Question

The solution of the differential equation x dx + y dy = x² y dy − y² x dx, is

विकल्प

• x² − 1 = C (1 + y²)

• x² + 1 = C (1 − y²)

• x³ − 1 = C (1 + y³)

• x³ + 1 = C (1 − y³)

Answer 1

x² − 1 = C (1 + y²)

 

We have,

x dx + y dy = x²y dy − y²x dx

$$\Rightarrow (x+x{y}^2)dx=(x^{2}y-y)dy$$

$$\Rightarrow \frac{x}{(x^2-1)}dx=\frac{y}{(1+y^2)}dy$$

$$\Rightarrow \frac{2x}{2(x^2-1)}dx=\frac{2y}{2(1+y^2)}dy$$

Integrating both sides, we get

$$\frac{1}{2}\int \frac{2y}{(1+y^2)}dy=\frac{1}{2}\int \frac{2x}{(x^2-1)}dx$$

$$\Rightarrow \frac{1}{2}\log \left|(1+y^2)\right|=\frac{1}{2}\log \left|(x^2-1)\right|-\frac{1}{2}\log \left|C\right|$$

$$\Rightarrow \log \left|(1+y^2)\right|=\log \left|(x^2-1)\right|-\log \left|C\right|$$

$$\Rightarrow \log \left|(1+y^2)\right|=\log \left|\left(\frac{x^2-1}{C}\right)\right|$$

$$\Rightarrow 1+y^2=\frac{x^2-1}{C}$$

$$\Rightarrow C(1+y^2)=x^2-1$$