en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1+x)y^'=4y,y(0)=2

Solution

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

Solution

y = 2 (1 + x)^{4}

Solution steps

(1 + x) y^{'} (x) = 4 y

Solve separable ODE:

y = 2 (1 + x)^{4}

y = 2 (1 + x)^{4}

Graph

Popular Examples

x^2dx-(x^3y^2+3y^2)dy=0

x^{2} d x - (x^{3} y^{2} + 3 y^{2}) d y = 0

(dv)/(dt)=\sqrt[3]{v}

\frac{d v}{d t} = 3 v

(dy)/(dt)=1-t+y^2-ty^2

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

y^{^{'}} + 3 x^{2} y = 5 x^{2}

(dz)/(dt)+5e^{t+z}=0

\frac{d z}{d t} + 5 e^{t + z} = 0

Frequently Asked Questions (FAQ)

What is the solution for (1+x)y^'=4y,y(0)=2 ?
The solution for (1+x)y^'=4y,y(0)=2 is y=2(1+x)^4