Solve 2*x^2-dy/1x=x^2+y^2

$$2 \cdot x ^ { 2 } - \frac { d y } { 1 x } = x ^ { 2 } + y ^ { 2 }$$

Solve for d

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

$$y\neq 0\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$2x^{2}x-dy=xx^{2}+xy^{2}$$

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

$$2x^{3}-dy=xx^{2}+xy^{2}$$

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

$$2x^{3}-dy=x^{3}+xy^{2}$$

Subtract 2x^{3} from both sides.

$$-dy=x^{3}+xy^{2}-2x^{3}$$

Combine x^{3} and -2x^{3} to get -x^{3}.

$$-dy=-x^{3}+xy^{2}$$

The equation is in standard form.

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

Divide both sides by -y.

$$\frac{\left(-y\right)d}{-y}=\frac{x\left(y-x\right)\left(x+y\right)}{-y}$$

Dividing by -y undoes the multiplication by -y.

$$d=\frac{x\left(y-x\right)\left(x+y\right)}{-y}$$

Divide x\left(y-x\right)\left(y+x\right) by -y.

$$d=-\frac{x\left(y-x\right)\left(x+y\right)}{y}$$