What is expand (3ydx+6dx-xdy=0) 1/(ydy) ?

The solution to expand (3ydx+6dx-xdy=0) 1/(ydy) is (3dx)/(dy)+(6dx)/(ydy)-x/y =0

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expand (3ydx+6dx-xdy=0) 1/(ydy)

expand

(3 y d x + 6 d x - x d y = 0) \frac{1}{y d y}

\frac{3 d x}{d y} + \frac{6 d x}{y d y} - \frac{x}{y} = 0

Solution steps

(3 y d x + 6 d x - x d y = 0) \frac{1}{y d y}

= \frac{1}{y d y} (3 y d x + 6 d x - 0 = x d y)

Distribute parentheses

= \frac{1}{y d y} \cdot 3 y d x + \frac{1}{y d y} \cdot 6 d x + \frac{1}{y d y} (- x d y = 0)

Apply minus-plus rules

+ (- a) = - a

3 \cdot \frac{1}{y d y} y d x + 6 \cdot \frac{1}{y d y} d x - \frac{1}{y d y} x d y = 0

Expand

3 \cdot \frac{1}{y d y} y d x + 6 \cdot \frac{1}{y d y} d x - \frac{1}{y d y} x d y : \frac{3 d x}{d y} + \frac{6 d x}{y d y} - \frac{x}{y}

\frac{3 d x}{d y} + \frac{6 d x}{y d y} - \frac{x}{y} = 0

\frac{( x + 4 ) ( 6 x - 1 )}{6}

\frac{1}{( x + h + 5 ) ^{2}}

e^{2 t - 10} h (t - 5)

π^{e^{- 1}} - arccos (- 1)

(1 - 2) x

What is expand (3ydx+6dx-xdy=0) 1/(ydy) ?
The solution to expand (3ydx+6dx-xdy=0) 1/(ydy) is (3dx)/(dy)+(6dx)/(ydy)-x/y =0