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

The solution for 2y^{''}+y^'-2y=0 is y=c_{1}e^{((-1+sqrt(17))t)/4}+c_{2}e^{((-1-sqrt(17))t)/4}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{\frac{( - 1 + 17 ) t}{4}} + c_{2} e^{\frac{( - 1 - 17 ) t}{4}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 1 + 17 ) t}{4}} + c_{2} e^{\frac{( - 1 - 17 ) t}{4}}

y = c_{1} e^{\frac{( - 1 + 17 ) t}{4}} + c_{2} e^{\frac{( - 1 - 17 ) t}{4}}

x \to 5 lim (cos (x))

y = 2 x^{3 π}

(5 t + 3 y + 3) d t - d y = 0

\frac{d}{d x} (x^{3} - 4 x^{2} + 5)

What is the solution for 2y^{''}+y^'-2y=0 ?
The solution for 2y^{''}+y^'-2y=0 is y=c_{1}e^{((-1+sqrt(17))t)/4}+c_{2}e^{((-1-sqrt(17))t)/4}