Differential Equations — Linear ODE & Extrema
DAILY_CHALLENGE
Grade None
Question:
Let $y=f(x)$ be the real valued function defined on the interval $(0,\infty)$, satisfying $y(1)=0$ and the differential equation
$$x\frac{dy}{dx}=y-x^3.$$
Then which of the following statements is (are) TRUE?
The function $f$ has a local minimum at $x=\dfrac{1}{\sqrt{3}}$
The function $f$ has a local maximum at $x=\dfrac{1}{\sqrt{3}}$
The function $f$ is increasing in the interval $(1,2)$
If $g(x)=4x^3-5x^2+\dfrac{3}{2}x$ for $x>0$, then the number of elements in the set $\{x\in(0,\infty):f(x)=g(x)\}$ is $2$
Step-by-Step Solution
Key Concept: After solving the linear ODE to get $f(x)=\frac{x(1-x^2)}{2}$, analysis is straightforward. The second derivative test gives a local maximum (not minimum) at the critical point.
**Step 1: Solve the ODE**
Rewrite: $\frac{dy}{dx}-\frac{y}{x}=-x^2$. Integrating factor $=1/x$. $\frac{d}{dx}(y/x)=-x$. $y/x=-x^2/2+C$. $y=-x^3/2+Cx$. Using $y(1)=0$: $C=1/2$. So $f(x)=\frac{x-x^3}{2}$.
**Step 2: Find critical points**
$f'(x)=\frac{1-3x^2}{2}=0 \Rightarrow x=\frac{1}{\sqrt{3}}$. $f''(x)=-3x<0$ at $x=1/\sqrt{3}$. Local MAXIMUM. (A) FALSE, (B) TRUE ✓
**Step 3: Check (C): monotonicity on $(1,2)$**
$f'(1)=\frac{1-3}{2}=-1<0$. $f$ is decreasing on $(1,2)$. (C) FALSE ✗
**Step 4: Check (D): solutions to $f(x)=g(x)$**
$\frac{x-x^3}{2}=4x^3-5x^2+\frac{3}{2}x$. Multiply by 2: $x-x^3=8x^3-10x^2+3x \Rightarrow 9x^3-10x^2+2x=0 \Rightarrow x(9x^2-10x+2)=0$. For $x>0$: $9x^2-10x+2=0 \Rightarrow x=\frac{10\pm\sqrt{28}}{18}=\frac{5\pm\sqrt{7}}{9}$. Both positive. 2 elements. (D) TRUE ✓
Correct Answer: B, D