Substitutions — Question 9

Question 9 – Solution

Strategy. Subtract a known particular solution so that the constant term cancels, then use a reciprocal on the remaining quadratic equation.

Step 1: Verify and subtract the particular solution. For $y_p=x$, the right-hand side is $x^2+(1-2x)x+x^2-x+1=1=y_p'$. Now write $y=x+z$. Substitution and cancellation give $$1+z'=1+z+z^2,\qquad \boxed{z'=z+z^2},\qquad z(0)=2.$$

Step 2: Make the remaining equation linear. For $z\ne 0$, $u=1/z$ gives $$u'=-\frac{z'}{z^2}=-u-1,\qquad u(0)=\frac 12.$$ Solving $u'+u=-1$ yields $u=Ce^{-x}-1$, and the initial value requires $C=3/2$.

Step 3: Recover the IVP and its interval. Thus $$\boxed{y=x+\frac 1{\frac 32e^{-x}-1},\qquad I=(-\infty,\ln(3/2))}.$$ The denominator is positive before its unique zero at $\ln(3/2)$ and gives $y(0)=2$. At that endpoint $y\to+\infty$, preventing a finite differentiable extension.

Step 4: Verify and account for the missing solution. Where $u\ne 0$, $z=1/u$ satisfies $z'=-u'/u^2=z+z^2$. Thus $y'=1+z+z^2$, which is exactly the original right side after $y=x+z$. The excluded case $z\equiv 0$ restores $\boxed{y=x}$, a global solution with a different initial value.

The original equation contains a nonzero term independent of $y$. A direct reciprocal $w=1/y$ would retain a quadratic term in $w$ and would not be linear. Subtracting the known solution first cancels that obstruction. This two-step substitution is the useful structure, not the reciprocal alone.