What is the solution for 3y^{''}-4y^'-4y=0,y(0)=-7,y^'(0)=-8 ?

The solution for 3y^{''}-4y^'-4y=0,y(0)=-7,y^'(0)=-8 is y=(-7+9/4)e^{2t}-9/4 e^{-(2t)/3}

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3y^{''}-4y^'-4y=0,y(0)=-7,y^'(0)=-8

3 y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = - 7, y^{^{'}} (0) = - 8

y = (- 7 + \frac{9}{4}) e^{2 t} - \frac{9}{4} e^{- \frac{2 t}{3}}

Solution steps

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

Solve linear ODE:

y = (- 7 + \frac{9}{4}) e^{2 t} - \frac{9}{4} e^{- \frac{2 t}{3}}

y = (- 7 + \frac{9}{4}) e^{2 t} - \frac{9}{4} e^{- \frac{2 t}{3}}

\int (e^{x} + x) d x

y = 4 x^{2} + 5 x + 1, (0, 1)

\frac{d}{d x} (lo g_{10} (lo g_{10} (x)))

\int sin^{2} (θ) cos (4 θ) d θ

n = 1 \sum \infty \frac{3 n !}{n ^{n}}

What is the solution for 3y^{''}-4y^'-4y=0,y(0)=-7,y^'(0)=-8 ?
The solution for 3y^{''}-4y^'-4y=0,y(0)=-7,y^'(0)=-8 is y=(-7+9/4)e^{2t}-9/4 e^{-(2t)/3}