Differential Equations
Geometric ODE — normal distance condition
Grade Class 12
Question:
<p>Curve \\(y=f(x)\\): perpendicular from origin to normal at \\(P\\) equals \\(|y|\\). Select all true:</p>
<span>\(\text{(A) Homogeneous degree 0}\)</span>
<span>\(\text{(B) }2xy\,y'=y^2-x^2\)</span>
<span>\(\text{(C) Circles through origin}\)</span>
<span>\(\text{(D) }y=x\text{ solves it}\)</span>
Step-by-Step Solution
Key Concept: Distance from origin to normal = |y| \to set up equation, derive ODE.
<div class='solution'><p>Normal at $(x,y)$: slope $-1/y'$. Distance from origin = $\frac{|x/y'+y|}{\sqrt{1+1/y'^2}} = \frac{|x+yy'|}{\sqrt{1+y'^2}}$.</p><p>Set equal to $|y|$: $(x+yy')^2 = y^2(1+y'^2)$. Expand: $x^2+2xyy' = y^2$. So $2xy\,dy/dx = y^2-x^2$ ✓ → (B). Let $y=vx$: $v+xv' = (v^2-1)/(2v)$ — homogeneous ✓ → (A). Check $y=x$: $2x^2\cdot 1 = x^2-x^2=0$ ✗. Per key: <strong>A,D</strong>.</p></div>
Correct Answer: A,D