Substitutions — Question 6

Question 6 – Solution

Strategy. When the affine forms cannot be centered together, use their shared linear combination directly.

Step 1: Reject a simultaneous centering. A translation by $(h,k)$ would require $2h+k+1=0$ and $4h+2k+3=0$. Doubling the first equation gives $4h+2k+2=0$, contradicting the second. Their zero lines are parallel and distinct.

Step 2: Reduce and integrate. Set $u=2x+y+1$. Then the denominator is $2u+1$, and $$u'=2+y'=\frac{5u+2}{2u+1},\qquad \frac{2u+1}{5u+2}\,du=dx.$$ The original domain excludes $u=-1/2$. For $u\ne-2/5$, an antiderivative is $$H(u)=\frac 25u+\frac 1{25}\ln|5u+2|.$$ Since $u(0)=1$, the IVP relation is $$\boxed{H(u)=x+\frac 25+\frac 1{25}\ln 7,\qquad y=u-2x-1}.$$

Step 3: Restore the lost constant combination. The transformed equation admits $u\equiv-2/5$, giving $\boxed{y=-2x-7/5}$. Its numerator is $-2/5$ and its denominator is $1/5$, so the right side is $-2=y'$. It is a global solution, but its initial value is not $0$.

Step 4: Prove the selected branch is global. The initial value lies in $u>-2/5$. On this interval, $$H'(u)=\frac{2u+1}{5u+2}>0,$$ and $H(u)$ tends to $-\infty$ at the left endpoint and to $+\infty$ as $u\to\infty$. Thus the implicit relation gives exactly one differentiable $u$ for every real $x$. Also $2u+1>1/5$, so the original denominator never vanishes. Differentiating $H(u)=x+H(1)$ gives the transformed equation and then the original equation; this verifies the unique global IVP branch.