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CUET · MATHS · PYQ PAPER 2025

The general solution of the differential equation \(\frac{x d y}{d x}+4 y=x^3,(x \neq 0)\) is :

  1. A \(y=\frac{x^3}{7}+c x^{-4}, c\) is arbitrary constant.
  2. B \(y=\frac{x^3}{7}+c x^{-7}, c\) is arbitrary constant.
  3. C \(y=\frac{x^7}{7}+c x^{-4}, c\) is arbitrary constant.
  4. D \(y=\frac{x^7}{7}+c x^{-3}, c\) is arbitrary constant.
Verified Solution

Answer & Solution

Correct Answer

(A) \(y=\frac{x^3}{7}+c x^{-4}, c\) is arbitrary constant.

Step-by-step Solution

Detailed explanation

\(\frac{d y}{d x}+\frac{4}{x} y=x^2\) \(P(x)=\frac{4}{x}, Q(x)=x^2\) \(I.F. = e^{\int \frac{4}{x} d x} = e^{4 \ln|x|} = e^{\ln x^4} = x^4\) \(y \cdot x^4 = \int x^2 \cdot x^4 d x\) \(y x^4 = \int x^6 d x\) \(y x^4 = \frac{x^7}{7} + c\) \(y = \frac{x^7}{7x^4} + \frac{c}{x^4}\)…
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