What is the solution for (2x-1)dx(3y+7)dy=0 ?

The solution for (2x-1)dx(3y+7)dy=0 is y=c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2x-1)dx(3y+7)dy=0

(2 x - 1) d x (3 y + 7) d y = 0

y = c_{1}

Solution steps

(2 x - 1) d x (3 y + 7) d y = 0

Let

y

be the dependent variable. Divide by

d x

(2 x - 1) (3 y + 7) \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

(2 x - 1) (3 y + 7) y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = 0

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 0 d x

\int 0 d x = c_{1}

y = c_{1}

\frac{d}{d r ( θ )} (r (θ) cos (θ))

\frac{\partial}{\partial y} (e^{7 yz})

x \to 4 + lim (\frac{7 x - 1}{x - 4})

y^{^{''}} - 8 y^{^{'}} = 64 t

\frac{d}{d x} (3 + 7 x)

What is the solution for (2x-1)dx(3y+7)dy=0 ?
The solution for (2x-1)dx(3y+7)dy=0 is y=c_{1}