en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+100y^'+2500y=30,y(0)=0,y^'(0)=2

Solution

y^{^{''}} + 100 y^{^{'}} + 2500 y = 30, y (0) = 0, y^{^{'}} (0) = 2

Solution

y = - \frac{3}{250} e^{- 50 t} + \frac{7 e ^{- 50 t} t}{5} + \frac{3}{250}

Solution steps

y^{''} (t) + 100 y^{'} (t) + 2500 y = 30

Solve linear ODE:

y = - \frac{3}{250} e^{- 50 t} + \frac{7 e ^{- 50 t} t}{5} + \frac{3}{250}

y = - \frac{3}{250} e^{- 50 t} + \frac{7 e ^{- 50 t} t}{5} + \frac{3}{250}

Graph

Popular Examples

linear (d^2x)/(dt^2)+x=6cos(2t),x(0)=3 linear

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t), x (0) = 3

1/2 y^{''}+2y=4tan(2t)-1/2 e^{2t}

\frac{1}{2} y^{^{''}} + 2 y = 4 tan (2 t) - \frac{1}{2} e^{2 t}

y^{''}+36y=12cos(6x)+18sin(6x)

y^{^{''}} + 36 y = 12 cos (6 x) + 18 sin (6 x)

y^{''}+4y^'+4y=2x+6,y(0)=1,y^'(0)= 1/2

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 6, y (0) = 1, y^{^{'}} (0) = \frac{1}{2}

y^{''}+y=2e^{-x}

y^{^{''}} + y = 2 e^{- x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+100y^'+2500y=30,y(0)=0,y^'(0)=2 ?
The solution for y^{''}+100y^'+2500y=30,y(0)=0,y^'(0)=2 is y=-3/250 e^{-50t}+(7e^{-50t}t)/5+3/250