What is the solution for 3y^{''}-7y^'+2y=0,y(0)=-3,y^'(0)=-8 ?

The solution for 3y^{''}-7y^'+2y=0,y(0)=-3,y^'(0)=-8 is y=(-3-6/5)e^{2t}+6/5 e^{t/3}

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3y^{''}-7y^'+2y=0,y(0)=-3,y^'(0)=-8

3 y^{^{''}} - 7 y^{^{'}} + 2 y = 0, y (0) = - 3, y^{^{'}} (0) = - 8

y = (- 3 - \frac{6}{5}) e^{2 t} + \frac{6}{5} e^{\frac{t}{3}}

Solution steps

3 y^{''} (t) - 7 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = (- 3 - \frac{6}{5}) e^{2 t} + \frac{6}{5} e^{\frac{t}{3}}

y = (- 3 - \frac{6}{5}) e^{2 t} + \frac{6}{5} e^{\frac{t}{3}}

\int sin (x) - 8 cos (x) d x

f (x) = cot (7 x)

x \to 0 lim \frac{( \frac{1}{( x + 4 )} ) - ( \frac{1}{4} )}{x}

5 x + 8

y^{^{'}} = \frac{x}{1 + x ^{2}}

What is the solution for 3y^{''}-7y^'+2y=0,y(0)=-3,y^'(0)=-8 ?
The solution for 3y^{''}-7y^'+2y=0,y(0)=-3,y^'(0)=-8 is y=(-3-6/5)e^{2t}+6/5 e^{t/3}