What is the solution for 14y^{''}+3y^'-2y=0 ?

The solution for 14y^{''}+3y^'-2y=0 is y=c_{1}e^{(2t)/7}+c_{2}e^{-t/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

14y^{''}+3y^'-2y=0

14 y^{^{''}} + 3 y^{^{'}} - 2 y = 0

y = c_{1} e^{\frac{2 t}{7}} + c_{2} e^{- \frac{t}{2}}

Solution steps

14 y^{''} (t) + 3 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{2 t}{7}} + c_{2} e^{- \frac{t}{2}}

y = c_{1} e^{\frac{2 t}{7}} + c_{2} e^{- \frac{t}{2}}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + 0.6 y^{^{'}} + 8 y = 0

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

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 26 y = 0

\frac{d ^{2} y}{d x ^{2}} + y = 0, y = cos (x) + sin (x)

What is the solution for 14y^{''}+3y^'-2y=0 ?
The solution for 14y^{''}+3y^'-2y=0 is y=c_{1}e^{(2t)/7}+c_{2}e^{-t/2}