What is the solution for y^{''}-9y^'+18y=0,y(0)=-5,y^'(0)=2 ?

The solution for y^{''}-9y^'+18y=0,y(0)=-5,y^'(0)=2 is y=(-5+32/3)e^{6t}-32/3 e^{3t}

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y^{''}-9y^'+18y=0,y(0)=-5,y^'(0)=2

y^{^{''}} - 9 y^{^{'}} + 18 y = 0, y (0) = - 5, y^{^{'}} (0) = 2

y = (- 5 + \frac{32}{3}) e^{6 t} - \frac{32}{3} e^{3 t}

Solution steps

y^{''} (t) - 9 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = (- 5 + \frac{32}{3}) e^{6 t} - \frac{32}{3} e^{3 t}

y = (- 5 + \frac{32}{3}) e^{6 t} - \frac{32}{3} e^{3 t}

f (x) = 9 x^{2} - x^{3}, (2, 28)

\frac{d}{d x} (\frac{1}{3 x ^{5}})

f (x) = (\frac{sin ( x )}{x})^{2}

\int 4 x^{\frac{2}{3}} d x

\frac{\partial}{\partial y} (3 x + 2 x y^{2})

What is the solution for y^{''}-9y^'+18y=0,y(0)=-5,y^'(0)=2 ?
The solution for y^{''}-9y^'+18y=0,y(0)=-5,y^'(0)=2 is y=(-5+32/3)e^{6t}-32/3 e^{3t}