CUET UG 2025 — Mathematics Calculus

Given differential equation: $\frac{dy}{dx} + \frac{3y}{x} = 0$dxdy​+x3y​=0 with initial condition $y(1) = 1$y(1)=1.

$$\frac{dy}{dx} = -\frac{3y}{x}$$dxdy​=−x3y​

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$$\frac{dy}{y} = -\frac{3}{x}dx$$ydy​=−x3​dx

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$$\int \frac{dy}{y} = \int -\frac{3}{x}dx$$∫ydy​=∫−x3​dx

$$\ln|y| = -3\ln|x| + C$$ln∣y∣=−3ln∣x∣+C

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Using the property $n\ln(x) = \ln(x^n)$nln(x)=ln(xn):

$$\ln|y| = \ln|x^{-3}| + C$$ln∣y∣=ln∣x−3∣+C

$$\ln|y| = \ln\left(\frac{1}{x^3}\right) + C$$ln∣y∣=ln(x31​)+C

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Taking exponential on both sides:

$$y = e^C \cdot \frac{1}{x^3}$$y=eC⋅x31​

Let $e^C = A$eC=A:

$$y = \frac{A}{x^3}$$y=x3A​

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Using the initial condition $y(1) = 1$y(1)=1:

$$1 = \frac{A}{1^3}$$1=13A​

$$1 = \frac{A}{1}$$1=1A​

$$A = 1$$A=1

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Therefore, the particular solution is:

$$y = \frac{1}{x^3}$$y=x31​