Solve 3x(4+y^2)dx+2x^2y(1-x)^2dy=0

$$3 x ( 4 + y ^ { 2 } ) d x + 2 x ^ { 2 } y ( 1 - x ) ^ { 2 } d y = 0$$

Solve for d

$$d=0$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$3x^{2}\left(4+y^{2}\right)d+2x^{2}y\left(1-x\right)^{2}dy=0$$

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

$$3x^{2}\left(4+y^{2}\right)d+2x^{2}y^{2}\left(1-x\right)^{2}d=0$$

Use the distributive property to multiply 3x^{2} by 4+y^{2}.

$$\left(12x^{2}+3x^{2}y^{2}\right)d+2x^{2}y^{2}\left(1-x\right)^{2}d=0$$

Use the distributive property to multiply 12x^{2}+3x^{2}y^{2} by d.

$$12x^{2}d+3x^{2}y^{2}d+2x^{2}y^{2}\left(1-x\right)^{2}d=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$12x^{2}d+3x^{2}y^{2}d+2x^{2}y^{2}\left(1-2x+x^{2}\right)d=0$$

Use the distributive property to multiply 2x^{2}y^{2} by 1-2x+x^{2}.

$$12x^{2}d+3x^{2}y^{2}d+\left(2x^{2}y^{2}-4x^{3}y^{2}+2y^{2}x^{4}\right)d=0$$

Use the distributive property to multiply 2x^{2}y^{2}-4x^{3}y^{2}+2y^{2}x^{4} by d.

$$12x^{2}d+3x^{2}y^{2}d+2x^{2}y^{2}d-4x^{3}y^{2}d+2y^{2}x^{4}d=0$$

Combine 3x^{2}y^{2}d and 2x^{2}y^{2}d to get 5x^{2}y^{2}d.

$$12x^{2}d+5x^{2}y^{2}d-4x^{3}y^{2}d+2y^{2}x^{4}d=0$$

Combine all terms containing d.

$$\left(12x^{2}+5x^{2}y^{2}-4x^{3}y^{2}+2y^{2}x^{4}\right)d=0$$

The equation is in standard form.

$$\left(5x^{2}y^{2}+12x^{2}+2y^{2}x^{4}-4y^{2}x^{3}\right)d=0$$

Divide 0 by 12x^{2}+5x^{2}y^{2}-4x^{3}y^{2}+2y^{2}x^{4}.

$$d=0$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$