What is the solution for 2x^{''}+4x^'-3x=0 ?

The solution for 2x^{''}+4x^'-3x=0 is x=c_{1}e^{((-2+sqrt(10))t)/2}+c_{2}e^{-((2+sqrt(10))t)/2}

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2x^{''}+4x^'-3x=0

2 x^{^{''}} + 4 x^{^{'}} - 3 x = 0

x = c_{1} e^{\frac{( - 2 + 10 ) t}{2}} + c_{2} e^{- \frac{( 2 + 10 ) t}{2}}

Solution steps

2 x^{''} (t) + 4 x^{'} (t) - 3 x = 0

Solve linear ODE:

x = c_{1} e^{\frac{( - 2 + 10 ) t}{2}} + c_{2} e^{- \frac{( 2 + 10 ) t}{2}}

x = c_{1} e^{\frac{( - 2 + 10 ) t}{2}} + c_{2} e^{- \frac{( 2 + 10 ) t}{2}}

What is the solution for 2x^{''}+4x^'-3x=0 ?
The solution for 2x^{''}+4x^'-3x=0 is x=c_{1}e^{((-2+sqrt(10))t)/2}+c_{2}e^{-((2+sqrt(10))t)/2}