2.4 Special Integrating Factors and Substitutions

Our toolbox already handles a good number of equation types. Still, an equation met in practice often fits none of them exactly, and then we have to bend it into a shape we know. Two ideas do most of that work.

• Repair the equation. If $Mdx+Ndy=0$ is not exact, multiply it by an integrating factor. We will find simple tests that tell us when a factor depending on $x$ alone, or on $y$ alone, exists, and a formula for it.
• Change the variables. A shift of the origin, or a new unknown, can turn an unfamiliar equation into a familiar one.

The condition for an integrating factor

Recall from Section 2.1 that $\mu (x,y)$ is an integrating factor of $Mdx+Ndy=0$ when $\mu Mdx+\mu Ndy=0$ is exact, i.e.

Expand both sides with the product rule:

This is a partial differential equation for $\mu$. Solving it in general is harder than the problem we started with. The trick is to guess the form of $\mu$, so that the equation collapses to something easy.

Integrating factors that depend only on $x$

Suppose $\mu =\mu (x)$. Then ${\mu }_y=0$ and ${\mu }_x={\mu }^{\prime }(x)$. The condition becomes

Look at the two sides. The left side depends only on $x$. So this can only work if the right side also depends only on $x$. When it does, call it $g(x)$; then ${\mu }^{\prime }/\mu =g(x)$ gives $\ln \mu =\int g(x)dx$, and

Conversely, if the right side really depends on $y$, no integrating factor of the form $\mu (x)$ exists.

Integrating factors that depend only on $y$

Now suppose $\mu =\mu (y)$. Then ${\mu }_x=0$ and ${\mu }_y={\mu }^{\prime }(y)$, and the condition becomes

The same reasoning applies with the roles of $x$ and $y$ exchanged. Notice that the numerator has the opposite order, $N_x-M_y$, and the denominator is $M$.

The exactness solver tries both quotients automatically when an equation is not exact. Enter polynomial $M$ and $N$ and see which test succeeds.

Worked examples: integrating factors

Integrating factors of the form $x^p{y}^q$

Sometimes neither quotient works, but the equation has polynomial coefficients. Then a factor $\mu =x^p{y}^q$ is worth trying. Multiply, impose exactness, and compare coefficients of matching monomials. This produces linear equations for $p$ and $q$. If they have a solution, you have found $\mu$; if not, try something else.

Why does comparing coefficients work? Multiplying a monomial $x^a{y}^b$ by $x^p{y}^q$ and then differentiating gives another monomial, times a number that depends linearly on $p$ or $q$. So both $(\mu M{)}_y$ and $(\mu N{)}_x$ are sums of monomials with coefficients linear in $p$ and $q$. Two such sums are equal for all $x$ and $y$ only if the coefficients of each monomial agree. When $M$ and $N$ each have two terms, as is typical, this gives two equations for two unknowns. A monomial that appears on only one side forces its coefficient to be zero, which is an extra condition and may make the system unsolvable.

Equations with linear numerator and denominator

Consider equations of the form

where the $a_i$, $b_i$, $c_i$ are constants. If $c_1=c_2=0$, this is homogeneous (Section 2.2). The constants $c_1$, $c_2$ spoil homogeneity. The cure depends on whether the two lines

intersect or are parallel.

Why the shift works

1. Because $h$ and $k$ are constants, $dx=dX$ and $dy=dY$, so $y^{\prime }=dY/dX$.
2. Substitute into the numerator: $$a_1(X+h)+b_1(Y+k)+c_1=a_{1}X+b_{1}Y+\underset{=0}{\underbrace{(a_{1}h+b_{1}k+c_1)}}.$$
3. The denominator works the same way. The constants vanish, and what remains is homogeneous of degree zero.

Geometrically, we move the origin to the point where the two lines cross. In the new coordinates both lines pass through the origin, and lines through the origin are exactly what homogeneous equations "see".

Why $z=a_{1}x+b_{1}y$ works

1. If $a_1{b}_2=a_2{b}_1$, the pairs $(a_1,b_1)$ and $(a_2,b_2)$ are proportional: $a_2=\lambda a_1$ and $b_2=\lambda b_1$ for some constant $\lambda$ (when $a_1$, $b_1$ are not both zero).
2. Then $a_{2}x+b_{2}y=\lambda z$, so the equation reads $y^{\prime }=\frac{z+c_1}{\lambda z+c_2}$.
3. Differentiate $z=a_{1}x+b_{1}y$: $z^{\prime }=a_1+b_1{y}^{\prime }$. Substituting, $$z^{\prime }=a_1+b_1\frac{z+c_1}{\lambda z+c_2}.$$ The right side depends on $z$ only, so the equation is separable.

If $b_1=0$ (and $a_1\ne 0$), parallelism forces $b_2=0$, and $y^{\prime }$ is a function of $x$ alone: just integrate. In practice, it is often quickest to let $z$ be the expression that appears in both numerator and denominator.

The solver below follows these steps for any choice of the six coefficients. Try both an intersecting and a parallel case.

Where did the line solution come from? In the parallel case the right side of the original equation is constant along every line $x-3y=\text{const}$. On one special line the slope given by the equation equals the slope of the line itself, $\frac{1}{3}$. That line is then a solution curve. Algebraically, it is the root of the separated $z$-equation, which we divided by. As always, dividing by an expression that can vanish means checking its zeros.

Choosing a method: a checklist

You now know seven techniques. Faced with a new first-order equation, it pays to test them in a sensible order, from the quickest check to the most laborious one. One equation may fit several types; then choose the method with the easiest integrals.

1. Separable? Can $y^{\prime }$ be written as $f(x)g(y)$? If yes, separate, and remember the constant solutions $g(y_0)=0$.
2. Linear in $y$? Can it be written as $y^{\prime }+P(x)y=Q(x)$? If yes, use $\mu =e^{\int Pdx}$.
3. Linear in $x$? Sometimes $dx/dy+P(y)x=Q(y)$ works when the $y$-form does not (Example 2.3.5).
4. Exact? Write it as $Mdx+Ndy=0$ and compare $M_y$ with $N_x$.
5. Bernoulli? Is it $y^{\prime }+Py=Q{y}^n$? Substitute $v=y^{1-n}$.
6. Homogeneous? Is $f(tx,ty)=f(x,y)$? Substitute $y=vx$.
7. Integrating factor of one variable? Compute $(M_y-N_x)/N$ and $(N_x-M_y)/M$.
8. Linear fraction or another substitution? Look for an expression, such as $ax+by+c$, that the right side really depends on.

If none of these work, the equation may have no solution in terms of elementary functions. That is normal, not a failure. Chapter 8 shows how to obtain accurate approximate solutions in such cases, and the slope fields of Chapter 1 always give a qualitative picture.