Solve d/dx^2+xdy/dx+xy=0

$$\frac { d } { d x ^ { 2 } } + x \frac { d y } { d x } + x y = 0$$

Solve for d

$$d\neq 0$$

$$x=-\frac{1}{\sqrt[3]{y}}\text{ and }y\neq 0$$

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^{2}.

$$d+x\frac{\mathrm{d}(y)}{\mathrm{d}x}dx^{2}+xydx^{2}=0$$

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

$$d+x^{3}\frac{\mathrm{d}(y)}{\mathrm{d}x}d+xydx^{2}=0$$

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

$$d+x^{3}\frac{\mathrm{d}(y)}{\mathrm{d}x}d+x^{3}yd=0$$

Combine all terms containing d.

$$\left(1+x^{3}\frac{\mathrm{d}(y)}{\mathrm{d}x}+x^{3}y\right)d=0$$

The equation is in standard form.

$$\left(yx^{3}+1\right)d=0$$

Divide 0 by 1+x^{3}y.

$$d=0$$

Variable d cannot be equal to 0.

$$d\in \emptyset$$

Solve for x

$$x=-\frac{1}{\sqrt[3]{y}}$$

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