Substitutions — Question 10

Question 10 – Solution

Strategy. The logarithmic independent variable turns multiplication of $x$ into translation of $t$ and removes the factor $x$ from the ratio equation.

Step 1: Derive and design the transformed equation. Writing $y=xv(\ln x)$ gives $y'=v+dv/dt$, so $$\frac{dv}{dt}=F(v)-v.$$ This polynomial has roots $1,3$ and degree at most two, hence equals $k(v-1)(v-3)$. Its value $-1$ at $v=2$ forces $k=1$. Therefore $$\boxed{F(v)=v^2-3v+3},\qquad v'=(v-1)(v-3).$$

Step 2: Solve the selected trajectory. The data give $v(0)=2$, between the equilibria. Separation yields $$\frac 12\ln\left|\frac{v-3}{v-1}\right|=t+C,\qquad C=0.$$ The ratio is negative on the selected branch, so $(v-3)/(v-1)=-e^{2t}$. Thus $$v=1+\frac 2{1+e^{2t}},\qquad \boxed{y=x+\frac{2x}{1+x^2},\quad x>0}.$$ Directly $dv/dt=-4e^{2t}/(1+e^{2t})^2=(v-1)(v-3)$, so the chain rule verifies the original equation and $y(1)=2$.

Step 3: Interpret the limits and equilibria. As $x\downarrow 0$, $t\to-\infty$ and $y/x\to 3$; as $x\to\infty$, $t\to\infty$ and $y/x\to 1$. The constant ratios give $\boxed{y=x,\ y=3x}$, both valid on $x>0$. They are straight lines, not constant original functions: constancy applies to the ratio in the new coordinates. Moreover, $y'=F(v)=(v-3/2)^2+3/4>0$, so the original $y$ increases even though its ratio $v$ decreases.