CUET UG 2025 — Mathematics Algebra

The differential equation is $x^2 \frac{dy}{dx} + xy = \log_e x$x2dxdy​+xy=loge​x.

The standard form of a linear differential equation is:

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

Dividing the entire equation by $x^2$x2:

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

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Comparing with the standard form $\frac{dy}{dx} + P(x) \cdot y = Q(x)$dxdy​+P(x)⋅y=Q(x):

$$P(x) = \frac{1}{x}$$P(x)=x1​

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The integrating factor formula is:

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

Substituting $P(x) = \frac{1}{x}$P(x)=x1​:

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

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Since $\displaystyle \int \frac{1}{x}dx = \log_e x$∫x1​dx=loge​x:

$$\text{I.F.} = e^{\log_e x}$$I.F.=eloge​x

Using the property $e^{\log_e x} = x$eloge​x=x:

$$\text{I.F.} = x$$I.F.=x

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Therefore, the integrating factor is $x$x.