CUET UG 2025 — Mathematics Calculus

To find the integrating factor, the differential equation must be in standard form:

$$\frac{dy}{dx} + P(x) \cdot y = Q(x)$$dxdy​+P(x)⋅y=Q(x)

The integrating factor is $\displaystyle I.F. = e^{\int P(x)dx}$I.F.=e∫P(x)dx

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Given equation:

$$(x \log_e x)\frac{dy}{dx} + y = 2\log_e x$$(xloge​x)dxdy​+y=2loge​x

Dividing by $(x \log_e x)$(xloge​x):

$$\frac{dy}{dx} + \frac{y}{x \log_e x} = \frac{2\log_e x}{x \log_e x}$$dxdy​+xloge​xy​=xloge​x2loge​x​

$$\frac{dy}{dx} + \frac{1}{x \log_e x} \cdot y = \frac{2}{x}$$dxdy​+xloge​x1​⋅y=x2​

Therefore, $P(x) = \frac{1}{x \log_e x}$P(x)=xloge​x1​

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The integrating factor is:

$$I.F. = e^{\int P(x)dx}$$I.F.=e∫P(x)dx

$$I.F. = e^{\int \frac{1}{x \log_e x}dx}$$I.F.=e∫xloge​x1​dx

To solve $\displaystyle \int \frac{1}{x \log_e x}dx$∫xloge​x1​dx, let $u = \log_e x$u=loge​x

Then $\frac{du}{dx} = \frac{1}{x}$dxdu​=x1​, so $du = \frac{1}{x}dx$du=x1​dx

$$\int \frac{1}{x \log_e x}dx = \int \frac{1}{u}du$$∫xloge​x1​dx=∫u1​du

$$= \log_e u$$=loge​u

$$= \log_e(\log_e x)$$=loge​(loge​x)

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$$I.F. = e^{\log_e(\log_e x)}$$I.F.=eloge​(loge​x)

$$I.F. = \log_e x$$I.F.=loge​x

Therefore, the integrating factor is $\log_e x$loge​x.