CUET UG 2025 — Mathematics Calculus

The differential equation is $x(1 + y^2)dx + y(1 + x^2)dy = 0$x(1+y2)dx+y(1+x2)dy=0

Rearranging to separate variables:

$$x(1 + y^2)dx = -y(1 + x^2)dy$$x(1+y2)dx=−y(1+x2)dy

$$\frac{x}{1 + x^2}dx = -\frac{y}{1 + y^2}dy$$1+x2x​dx=−1+y2y​dy

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Integrating both sides:

$$\int \frac{x}{1 + x^2}dx = -\int \frac{y}{1 + y^2}dy$$∫1+x2x​dx=−∫1+y2y​dy

For the left side, let $u = 1 + x^2$u=1+x2, then $du = 2x \, dx$du=2xdx:

$$\int \frac{x}{1 + x^2}dx = \frac{1}{2}\ln|1 + x^2|$$∫1+x2x​dx=21​ln∣1+x2∣

For the right side, let $v = 1 + y^2$v=1+y2, then $dv = 2y \, dy$dv=2ydy:

$$-\int \frac{y}{1 + y^2}dy = -\frac{1}{2}\ln|1 + y^2|$$−∫1+y2y​dy=−21​ln∣1+y2∣

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Combining the results:

$$\frac{1}{2}\ln|1 + x^2| = -\frac{1}{2}\ln|1 + y^2| + C_1$$21​ln∣1+x2∣=−21​ln∣1+y2∣+C1​

$$\ln|1 + x^2| = -\ln|1 + y^2| + C_2$$ln∣1+x2∣=−ln∣1+y2∣+C2​

$$\ln|1 + x^2| + \ln|1 + y^2| = C_2$$ln∣1+x2∣+ln∣1+y2∣=C2​

$$\ln|(1 + x^2)(1 + y^2)| = C_2$$ln∣(1+x2)(1+y2)∣=C2​

$$(1 + x^2)(1 + y^2) = e^{C_2}$$(1+x2)(1+y2)=eC2​

Since $e^{C_2}$eC2​ is an arbitrary constant, let $C = e^{C_2}$C=eC2​:

$$(1 + x^2)(1 + y^2) = C$$(1+x2)(1+y2)=C