Exact Equations — Question 6

Question 6 – Solution

Strategy. Derive the integrating-factor condition from exactness rather than importing a formula, and return to the original equation at a zero of the chosen factor.

Step 1: Derive the factor. Originally $M_y=2$ while $N_x=1$. For $\mu=\mu(x)$, exactness requires $$(\mu M)_y=2\mu=(\mu x)_x=x\mu'+\mu.$$ Thus $x\mu'=\mu$. On $x>0$, its nonzero solutions are $\mu=Kx$, $K\ne 0$. Choose $\boxed{\mu=x}$, which is nonzero throughout the stated half-plane.

Step 2: Integrate the exact form. The multiplied coefficients are $2xy+x^2$ and $x^2$. Integrating in $y$ gives $$F=x^2y+g(x),\qquad F_x=2xy+g'(x)=2xy+x^2.$$ Hence $g=x^3/3$ is a choice, giving the family $\boxed{x^2y+x^3/3=C}$.

Step 3: Apply and verify the data. At $(1,2)$, $C=7/3$, so $$\boxed{y=\frac 7{3x^2}-\frac x3,\qquad x>0}.$$ Its derivative is $-14/(3x^3)-1/3$, and substitution gives $$2y+x+xy'=\frac{14}{3x^2}-\frac{2x}{3}+x -\frac{14}{3x^2}-\frac x3=0.$$ The initial value is also $2$.

Step 4: Check extension in the original equation. The selected solution tends to $+\infty$ as $x\downarrow 0$, so it cannot extend continuously through $0$. In the general family, $y=C/x^2-x/3$, a finite extension requires $C=0$. This member extends as $\boxed{y=-x/3}$ on all of $\mathbb R$ and satisfies the original equation even at $0$, where the equation requires $2y(0)=0$. The multiplied form alone gives $0=0$ there because its factor vanishes; that weakened condition must not replace the original one.