CUET UG 2025 — Mathematics Calculus

The given differential equation is:

$$\frac{dy}{dx} = xy + x + y + 1$$dxdy​=xy+x+y+1

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Factoring the right side by grouping terms:

$$\frac{dy}{dx} = (xy + x) + (y + 1)$$dxdy​=(xy+x)+(y+1)

$$\frac{dy}{dx} = x(y + 1) + 1(y + 1)$$dxdy​=x(y+1)+1(y+1)

$$\frac{dy}{dx} = (y + 1)(x + 1)$$dxdy​=(y+1)(x+1)

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Separating the variables:

$$\frac{dy}{y + 1} = (x + 1) dx$$y+1dy​=(x+1)dx

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Integrating both sides:

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

For the left side:

$$\int \frac{dy}{y + 1} = \log_e|y + 1|$$∫y+1dy​=loge​∣y+1∣

For the right side:

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

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The general solution is:

$$\log_e|y + 1| = \frac{1}{2}x^2 + x + C$$loge​∣y+1∣=21​x2+x+C

where $C$C is the constant of integration.