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

The solution for 2x^{''}+3x^'+7x=0 is x=e^{-(3t)/4}(c_{1}cos((sqrt(47)t)/4)+c_{2}sin((sqrt(47)t)/4))

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

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

x = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{47 t}{4}) + c_{2} sin (\frac{47 t}{4}))

Solution steps

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

Solve linear ODE:

x = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{47 t}{4}) + c_{2} sin (\frac{47 t}{4}))

x = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{47 t}{4}) + c_{2} sin (\frac{47 t}{4}))

What is the solution for 2x^{''}+3x^'+7x=0 ?
The solution for 2x^{''}+3x^'+7x=0 is x=e^{-(3t)/4}(c_{1}cos((sqrt(47)t)/4)+c_{2}sin((sqrt(47)t)/4))