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

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

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2x^{''}+7x^'+3x=0,x(0)=3,x^'(0)=0

2 x^{^{''}} + 7 x^{^{'}} + 3 x = 0, x (0) = 3, x^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

(1 + ln (x) + \frac{y}{x}) d x = (8 - ln (x)) d y

\frac{d}{d x} (ln (sinh (2 x)))

\int \frac{e ^{x}}{1 + 9 e ^{2 x}} d x

x \to 3 lim (2 e^{x - 3})

\frac{s}{( s - 1 ) ^{2} + 4}

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