Differential Equations

Concept:Solve the first-order linear differential equation to find $f(x)$, then analyse its derivative and intersection with $g(x)$.Explanation:The given differential equation is $x\frac{dy}{dx} = y - x^3$.Rewrite as $x\,dy - y\,dx = -x^3\,dx$.Divide by $x^2$: $\frac{x\,dy - y\,dx}{x^2} = -x\,dx$.The left side equals $d\!\left(\frac{y}{x}\right)$.Integrate: $\frac{y}{x} = -\frac{x^2}{2} + C$.Multiply by $x$: $y = -\frac{x^3}{2} + Cx$.Use $y(1)=0$: $0 = -\frac12 + C \Rightarrow C = \frac12$.Thus $f(x) = \frac{x}{2} - \frac{x^3}{2}$.Differentiate: $f'(x) = \frac12 - \frac{3x^2}{2} = \frac12(1-3x^2)$.Critical points: $x = \pm\frac{1}{\sqrt3}$. Only $x=\frac{1}{\sqrt3}$ lies in $(0,\infty)$.For $x < \frac{1}{\sqrt3}$, $f'(x)>0$; for $x>\frac{1}{\sqrt3}$, $f'(x)<0$.Hence $f$ has a local maximum at $x=\frac{1}{\sqrt3}$ (option B true, A false).For $x\in(1,2)$, $f'(x)$ is negative (since $1>\frac{1}{\sqrt3}$), so $f$ is decreasing – option C false.Now $g(x)=4x^3-5x^2+\frac32x$.Solve $f(x)=g(x)$: $\frac{x}{2}-\frac{x^3}{2}=4x^3-5x^2+\frac32x$.Multiply by 2: $x - x^3 = 8x^3 -10x^2 + 3x \Rightarrow 0 = 9x^3 -10x^2 +2x$.Factor: $x(9x^2 -10x +2)=0$.$x=0$ is not in $(0,\infty)$.Quadratic discriminant: $\Delta = 100 - 72 = 28 >0$, roots $x = \frac{5\pm\sqrt7}{9}$, both positive.Thus two solutions in $(0,\infty)$ – option D true.Answer:Options B and D are true.