Exercises — Integrating factors for non-exact equations

$M=2xy,N=x^2$. Differentiate: $M_y=2x$ (treat $x$ as frozen, derivative of $2xy$ in $y$ is $2x$), $N_x=2x$. Since $M_y=N_x$, it is exact.

Find $F$: $F=\int 2xydx=x^{2}y+\varphi (y)$. Here $\varphi (y)$ is the undetermined function defined above (the "$x$-constant" that may still depend on $y$). Then $F_y=x^2+{\varphi }^{\prime }(y)$ must equal $N=x^2$, so ${\varphi }^{\prime }=0$, i.e. $\varphi$ is a genuine constant, absorbed into $C$. $\boxed{x^{2}y=C.}$

$M=y,N=-x$. $M_y=1$, $N_x=-1$. Not exact ($1\ne -1$).

$x$-test: $\frac{M_y-N_x}{N}=\frac{1-(-1)}{-x}=\frac{2}{-x}=-\frac{2}{x}$ — pure $x$. ✓ (Case 1 works.)

$y$-test: $\frac{N_x-M_y}{M}=\frac{-1-1}{y}=-\frac{2}{y}$ — pure $y$. ✓ (Case 2 ALSO works.)

Both survive here, a rare happy case. Either factor solves it; the $x$-factor gives $\mu =e^{\int (-2/x)dx}=e^{-2\ln |x|}=|x{|}^{-2}=x^{-2}$ (away from $x=0$).

$M=3x^2+y,N=x$. $M_y=1,N_x=1$. Wait — equal! It is already exact. $F=\int (3x^2+y)dx=x^3+xy+\varphi (y)$ (with $\varphi (y)$ the undetermined $x$-constant); $F_y=x+{\varphi }^{\prime }(y)=x\Rightarrow {\varphi }^{\prime }=0$. $\boxed{x^3+xy=C.}$ (Lesson: always test BEFORE hunting a factor — sometimes there is nothing to fix.)

$M=x^2+y^2+x,N=xy$. $M_y=2y,N_x=y$. Not exact.

$x$-test: $\frac{M_y-N_x}{N}=\frac{2y-y}{xy}=\frac{y}{xy}=\frac{1}{x}$ — pure $x$. ✓

$\mu =e^{\int (1/x)dx}=e^{\ln |x|}=|x|=x$ on $x>0$ (drop the bars; the sign cancels from exactness).

Multiply: $M^{\ast }=x(x^2+y^2+x)=x^3+x{y}^2+x^2$, $N^{\ast }=x^{2}y$. Re-test: $M_y^{\ast }=2xy$, $N_x^{\ast }=2xy$. ✓

$F=\int (x^3+x{y}^2+x^2)dx=\frac{1}{4}{x}^4+\frac{1}{2}{x}^2{y}^2+\frac{1}{3}{x}^3+\varphi (y)$. $F_y=x^{2}y+{\varphi }^{\prime }(y)$ must equal $N^{\ast }=x^{2}y$, so ${\varphi }^{\prime }=0$. $\boxed{\frac{1}{4}{x}^4+\frac{1}{3}{x}^3+\frac{1}{2}{x}^2{y}^2=C.}$

$M=2xy,N=y^2-x^2$. $M_y=2x,N_x=-2x$. Not exact.

$x$-test: $\frac{2x-(-2x)}{y^2-x^2}=\frac{4x}{y^2-x^2}$ — has $y$. ✗

$y$-test: $\frac{N_x-M_y}{M}=\frac{-2x-2x}{2xy}=\frac{-4x}{2xy}=\frac{-2}{y}$ — pure $y$. ✓

$\mu =e^{\int (-2/y)dy}=e^{-2\ln |y|}=|y{|}^{-2}=y^{-2}$ (valid on $y\ne 0$; sign of $y$ cancels).

Multiply: $M^{\ast }=\frac{2x}{y}$, $N^{\ast }=\frac{y^2-x^2}{y^2}=1-\frac{x^2}{y^2}$. Re-test: $M_y^{\ast }=-\frac{2x}{y^2}$, $N_x^{\ast }=-\frac{2x}{y^2}$. ✓

$F=\int \frac{2x}{y}dx=\frac{x^2}{y}+\varphi (y)$; $F_y=-\frac{x^2}{y^2}+{\varphi }^{\prime }(y)=1-\frac{x^2}{y^2}$ $\Rightarrow {\varphi }^{\prime }=1\Rightarrow \varphi =y$. $\boxed{\frac{x^2}{y}+y=C.}$

$M=3xy+y^2,N=x^2+xy$. $M_y=3x+2y,N_x=2x+y$.

