Solve yx^2dx-(x^3+y^3)dy=0,y(1)=2

$$y x ^ { 2 } d x - ( x ^ { 3 } + y ^ { 3 } ) d y = 0 , y ( 1 ) = 2$$

Solve for x, y (complex solution)

$$x\in \mathrm{C}$$

$$y=2$$

$$d=0$$

Solution Steps

Consider the second equation. Divide both sides by 1.

$$y=2$$

Consider the first equation. Insert the known values of variables into the equation.

$$2x^{2}dx-\left(x^{3}+2^{3}\right)d\times 2=0$$

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

$$2x^{3}d-\left(x^{3}+2^{3}\right)d\times 2=0$$

Calculate 2 to the power of 3 and get 8.

$$2x^{3}d-\left(x^{3}+8\right)d\times 2=0$$

Use the distributive property to multiply x^{3}+8 by d.

$$2x^{3}d-\left(x^{3}d+8d\right)\times 2=0$$

Use the distributive property to multiply x^{3}d+8d by 2.

$$2x^{3}d-\left(2x^{3}d+16d\right)=0$$

To find the opposite of 2x^{3}d+16d, find the opposite of each term.

$$2x^{3}d-2x^{3}d-16d=0$$

Combine 2x^{3}d and -2x^{3}d to get 0.

$$-16d=0$$

Divide both sides by -16. Zero divided by any non-zero number gives zero.

$$d=0$$

The system is now solved.

$$d=0 y=2$$

Solve for x, y

$$x\in \mathrm{R}$$

$$y=2$$

$$d=0$$

Solution Steps

Consider the second equation. Divide both sides by 1.

$$y=2$$

Consider the first equation. Insert the known values of variables into the equation.

$$2x^{2}dx-\left(x^{3}+2^{3}\right)d\times 2=0$$

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

$$2x^{3}d-\left(x^{3}+2^{3}\right)d\times 2=0$$

Calculate 2 to the power of 3 and get 8.

$$2x^{3}d-\left(x^{3}+8\right)d\times 2=0$$

Use the distributive property to multiply x^{3}+8 by d.

$$2x^{3}d-\left(x^{3}d+8d\right)\times 2=0$$

Use the distributive property to multiply x^{3}d+8d by 2.

$$2x^{3}d-\left(2x^{3}d+16d\right)=0$$

To find the opposite of 2x^{3}d+16d, find the opposite of each term.

$$2x^{3}d-2x^{3}d-16d=0$$

Combine 2x^{3}d and -2x^{3}d to get 0.

$$-16d=0$$

Divide both sides by -16. Zero divided by any non-zero number gives zero.

$$d=0$$

The system is now solved.

$$d=0 y=2$$