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

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

Solve for d

$$d\neq 0$$

$$\left(y=0\text{ and }x\neq 0\right)\text{ or }x=\frac{1}{4}$$

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.

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

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

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

Combine x^{3}dy and 3x^{3}yd to get 4x^{3}dy.

$$4x^{3}dy=xydx$$

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

$$4x^{3}dy=x^{2}yd$$

Subtract x^{2}yd from both sides.

$$4x^{3}dy-x^{2}yd=0$$

Reorder the terms.

$$4dyx^{3}-dyx^{2}=0$$

Combine all terms containing d.

$$\left(4yx^{3}-yx^{2}\right)d=0$$

Divide 0 by 4yx^{3}-yx^{2}.

$$d=0$$

Variable d cannot be equal to 0.

$$d\in \emptyset$$

Solve for x

$$\left\{\begin{matrix}x=\frac{1}{4}\text{, }&d\neq 0\\x\neq 0\text{, }&y=0\text{ and }d\neq 0\end{matrix}\right.$$