$x$-test: $\frac{M_y-N_x}{N}=\frac{(3x+2y)-(2x+y)}{x^2+xy}=\frac{x+y}{x(x+y)}=\frac{1}{x}$. The $(x+y)$ cancels — pure $x$. ✓ So $\mu (x)=e^{\int (1/x)dx}=e^{\ln |x|}=x$ (on $x>0$) exists.

$y$-test: $\frac{N_x-M_y}{M}=\frac{(2x+y)-(3x+2y)}{3xy+y^2}=\frac{-x-y}{y(3x+y)}$. No cancellation: $-x-y$ over $y(3x+y)$ still carries $x$. ✗ So no $\mu (y)$.

Why: a $y$-only factor requires the ratio's $x$'s to all cancel — but here the numerator's $x$ and the denominator's $x$ live in different combinations ($-x-y$ vs $3x+y$), so nothing cleans up. This is Worked Example 1 from the parent, giving $x^{3}y+\frac{1}{2}{x}^2{y}^2=C$.

$M=y^2+2xy,N=-x^2$. $M_y=2y+2x,N_x=-2x$. Not exact.

$x$-test: $\frac{(2y+2x)-(-2x)}{-x^2}=\frac{2y+4x}{-x^2}$ — has $y$. ✗

$y$-test: $\frac{N_x-M_y}{M}=\frac{-2x-(2y+2x)}{y^2+2xy}=\frac{-4x-2y}{y(y+2x)}=\frac{-2(2x+y)}{y(y+2x)}=\frac{-2}{y}$. The $(2x+y)$ cancels — pure $y$. ✓

$\mu =e^{\int (-2/y)dy}=e^{-2\ln |y|}=y^{-2}$ (on $y\ne 0$).

Multiply: $M^{\ast }=\frac{y^2+2xy}{y^2}=1+\frac{2x}{y}$, $N^{\ast }=-\frac{x^2}{y^2}$. Re-test: $M_y^{\ast }=-\frac{2x}{y^2}$, $N_x^{\ast }=-\frac{2x}{y^2}$. ✓

$F=\int \left(1+\frac{2x}{y}\right)dx=x+\frac{x^2}{y}+\varphi (y)$; $F_y=-\frac{x^2}{y^2}+{\varphi }^{\prime }(y)=-\frac{x^2}{y^2}\Rightarrow {\varphi }^{\prime }=0$. $\boxed{x+\frac{x^2}{y}=C.}$

Rewrite: $dy=\left(x-\frac{y}{x}\right)dx$, i.e. $\left(\frac{y}{x}-x\right)dx+dy=0$. So $M=\frac{y}{x}-x,N=1$. $M_y=\frac{1}{x},N_x=0$. Not exact.

$x$-test: $\frac{M_y-N_x}{N}=\frac{1/x-0}{1}=\frac{1}{x}$ — pure $x$. ✓ $\mu =e^{\int (1/x)dx}=e^{\ln |x|}=x$ (on $x>0$).

Linear-ODE factor: $P(x)=\frac{1}{x}$, so $e^{\int Pdx}=e^{\ln |x|}=x$. Same factor. ✓

Multiply: $M^{\ast }=y-x^2,N^{\ast }=x$. Re-test: $M_y^{\ast }=1=N_x^{\ast }$. ✓ $F=\int (y-x^2)dx=xy-\frac{1}{3}{x}^3+\varphi (y)$; $F_y=x+{\varphi }^{\prime }=x\Rightarrow {\varphi }^{\prime }=0$. $\boxed{xy-\frac{1}{3}{x}^3=C⟺y=\frac{1}{3}{x}^2+\frac{C}{x}.}$

Factor. $M=y,N=2x-y{e}^y$, $M_y=1,N_x=2$; not exact. $x$-test $\frac{1-2}{2x-y{e}^y}$ keeps $x$. ✗ $y$-test $\frac{N_x-M_y}{M}=\frac{2-1}{y}=\frac{1}{y}$ — pure $y$. ✓ $\mu =e^{\int (1/y)dy}=e^{\ln |y|}=y$ (on $y\ne 0$).

Multiply. $M^{\ast }=y^2,N^{\ast }=2xy-y^2{e}^y$. Re-test: $M_y^{\ast }=2y=N_x^{\ast }$. ✓

Build $F$. $F=\int y^{2}dx=x{y}^2+\varphi (y)$. Differentiate in $y$ and match $N^{\ast }$: $F_y=2xy+{\varphi }^{\prime }(y)=2xy-y^2{e}^y⟹{\varphi }^{\prime }(y)=-y^2{e}^y.$

Integrate ${\varphi }^{\prime }=-y^2{e}^y$ by parts (why parts? the integrand is a polynomial $\times$ exponential — parts peels one power of $y$ off per pass, and $e^y$ never gets worse). Use $\int udv=uv-\int vdu$.

