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}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}}

y^{^{''''}} - 13 y^{^{''}} + 36 y = 0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 0

9 y^{^{''}} - 16 y = 0

3 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + y = 0

y^{^{''}} + y^{^{'}} + 12 y = 0

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}