Solve (5x^3+12x^2+6y^2)dx+6xydy=0

$$( 5 x ^ { 3 } + 12 x ^ { 2 } + 6 y ^ { 2 } ) d x + 6 x y d y = 0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&\left(x\leq -\frac{12}{5}\text{ and }|y|=-\frac{\sqrt{-15x-36}x}{6}\right)\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$\left(5x^{3}+12x^{2}+6y^{2}\right)dx+6xy^{2}d=0$$

Use the distributive property to multiply 5x^{3}+12x^{2}+6y^{2} by d.

$$\left(5x^{3}d+12x^{2}d+6y^{2}d\right)x+6xy^{2}d=0$$

Use the distributive property to multiply 5x^{3}d+12x^{2}d+6y^{2}d by x.

$$5dx^{4}+12dx^{3}+6y^{2}dx+6xy^{2}d=0$$

Combine 6y^{2}dx and 6xy^{2}d to get 12y^{2}dx.

$$5dx^{4}+12dx^{3}+12y^{2}dx=0$$

Combine all terms containing d.

$$\left(5x^{4}+12x^{3}+12y^{2}x\right)d=0$$

The equation is in standard form.

$$\left(5x^{4}+12x^{3}+12xy^{2}\right)d=0$$

Divide 0 by 5x^{4}+12x^{3}+12y^{2}x.

$$d=0$$