The solution of the differential equation \(\displaystyle\frac{dy}{dx}-\frac{y}{x}=2 \log_ex\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(y=x((\log_ex)^2 + C)\): C is an arbitrary constant
This is a linear differential equation \(\frac{dy}{dx} + Py = Q\) with \(P = -\frac{1}{x}\) and \(Q = 2\log_e x\).
\( \displaystyle \text{I.F.} = e^{\int -\frac{1}{x}\,dx} = e^{-\log_e x} = \frac{1}{x} \)
The solution is \(y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx\):
\( \displaystyle \frac{y}{x} = \int \frac{2\log_e x}{x}\,dx \)
Putting \(t = \log_e x\), \(dt = \frac{dx}{x}\):
\( \displaystyle \frac{y}{x} = \int 2t\,dt = t^2 + C \)
\( \displaystyle \frac{y}{x} = (\log_e x)^2 + C \)
\( y = x\left((\log_e x)^2 + C\right) \)