Solve x^3y/dx+3x^2y=xy

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

Solve for d

$$\left\{\begin{matrix}d=-\frac{x}{3x-1}\text{, }&x\neq 0\text{ and }x\neq \frac{1}{3}\\d\neq 0\text{, }&y=0\text{ and }x\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable d cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by dx.

$$x^{3}y+3x^{2}ydx=xydx$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{3}y+3x^{3}yd=xydx$$

Multiply x and x to get x^{2}.

$$x^{3}y+3x^{3}yd=x^{2}yd$$

Subtract x^{2}yd from both sides.

$$x^{3}y+3x^{3}yd-x^{2}yd=0$$

Subtract x^{3}y from both sides. Anything subtracted from zero gives its negation.

$$3x^{3}yd-x^{2}yd=-x^{3}y$$

Combine all terms containing d.

$$\left(3x^{3}y-x^{2}y\right)d=-x^{3}y$$

The equation is in standard form.

$$\left(3yx^{3}-yx^{2}\right)d=-yx^{3}$$

Divide both sides by 3x^{3}y-x^{2}y.

$$\frac{\left(3yx^{3}-yx^{2}\right)d}{3yx^{3}-yx^{2}}=-\frac{yx^{3}}{3yx^{3}-yx^{2}}$$

Dividing by 3x^{3}y-x^{2}y undoes the multiplication by 3x^{3}y-x^{2}y.

$$d=-\frac{yx^{3}}{3yx^{3}-yx^{2}}$$

Divide -x^{3}y by 3x^{3}y-x^{2}y.

$$d=-\frac{x}{3x-1}$$

Variable d cannot be equal to 0.

$$d=-\frac{x}{3x-1}\text{, }d\neq 0$$

Solve for x

$$\left\{\begin{matrix}x=\frac{d}{3d+1}\text{, }&d\neq 0\text{ and }d\neq -\frac{1}{3}\\x\neq 0\text{, }&y=0\text{ and }d\neq 0\end{matrix}\right.$$