What is the solution for y^{''}-6y^'+8y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}-6y^'+8y=0,y(0)=2,y^'(0)=1 is y=(2-7/2)e^{4t}+7/2 e^{2t}

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y^{''}-6y^'+8y=0,y(0)=2,y^'(0)=1

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = (2 - \frac{7}{2}) e^{4 t} + \frac{7}{2} e^{2 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = (2 - \frac{7}{2}) e^{4 t} + \frac{7}{2} e^{2 t}

y = (2 - \frac{7}{2}) e^{4 t} + \frac{7}{2} e^{2 t}

x \to - 1 - lim (\frac{x}{x + 1})

\int \frac{5 x + 4}{x ^{3}} d x

\int e^{x^{3} - 3 x^{2}} d y

x \to 5 lim (\frac{6 x}{x - 5})

\int \frac{1}{5 x} d x

What is the solution for y^{''}-6y^'+8y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}-6y^'+8y=0,y(0)=2,y^'(0)=1 is y=(2-7/2)e^{4t}+7/2 e^{2t}