CUET UG 2025 — Mathematics Calculus

The differential equation $\frac{dy}{dx} = (1 + x^2)(1 + y^2)$dxdy​=(1+x2)(1+y2) is a separable differential equation.

The right side is a product of a function of $x$x and a function of $y$y.

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Separating the variables:

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

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Integrating both sides:

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

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For the left side:

$$\int \frac{dy}{(1 + y^2)} = \tan^{-1}y$$∫(1+y2)dy​=tan−1y

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For the right side:

$$\int (1 + x^2)dx = \int 1 \cdot dx + \int x^2 dx$$∫(1+x2)dx=∫1⋅dx+∫x2dx

$$= x + \frac{x^3}{3}$$=x+3x3​

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Combining both sides:

$$\tan^{-1}y = x + \frac{x^3}{3} + C$$tan−1y=x+3x3​+C

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Rearranging:

$$\tan^{-1}y - x - \frac{x^3}{3} = C$$tan−1y−x−3x3​=C

Therefore, the solution is $\tan^{-1} y - x - \frac{x^3}{3} = C$tan−1y−x−3x3​=C