CUET UG 2025 — Mathematics Calculus

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

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$$ydx = (x + 2y^2)dy$$ydx=(x+2y2)dy

$$\frac{dx}{dy} = \frac{x + 2y^2}{y}$$dydx​=yx+2y2​

$$\frac{dx}{dy} = \frac{x}{y} + 2y$$dydx​=yx​+2y

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$$\frac{dx}{dy} - \frac{x}{y} = 2y$$dydx​−yx​=2y

This is in the standard linear form $\frac{dx}{dy} + P(y) \cdot x = Q(y)$dydx​+P(y)⋅x=Q(y) where $P(y) = -\frac{1}{y}$P(y)=−y1​ and $Q(y) = 2y$Q(y)=2y

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The integrating factor is:

$$IF = e^{\int P(y)dy}$$IF=e∫P(y)dy

$$IF = e^{\int -\frac{1}{y}dy}$$IF=e∫−y1​dy

$$IF = e^{-\ln y}$$IF=e−lny

$$IF = e^{\ln y^{-1}}$$IF=elny−1

$$IF = \frac{1}{y}$$IF=y1​

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$$\frac{1}{y} \cdot \frac{dx}{dy} - \frac{1}{y} \cdot \frac{x}{y} = \frac{1}{y} \cdot 2y$$y1​⋅dydx​−y1​⋅yx​=y1​⋅2y

$$\frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} = 2$$y1​dydx​−y2x​=2

The left side is the derivative of $\frac{x}{y}$yx​:

$$\frac{d}{dy}\left(\frac{x}{y}\right) = 2$$dyd​(yx​)=2

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$$\int \frac{d}{dy}\left(\frac{x}{y}\right)dy = \int 2 \, dy$$∫dyd​(yx​)dy=∫2dy

$$\frac{x}{y} = 2y + C$$yx​=2y+C

$$x = 2y^2 + Cy$$x=2y2+Cy

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Therefore, the general solution is $x = 2y^2 + Cy$x=2y2+Cy where $C$C is the constant of integration.