What is the solution for 2y^{''}-4y^'-5y=0 ?

The solution for 2y^{''}-4y^'-5y=0 is y=c_{1}e^{((2+sqrt(14))t)/2}+c_{2}e^{((2-sqrt(14))t)/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-4y^'-5y=0

2 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

y = c_{1} e^{\frac{( 2 + 14 ) t}{2}} + c_{2} e^{\frac{( 2 - 14 ) t}{2}}

Solution steps

2 y^{''} (t) - 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 2 + 14 ) t}{2}} + c_{2} e^{\frac{( 2 - 14 ) t}{2}}

y = c_{1} e^{\frac{( 2 + 14 ) t}{2}} + c_{2} e^{\frac{( 2 - 14 ) t}{2}}

y^{^{''}} + 9 y = 0, y^{^{'}} (0) = 0

y^{^{''''}} + 2 y^{^{'''}} - 5 y^{^{''}} - 5 y^{^{'}} + \frac{1}{2} y = 0

\frac{d ^{2} y}{d x ^{2}} + 12 \cdot \frac{d y}{d x} + 27 y = 0

y^{^{'''}} - 7 y^{^{''}} + 10 y^{^{'}} = 0

2 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

What is the solution for 2y^{''}-4y^'-5y=0 ?
The solution for 2y^{''}-4y^'-5y=0 is y=c_{1}e^{((2+sqrt(14))t)/2}+c_{2}e^{((2-sqrt(14))t)/2}