Solve x^2+partialy/dx=1

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

Solve for d

$$\left\{\begin{matrix}d=-\frac{y∂}{x\left(x^{2}-1\right)}\text{, }&y\neq 0\text{ and }∂\neq 0\text{ and }x\neq 0\text{ and }|x|\neq 1\\d\neq 0\text{, }&\left(y=0\text{ or }∂=0\right)\text{ and }|x|=1\end{matrix}\right.$$

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.

$$dxx^{2}+∂y=dx$$

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

$$dx^{3}+∂y=dx$$

Subtract dx from both sides.

$$dx^{3}+∂y-dx=0$$

Subtract ∂y from both sides. Anything subtracted from zero gives its negation.

$$dx^{3}-dx=-∂y$$

Reorder the terms.

$$dx^{3}-dx=-y∂$$

Combine all terms containing d.

$$\left(x^{3}-x\right)d=-y∂$$

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

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

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

$$d=-\frac{y∂}{x^{3}-x}$$

Divide -y∂ by x^{3}-x.

$$d=-\frac{y∂}{x\left(x^{2}-1\right)}$$

Variable d cannot be equal to 0.

$$d=-\frac{y∂}{x\left(x^{2}-1\right)}\text{, }d\neq 0$$