CUET UG 2025 — Mathematics Arithmetic

Given:

$$\frac{dy}{dx} = \sqrt{1+x^2+y^2+x^2y^2}$$dxdy​=1+x2+y2+x2y2​

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The expression under the square root can be factored:

$$1+x^2+y^2+x^2y^2 = (1+x^2) + y^2(1+x^2)$$1+x2+y2+x2y2=(1+x2)+y2(1+x2)

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

The equation becomes:

$$\frac{dy}{dx} = \sqrt{(1+x^2)(1+y^2)}$$dxdy​=(1+x2)(1+y2)​

$$\frac{dy}{dx} = \sqrt{1+x^2} \cdot \sqrt{1+y^2}$$dxdy​=1+x2​⋅1+y2​

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Separating variables:

$$\frac{dy}{\sqrt{1+y^2}} = \sqrt{1+x^2} \, dx$$1+y2​dy​=1+x2​dx

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Integrating both sides:

$$\int \frac{dy}{\sqrt{1+y^2}} = \int \sqrt{1+x^2} \, dx$$∫1+y2​dy​=∫1+x2​dx

$$\log|y + \sqrt{1+y^2}| = \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log|x + \sqrt{1+x^2}| + C$$log∣y+1+y2​∣=2x​1+x2​+21​log∣x+1+x2​∣+C

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Rearranging:

$$\log|y + \sqrt{1+y^2}| - \frac{1}{2}\log|x + \sqrt{1+x^2}| = \frac{x}{2}\sqrt{1+x^2} + C$$log∣y+1+y2​∣−21​log∣x+1+x2​∣=2x​1+x2​+C

Using logarithm properties $\log a - \log b = \log\frac{a}{b}$loga−logb=logba​ and $\frac{1}{2}\log a = \log\sqrt{a}$21​loga=loga​:

$$\log\left|\frac{y + \sqrt{1+y^2}}{\sqrt{x + \sqrt{1+x^2}}}\right| = \frac{x}{2}\sqrt{1+x^2} + C$$log​x+1+x2​​y+1+y2​​​=2x​1+x2​+C

Therefore:

$$\log\left|\frac{y+\sqrt{1+y^2}}{\sqrt{x+\sqrt{1+x^2}}}\right| = \frac{x}{2}\sqrt{1+x^2} + C$$log​x+1+x2​​y+1+y2​​​=2x​1+x2​+C