en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-(2y)/(2x+100)

Solution

y^{^{'}} = - \frac{2 y}{2 x + 100}

Solution

y = \frac{c _{1}}{x + 50}

Solution steps

y^{'} (x) = - \frac{2 y}{2 x + 100}

Solve separable ODE:

y = \frac{c _{1}}{x + 50}

y = \frac{c _{1}}{x + 50}

Graph

Popular Examples

((x^2-9))/x*(dy)/(dx)+y=0

\frac{( x ^{2} - 9 )}{x} \cdot \frac{d y}{d x} + y = 0

(dy)/(dx)=(x^2+7x+3)/(y^2)

\frac{d y}{d x} = \frac{x ^{2} + 7 x + 3}{y ^{2}}

(xy^2-x)dx+(2x^2+4)dy=0

(x y^{2} - x) d x + (2 x^{2} + 4) d y = 0

(dy)/(dx)=(3x^2-e^x)/(2y-5)

\frac{d y}{d x} = \frac{3 x ^{2} - e ^{x}}{2 y - 5}

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^'=-(2y)/(2x+100) ?
The solution for y^'=-(2y)/(2x+100) is y=(c_{1})/(x+50)