Solve partialy/dx=2x+3

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

Solve for d

$$\left\{\begin{matrix}d=\frac{y∂}{x\left(2x+3\right)}\text{, }&y\neq 0\text{ and }∂\neq 0\text{ and }x\neq -\frac{3}{2}\text{ and }x\neq 0\\d\neq 0\text{, }&\left(∂=0\text{ or }y=0\right)\text{ and }x=-\frac{3}{2}\end{matrix}\right.$$

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.

$$∂y=2xdx+dx\times 3$$

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

$$∂y=2x^{2}d+dx\times 3$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}d+dx\times 3=∂y$$

Combine all terms containing d.

$$\left(2x^{2}+x\times 3\right)d=∂y$$

The equation is in standard form.

$$\left(2x^{2}+3x\right)d=y∂$$

Divide both sides by 2x^{2}+3x.

$$\frac{\left(2x^{2}+3x\right)d}{2x^{2}+3x}=\frac{y∂}{2x^{2}+3x}$$

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

$$d=\frac{y∂}{2x^{2}+3x}$$

Divide ∂y by 2x^{2}+3x.

$$d=\frac{y∂}{x\left(2x+3\right)}$$

Variable d cannot be equal to 0.

$$d=\frac{y∂}{x\left(2x+3\right)}\text{, }d\neq 0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{d\left(8y∂+9d\right)}-3d}{4d}\text{, }&\left(arg(d)\geq \pi \text{ and }d\neq 0\right)\text{ or }\left(y\neq 0\text{ and }∂\neq 0\text{ and }d\neq 0\right)\\x=-\frac{\sqrt{d\left(8y∂+9d\right)}+3d}{4d}\text{, }&\left(arg(d)<\pi \text{ and }d\neq 0\right)\text{ or }\left(y\neq 0\text{ and }∂\neq 0\text{ and }d\neq 0\right)\end{matrix}\right.$$