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

The solution for 2x^{''}+7x^'+3x=0 is x=c_{1}e^{-t/2}+c_{2}e^{-3t}

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2x^{''}+7x^'+3x=0

2 x^{^{''}} + 7 x^{^{'}} + 3 x = 0

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- 3 t}

Solution steps

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

Solve linear ODE:

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- 3 t}

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- 3 t}

t an g e n t f (x) = e^{5 x} cos (π x), at x = 0

d er i v a t i v e f (x) = x^{\frac{3}{2}}

d er i v a t i v e \frac{x ^{3}}{x ^{2} + 12}

y y^{^{'}} = 8 sin (x)

\frac{d}{d x} (\frac{x}{( x - 8 ) ^{2}})

What is the solution for 2x^{''}+7x^'+3x=0 ?
The solution for 2x^{''}+7x^'+3x=0 is x=c_{1}e^{-t/2}+c_{2}e^{-3t}