en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y+20y^'+169y=0

Solution

4 y + 20 y^{^{'}} + 169 y = 0

Solution

y = e^{\frac{- 173 t + c _{1}}{20}}

Solution steps

4 y + 20 y^{'} (t) + 169 y = 0

Solve separable ODE:

y = e^{\frac{- 173 t + c _{1}}{20}}

y = e^{\frac{- 173 t + c _{1}}{20}}

Popular Examples

y^'+(n^2-1)y= 2/n (-1)^n

y^{^{'}} + (n^{2} - 1) y = \frac{2}{n} (- 1)^{n}

y^'=(e^{2x})/(cos(2y))

y^{^{'}} = \frac{e ^{2 x}}{cos ( 2 y )}

(dy)/(dx)=(y^2)/(1+x^2),y(0)=1

\frac{d y}{d x} = \frac{y ^{2}}{1 + x ^{2}}, y (0) = 1

(dy)/(dt)=100-y,y(0)=30

\frac{d y}{d t} = 100 - y, y (0) = 30

y^{^{'}} = y^{2} x^{7}

Frequently Asked Questions (FAQ)

What is the solution for 4y+20y^'+169y=0 ?
The solution for 4y+20y^'+169y=0 is y=e^{(-173t+c_{1})/(20)}