CUET UG 2025 — Mathematics Calculus

Rearranging the given equation into standard linear form:

$$\dfrac{dy}{dx} - xy = e^{x^2/2}$$dxdy​−xy=ex2/2

This is a first-order linear differential equation of the form $\dfrac{dy}{dx} + P(x)\,y = Q(x)$dxdy​+P(x)y=Q(x), where $P(x) = -x$P(x)=−x and $Q(x) = e^{x^2/2}$Q(x)=ex2/2.

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Finding the Integrating Factor:

$$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int -x\,dx} = e^{-x^2/2}$$I.F.=e∫P(x)dx=e∫−xdx=e−x2/2

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Using the solution formula $\displaystyle y \times \text{I.F.} = \int Q(x) \times \text{I.F.}\,dx + C$y×I.F.=∫Q(x)×I.F.dx+C, we get:

$$y \cdot e^{-x^2/2} = \int e^{x^2/2} \cdot e^{-x^2/2}\,dx + C$$y⋅e−x2/2=∫ex2/2⋅e−x2/2dx+C

Since $e^{x^2/2} \cdot e^{-x^2/2} = e^0 = 1$ex2/2⋅e−x2/2=e0=1, this simplifies to:

$$y \cdot e^{-x^2/2} = \int 1\,dx + C$$y⋅e−x2/2=∫1dx+C

$$y \cdot e^{-x^2/2} = x + C$$y⋅e−x2/2=x+C

Multiplying both sides by $e^{x^2/2}$ex2/2:

$$y = (x + C)\,e^{x^2/2}$$y=(x+C)ex2/2

This is the general solution.

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Applying the initial condition $y(0) = 1$y(0)=1:

$$1 = (0 + C)\,e^{0}$$1=(0+C)e0

$$C = 1$$C=1

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The particular solution is:

$$\boxed{y = (x + 1)\,e^{x^2/2}}$$y=(x+1)ex2/2​