Solve 3ay^2dy=ay^3+c

$$3 a y ^ { 2 } d y = a y ^ { 3 } + c$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{c}{\left(3d-1\right)y^{3}}\text{, }&y\neq 0\text{ and }d\neq \frac{1}{3}\\a\in \mathrm{C}\text{, }&\left(y=0\text{ or }d=\frac{1}{3}\right)\text{ and }c=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$3ay^{3}d=ay^{3}+c$$

Subtract ay^{3} from both sides.

$$3ay^{3}d-ay^{3}=c$$

Reorder the terms.

$$3ady^{3}-ay^{3}=c$$

Combine all terms containing a.

$$\left(3dy^{3}-y^{3}\right)a=c$$

Divide both sides by 3dy^{3}-y^{3}.

$$\frac{\left(3dy^{3}-y^{3}\right)a}{3dy^{3}-y^{3}}=\frac{c}{3dy^{3}-y^{3}}$$

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

$$a=\frac{c}{3dy^{3}-y^{3}}$$

Divide c by 3dy^{3}-y^{3}.

$$a=\frac{c}{\left(3d-1\right)y^{3}}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{c}{\left(3d-1\right)y^{3}}\text{, }&y\neq 0\text{ and }d\neq \frac{1}{3}\\a\in \mathrm{R}\text{, }&\left(y=0\text{ or }d=\frac{1}{3}\right)\text{ and }c=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$3ay^{3}d=ay^{3}+c$$

Subtract ay^{3} from both sides.

$$3ay^{3}d-ay^{3}=c$$

Reorder the terms.

$$3ady^{3}-ay^{3}=c$$

Combine all terms containing a.

$$\left(3dy^{3}-y^{3}\right)a=c$$

Divide both sides by 3dy^{3}-y^{3}.

$$\frac{\left(3dy^{3}-y^{3}\right)a}{3dy^{3}-y^{3}}=\frac{c}{3dy^{3}-y^{3}}$$

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

$$a=\frac{c}{3dy^{3}-y^{3}}$$

Divide c by 3dy^{3}-y^{3}.

$$a=\frac{c}{\left(3d-1\right)y^{3}}$$

Solve for c

$$c=a\left(3d-1\right)y^{3}$$