The particular solution of the differential equation \( \displaystyle\frac{dy}{dx}=x(2\log x+1) \) , given that \( y=0 \) when \( x=2 \) , is:
( consider \( \log x=\log_e x \) )
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( y=x^2\log x-4\log 2 \)
The variables are already separate: \( dy=(2x\log x+x)\,dx \). The term \( 2x\log x \) is integrated by parts, with \( u=\log x \), \( dv=2x\,dx \):
\( \displaystyle \int 2x\log x\,dx \)
\( \displaystyle =x^2\log x-\int x^2\cdot\frac{1}{x}\,dx \)
\( \displaystyle =x^2\log x-\frac{x^2}{2} \)
Adding \( \int x\,dx=\frac{x^2}{2} \):
\( \displaystyle y=x^2\log x-\frac{x^2}{2}+\frac{x^2}{2}+C \)
\( \displaystyle y=x^2\log x+C \)
Putting y = 0 when x = 2:
\( \displaystyle 0=4\log 2+C\ \Rightarrow\ C=-4\log 2 \)
\( \displaystyle y=x^2\log x-4\log 2 \)
Shortcut: By the product rule, \( \frac{d}{dx}(x^2\log x)=2x\log x+x \), exactly the right side; so \( y=x^2\log x+C \), and only option (3) gives y = 0 at x = 2.