CUET UG 2025 — Mathematics Calculus

Given the differential equation $xdy = (2x^2 + 1)dx$xdy=(2x2+1)dx, where $x \neq 0$x=0, with the initial condition $y = 1$y=1 when $x = 1$x=1.

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Divide both sides by $x$x:

$$dy = \frac{2x^2 + 1}{x}dx$$dy=x2x2+1​dx

Split the fraction:

$$dy = \left(\frac{2x^2}{x} + \frac{1}{x}\right)dx$$dy=(x2x2​+x1​)dx

$$dy = \left(2x + \frac{1}{x}\right)dx$$dy=(2x+x1​)dx

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Integrate both sides:

$$\int dy = \int \left(2x + \frac{1}{x}\right)dx$$∫dy=∫(2x+x1​)dx

$$y = x^2 + \log|x| + C$$y=x2+log∣x∣+C

where $C$C is the constant of integration.

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Using the initial condition $y = 1$y=1 when $x = 1$x=1:

$$1 = (1)^2 + \log|1| + C$$1=(1)2+log∣1∣+C

$$1 = 1 + 0 + C$$1=1+0+C

$$C = 0$$C=0

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Therefore, the particular solution is:

$$y = x^2 + \log|x|$$y=x2+log∣x∣