CUET UG 2025 — Mathematics Calculus

Given differential equation: $x(1 + y^2)dx - y(1 + x^2)dy = 0$x(1+y2)dx−y(1+x2)dy=0

Initial condition: When $x = 1$x=1, $y = 0$y=0

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

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

$$\int \frac{x}{1 + x^2}dx = \frac{1}{2}\int \frac{1}{u}du = \frac{1}{2}\ln|1 + x^2|$$∫1+x2x​dx=21​∫u1​du=21​ln∣1+x2∣

Similarly:

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

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$$\frac{1}{2}\ln(1 + x^2) = \frac{1}{2}\ln(1 + y^2) + C$$21​ln(1+x2)=21​ln(1+y2)+C

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

$$\ln(1 + x^2) - \ln(1 + y^2) = 2C$$ln(1+x2)−ln(1+y2)=2C

$$\ln\left(\frac{1 + x^2}{1 + y^2}\right) = 2C$$ln(1+y21+x2​)=2C

$$\frac{1 + x^2}{1 + y^2} = e^{2C} = A$$1+y21+x2​=e2C=A

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

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When $x = 1$x=1, $y = 0$y=0:

$$(1 + 1^2) = A(1 + 0^2)$$(1+12)=A(1+02)

$$2 = A$$2=A

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