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

The solution to expand (3ydx+6dx-xdy=0)(dy)/(xdx) is (3ydy)/x+(6dy)/x-(dy^2)/(dx)=0

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expand (3ydx+6dx-xdy=0)(dy)/(xdx)

expand

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

\frac{3 y d y}{x} + \frac{6 d y}{x} - \frac{d y ^{2}}{d x} = 0

Solution steps

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

Expand

3 \cdot \frac{d y}{x d x} y d x + 6 \cdot \frac{d y}{x d x} d x - \frac{d y}{x d x} x d y : \frac{3 y d y}{x} + \frac{6 d y}{x} - \frac{d y ^{2}}{d x}

\frac{3 y d y}{x} + \frac{6 d y}{x} - \frac{d y ^{2}}{d x} = 0

2 (x + 1) (ln (x + 1)) + x + 1

\frac{d}{d θ} ((x) (c + v (x, θ) cos (θ)))

2 x y^{3} d x + 3 (x^{2} + 1) y^{2} d y

A (3, 4) - B (1, - 1) - C (- 5, 3)

(1 - y) y^{'} (t) + 2 y

What is expand (3ydx+6dx-xdy=0)(dy)/(xdx) ?
The solution to expand (3ydx+6dx-xdy=0)(dy)/(xdx) is (3ydy)/x+(6dy)/x-(dy^2)/(dx)=0