• Pass 1: $u=y^2,dv=e^{y}dy\Rightarrow du=2ydy,v=e^y$. $\int y^2{e}^{y}dy=y^2{e}^y-\int 2y{e}^{y}dy.$
• Pass 2: $u=2y,dv=e^{y}dy\Rightarrow du=2dy,v=e^y$. $\int 2y{e}^{y}dy=2y{e}^y-\int 2e^{y}dy=2y{e}^y-2e^y.$
• Combine: $\int y^2{e}^{y}dy=y^2{e}^y-(2y{e}^y-2e^y)=e^y(y^2-2y+2).$ Therefore $\varphi (y)=-\int y^2{e}^{y}dy=-e^y(y^2-2y+2).$

Assemble. $\boxed{x{y}^2-e^y(y^2-2y+2)=C.}$ Check (differentiate the boxed $F$): $F_x=y^2=M^{\ast }$ ✓; and $F_y=2xy-[e^y(y^2-2y+2)+e^y(2y-2)]=2xy-y^2{e}^y=N^{\ast }$ ✓ (the two exponential pieces cancel to leave exactly $-y^2{e}^y$, confirming the parts computation).

$M=2y-6x,N=3x-4x^2{y}^{-1}$. Compute $M_y=2$, $N_x=3-8x{y}^{-1}$.

Confirm both pure tests fail: $\frac{M_y-N_x}{N}=\frac{2-3+8x{y}^{-1}}{3x-4x^2{y}^{-1}}=\frac{-1+8x/y}{\cdots }$ — has $y$. ✗ $\frac{N_x-M_y}{M}=\frac{1-8x/y}{2y-6x}$ — has $x$. ✗

Try $\mu =x^a{y}^b$. Then $\mu M=x^a{y}^b(2y-6x)=2x^a{y}^{b+1}-6x^{a+1}{y}^b$ and $\mu N=x^a{y}^b(3x-4x^2{y}^{-1})=3x^{a+1}{y}^b-4x^{a+2}{y}^{b-1}$.

Exactness ${\partial }_y(\mu M)={\partial }_x(\mu N)$: LHS $=2(b+1)x^a{y}^b-6b{x}^{a+1}{y}^{b-1}$. RHS $=3(a+1)x^a{y}^b-4(a+2)x^{a+1}{y}^{b-1}$.

Match the two independent power-blocks:

• $x^a{y}^b$: $2(b+1)=3(a+1)$.
• $x^{a+1}{y}^{b-1}$: $-6b=-4(a+2)\Rightarrow 6b=4a+8\Rightarrow 3b=2a+4$.

From the first: $2b+2=3a+3\Rightarrow 2b=3a+1$. Combine with $3b=2a+4$: multiply first by 3, second by 2: $6b=9a+3$ and $6b=4a+8\Rightarrow 9a+3=4a+8\Rightarrow 5a=5\Rightarrow a=1$, then $2b=3+1=4\Rightarrow b=2$. So $\boxed{\mu =x{y}^2}$.

Multiply original by $x{y}^2$: $M^{\ast }=x{y}^2(2y-6x)=2x{y}^3-6x^2{y}^2$, $N^{\ast }=x{y}^2(3x-4x^2{y}^{-1})=3x^2{y}^2-4x^{3}y$. Re-test: $M_y^{\ast }=6x{y}^2-12x^{2}y$, $N_x^{\ast }=6x{y}^2-12x^{2}y$. ✓

$F=\int (2x{y}^3-6x^2{y}^2)dx=x^2{y}^3-2x^3{y}^2+\varphi (y)$; $F_y=3x^2{y}^2-4x^{3}y+{\varphi }^{\prime }(y)=3x^2{y}^2-4x^{3}y\Rightarrow {\varphi }^{\prime }=0$. $\boxed{x^2{y}^3-2x^3{y}^2=C.}$

Exactness demands ${\partial }_y(x^a{y}^{b}M)={\partial }_x(x^a{y}^{b}N)$. Product rule: $x^a{y}^b\left(b\frac{M}{y}+M_y\right)=x^a{y}^b\left(a\frac{N}{x}+N_x\right).$ Cancel $x^a{y}^b\ne 0$: $\frac{b}{y}M+M_y=\frac{a}{x}N+N_x⟹b\frac{M}{y}-a\frac{N}{x}=N_x-M_y.$ When $M,N$ are polynomials, each distinct monomial in $x,y$ gives one linear equation in $a,b$. Two independent monomials ⇒ a $2\times 2$ linear system ⇒ unique $(a,b)$ when it is consistent. That is exactly the pair of equations solved in Ex 5.1. $■$