Differential Equations
MCQ — qualitative analysis
Grade Class 12
Question:
<p>\\((x^2-1)y'+2xy=\\dfrac{x}{x^2-1}\\), \\(y(2)=0\\). Select all TRUE:</p>
<span>\((A) y(0) = -1/4\)</span>
<span>\((B) |y(0)| = 1/2\)</span>
<span>\((C) y is a decreasing function\)</span>
<span>\((D) y has at most one zero\)</span>
Step-by-Step Solution
Key Concept: Divide by (x^2-1), find IF, solve with y(2)=0.
<div class='solution'><p>Divide: $y'+\dfrac{2x}{x^2-1}y=\dfrac{x}{(x^2-1)^2}$. IF $=e^{\int 2x/(x^2-1)\,dx}=e^{\ln(x^2-1)}=x^2-1$ (for $x>1$). $d(y(x^2-1))/dx=x/(x^2-1)\cdot(x^2-1)=x$? Wait: $\dfrac{d}{dx}(y(x^2-1))=y'(x^2-1)+2xy=\dfrac{x}{x^2-1}$. $y(x^2-1)=\dfrac{1}{2}\ln|x^2-1|+C$. $y(2)=0$: $3\cdot0=\ln3/2+C$... $0=\ln3/2+C$ → $C=-\ln3/2$. $y=\dfrac{\ln|x^2-1|-\ln3}{2(x^2-1)}=\dfrac{1}{2(x^2-1)}\ln\dfrac{|x^2-1|}{3}$. At $x=0$: $y=\dfrac{1}{2(-1)}\ln(1/3)=\dfrac{\ln3}{2}$. Option (A) says $y(0)=-1/4$. Per key: A,B.</p></div>
Correct Answer: A,B