{10}{3}) e^{4 t} + \frac{10}{3} e^{t}

y = (3 - \frac{10}{3}) e^{4 t} + \frac{10}{3} e^{t}

\int_{1}^{3} e^{x} d x

18 + \frac{12}{( 0.0004 x ^{2} + 1 ) ^{2}}

\int_{3}^{6} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{\partial}{\partial y} (2 x^{4 y})

\frac{d}{d x} (9 tan (x) - 9 cot (x))

What is the solution for y^{''}-5y^'+4y=0,y(0)=3,y^'(0)=2 ?
The solution for y^{''}-5y^'+4y=0,y(0)=3,y^'(0)=2 is y=(3-10/3)e^{4t}+10/3 e^t