en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2+x)y^'-y=0

Solution

(2 + x) y^{^{'}} - y = 0

Solution

y = 2 e^{c_{1}} + e^{c_{1}} x

Solution steps

(2 + x) y^{'} (x) - y = 0

Solve separable ODE:

y = 2 e^{c_{1}} + e^{c_{1}} x

y = 2 e^{c_{1}} + e^{c_{1}} x

Graph

Popular Examples

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

x (1 + y^{2}) d x - y (1 + x^{2}) d y = 0

(dx)/(dt)=125-x/(100)

\frac{d x}{d t} = 125 - \frac{x}{100}

(dQ)/(dt)=4Q^2ln(t),Q(1)=-1

\frac{d Q}{d t} = 4 Q^{2} ln (t), Q (1) = - 1

(dp)/(dt)=t^2p-p+t^2-1

\frac{d p}{d t} = t^{2} p - p + t^{2} - 1

x d y - y d x = 2 d x

Frequently Asked Questions (FAQ)

What is the solution for (2+x)y^'-y=0 ?
The solution for (2+x)y^'-y=0 is y=2e^{c_{1}}+e^{c_{1}}x