Solve (y^2-2xy)(8x+x^2dy)=θ

$$( y ^ { 2 } - 2 x y ) ( 8 x + x ^ { 2 } d y ) = \theta$$

Solve for d

$$\left\{\begin{matrix}d=-\frac{8xy^{2}-16yx^{2}-\theta }{\left(y-2x\right)\left(xy\right)^{2}}\text{, }&x\neq 0\text{ and }y\neq 2x\text{ and }y\neq 0\\d\in \mathrm{R}\text{, }&\left(x=0\text{ or }y=2x\text{ or }y=0\right)\text{ and }\theta =0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply y^{2}-2xy by 8x+x^{2}dy.

$$8y^{2}x+x^{2}dy^{3}-16x^{2}y-2dx^{3}y^{2}=\theta$$

Subtract 8y^{2}x from both sides.

$$x^{2}dy^{3}-16x^{2}y-2dx^{3}y^{2}=\theta -8y^{2}x$$

Add 16x^{2}y to both sides.

$$x^{2}dy^{3}-2dx^{3}y^{2}=\theta -8y^{2}x+16x^{2}y$$

Combine all terms containing d.

$$\left(x^{2}y^{3}-2x^{3}y^{2}\right)d=\theta -8y^{2}x+16x^{2}y$$

The equation is in standard form.

$$\left(x^{2}y^{3}-2y^{2}x^{3}\right)d=\theta +16yx^{2}-8xy^{2}$$

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

$$\frac{\left(x^{2}y^{3}-2y^{2}x^{3}\right)d}{x^{2}y^{3}-2y^{2}x^{3}}=\frac{\theta +16yx^{2}-8xy^{2}}{x^{2}y^{3}-2y^{2}x^{3}}$$

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

$$d=\frac{\theta +16yx^{2}-8xy^{2}}{x^{2}y^{3}-2y^{2}x^{3}}$$

Divide \theta -8y^{2}x+16x^{2}y by x^{2}y^{3}-2x^{3}y^{2}.

$$d=\frac{\theta +16yx^{2}-8xy^{2}}{\left(y-2x\right)\left(xy\right)^{2}}$$