What is the solution for 3y^{''}+13y^'+4=0,y(0)=0,y^'(0)=0 ?

The solution for 3y^{''}+13y^'+4=0,y(0)=0,y^'(0)=0 is y= 12/169-12/169 e^{-(13t)/3}-(4t)/(13)

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3y^{''}+13y^'+4=0,y(0)=0,y^'(0)=0

3 y^{^{''}} + 13 y^{^{'}} + 4 = 0, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{12}{169} - \frac{12}{169} e^{- \frac{13 t}{3}} - \frac{4 t}{13}

Solution steps

3 y^{''} (t) + 13 y^{'} (t) + 4 = 0

Solve linear ODE:

y = \frac{12}{169} - \frac{12}{169} e^{- \frac{13 t}{3}} - \frac{4 t}{13}

y = \frac{12}{169} - \frac{12}{169} e^{- \frac{13 t}{3}} - \frac{4 t}{13}

\frac{d}{d x} (3 (7 - 3 x^{3})^{3})

- (1 + y^{2})^{\frac{1}{2}} d x + 4 x y d y = 0

\frac{d}{d x} (1 + csc (x))

x \to - \infty lim (- x e^{x})

y = x^{2} - 4 x, y = 2 x - x^{2}

What is the solution for 3y^{''}+13y^'+4=0,y(0)=0,y^'(0)=0 ?
The solution for 3y^{''}+13y^'+4=0,y(0)=0,y^'(0)=0 is y= 12/169-12/169 e^{-(13t)/3}-(4t)/(13)