en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2x-1)y^'+2y=0

Solution

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

Solution

y = \frac{c _{1}}{2 x - 1}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{2 x - 1}

y = \frac{c _{1}}{2 x - 1}

Popular Examples

(dy}{dt}=4t(sin(y))^2,y(-1)=\frac{5pi)/4

\frac{d y}{d t} = 4 t (sin (y))^{2}, y (- 1) = \frac{5 π}{4}

y=(x^2(2+x))/((1+x)^2)y^'

y = \frac{x ^{2} ( 2 + x )}{( 1 + x ) ^{2}} y^{^{'}}

(dx)/(dt)=12-9/25 x

\frac{d x}{d t} = 12 - \frac{9}{25} x

[3xcos(3x)+sin(3x)-3]dx+(2y+5)dy=0

[3 x cos (3 x) + sin (3 x) - 3] d x + (2 y + 5) d y = 0

y^'=(sqrt(1-y^2))/(sqrt(1-x^2))

y^{^{'}} = \frac{1 - y ^{2}}{1 - x ^{2}}

Frequently Asked Questions (FAQ)

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