en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/2 y^'+10y=12

Solution

\frac{1}{2} y^{^{'}} + 10 y = 12

Solution

y = - \frac{e ^{- 20 t - c_{1}}}{5} + \frac{6}{5}

Solution steps

\frac{1}{2} y^{'} (t) + 10 y = 12

Solve separable ODE:

y = - \frac{e ^{- 20 t - c_{1}}}{5} + \frac{6}{5}

y = - \frac{e ^{- 20 t - c_{1}}}{5} + \frac{6}{5}

Graph

Popular Examples

X d Y - Y d X = 0

(dr)/(ds)=e^{r+s},r(1)=0

\frac{d r}{d s} = e^{r + s}, r (1) = 0

(dp}{dt}=kp(t)ln(\frac{4.5)/p)

\frac{d p}{d t} = k p (t) ln (\frac{4.5}{p})

y^'=(sqrt(x)+2)sqrt(y)

y^{^{'}} = (x + 2) y

(dy)/(dx)=4sqrt(yx),y(1)=6

\frac{d y}{d x} = 4 y x, y (1) = 6

Frequently Asked Questions (FAQ)

What is the solution for 1/2 y^'+10y=12 ?
The solution for 1/2 y^'+10y=12 is y=-(e^{-20t-c_{1}})/5+6/5