Bernoulli Differential Equations — Question 6

Question 6 – Solution

Strategy. The reciprocal transformation turns coefficient recovery into two linear identities; their difference eliminates the forcing term.

Step 1: Recover the coefficients. Each transformed solution satisfies $v'-pv=-q$. Here $v_1=1+x$ and $v_2=1+2x$, so $$1-p(1+x)=-q,\qquad 2-p(1+2x)=-q.$$ Subtracting gives $1-px=0$. Since $x>0$, this fixes $p=1/x$ uniquely. Substituting back gives $q=1/x$. Thus $$\boxed{y'+\frac yx=\frac{y^2}{x}}.$$

Step 2: Verify the given data. For $y=1/(1+kx)$, with $k=1$ or $2$, $$y'+\frac yx=-\frac{k}{(1+kx)^2}+\frac 1{x(1+kx)} =\frac 1{x(1+kx)^2}=\frac{y^2}{x}.$$ Both functions are positive and defined throughout $x>0$.

Step 3: Construct the new solution. The transformed equation is $v'-v/x=-1/x$. Multiplying by $1/x$ gives $(v/x)'=-1/x^2$, hence $v=1+Cx$. Since $y(1)=2$, $v(1)=1/2$, so $C=-1/2$. Therefore $$\boxed{y=\frac 1{1-x/2},\qquad I=(0,2)}.$$ This is the maximal interval within the specified domain through $1$: the solution has a pole at $2$, and the coefficients are undefined at $0$.

Step 4: Explain the advantage and restore zero. Subtracting the transformed equations gives the homogeneous linear relation $\Delta v'-p\Delta v=0$. Subtracting the original equations leaves $q(y_2^2-y_1^2)$ and does not give that linear elimination. The function $\boxed{y\equiv 0}$ satisfies the recovered equation but is excluded by the reciprocal transformation.