What is the solution for 20y^{''}+11y^'-3y=0 ?

The solution for 20y^{''}+11y^'-3y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(3t)/4}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

20y^{''}+11y^'-3y=0

20 y^{^{''}} + 11 y^{^{'}} - 3 y = 0

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{3 t}{4}}

Solution steps

20 y^{''} (t) + 11 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{3 t}{4}}

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{3 t}{4}}

\frac{d}{d y} (\frac{z}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

x \to 1 lim (\frac{x ^{2} - 3 x + 2}{x ^{3} - 1})

\int \frac{3}{x ^{2} 25 - x ^{2}} d x

f (x) = \frac{2}{x + 2}

\frac{\partial}{\partial y} ((x^{2} + y) e^{\frac{y}{2}})

What is the solution for 20y^{''}+11y^'-3y=0 ?
The solution for 20y^{''}+11y^'-3y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(3t)/4}