The Integrating Factor

An equation that will not separate

A first-order linear equation has the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$dy/dx + P(x)y = Q(x), where P and Q are functions of x alone. An example is $\frac{dy}{dx}+\frac{2y}{x}=x$dy/dx + 2y/x = x.

Separating the variables needs every y on one side with dy and every x on the other with dx. Here $\frac{dy}{dx}=x-\frac{2y}{x}$dy/dx = x − 2y/x, and the right side is a difference, not a product of a function of x and a function of y. The y and the x will not come apart, so another method is needed.

Make the left side one derivative

The left side looks almost like the product rule. For a function I of x, the product rule gives $\frac{d}{dx}\left(Iy\right)=I\frac{dy}{dx}+I^{\prime }y$d/dx (I y) = I dy/dx + I' y.

Multiply the equation through by I and its left side becomes $I\frac{dy}{dx}+IPy$I dy/dx + I P y. That is exactly $\frac{d}{dx}\left(Iy\right)$d/dx (I y) when the second terms agree, that is when I' = I P.

Finding the factor

The condition I' = I P is itself a separable equation. Divide by I to get I'/I = P, and integrate both sides: $\ln I=\int Pdx$ln I = ∫P dx. So $I=e^{\int Pdx}$I = e^(∫P dx). This I is called the integrating factor.

No constant is needed in $\int Pdx$∫P dx. Adding a constant k multiplies I by $e^k$eᵏ, and multiplying the whole equation by a constant does not change its solutions, so the simplest choice will do.

Multiply through and integrate

Multiply every term by I: $I\frac{dy}{dx}+IPy=IQ$I dy/dx + I P y = I Q. The left side is now the single derivative $\frac{d}{dx}\left(Iy\right)$d/dx (I y), so the equation reads $\frac{d}{dx}\left(Iy\right)=IQ$d/dx (I y) = I Q.

Integrate both sides: $Iy=\int IQdx+c$I y = ∫ I Q dx + c. Then divide by I to find y. There is one constant, and it arrives when the right side is integrated, before the division.

The equation must be in this form first, with $\frac{dy}{dx}$dy/dx standing alone. For $x\frac{dy}{dx}+2y=x^2$x dy/dx + 2y = x², divide every term by x to get $\frac{dy}{dx}+\frac{2y}{x}=x$dy/dx + 2y/x = x, and only then read off $P=\frac{2}{x}$P = 2/x.

An example

Solve $\frac{dy}{dx}+\frac{2y}{x}=x$dy/dx + 2y/x = x for x > 0. Here $P=\frac{2}{x}$P = 2/x, so $\int Pdx=2\ln x$∫P dx = 2 ln x, and the integrating factor is $I=e^{2\ln x}=e^{\ln x^2}=x^2$I = e^(2 ln x) = e^(ln x²) = x².

Multiply through by $x^2$x²: $x^2\frac{dy}{dx}+2xy=x^3$x² dy/dx + 2xy = x³. The left side is $\frac{d}{dx}\left(x^{2}y\right)$d/dx (x²y), so $x^{2}y=\int x^{3}dx=\frac{x^4}{4}+c$x²y = ∫x³ dx = x⁴/4 + c, and dividing by $x^2$x² gives $y=\frac{x^2}{4}+\frac{c}{x^2}$y = x²/4 + c/x².

Check it. With $y=\frac{x^2}{4}+\frac{c}{x^2}$y = x²/4 + c/x², $\frac{dy}{dx}=\frac{x}{2}-\frac{2c}{x^3}$dy/dx = x/2 − 2c/x³ and $\frac{2y}{x}=\frac{x}{2}+\frac{2c}{x^3}$2y/x = x/2 + 2c/x³. Adding them, the c terms cancel and the sum is x, as the equation says.

A known value fixes c. If y = 1 when x = 1, then $1=\frac{1}{4}+c$1 = 1/4 + c, so $c=\frac{3}{4}$c = 3/4 and $y=\frac{x^2}{4}+\frac{3}{4x^2}$y = x²/4 + 3/4x². At x = 2 this gives $y=1+\frac{3}{16}=1.1875$y = 1 + 3/16 = 1.1875.

Three solutions of $\frac{dy}{dx}+\frac{2y}{x}=x$dy/dx + 2y/x = x. The gold curve is $y=\frac{x^2}{4}+\frac{3}{4x^2}$y = x²/4 + 3/4x², the one through (1, 1). The dashed curves have c = 0, which is $y=\frac{x^2}{4}$y = x²/4, and c = −1. Each value of c gives a different curve, and all of them come close to $y=\frac{x^2}{4}$y = x²/4 as x grows, because $\frac{c}{x^2}$c/x² shrinks.

A second example: the factor $e^x$eˣ

Solve $\frac{dy}{dx}+y=x$dy/dx + y = x. Here P = 1, so $\int Pdx=x$∫P dx = x and the integrating factor is $e^x$eˣ.

Multiply through: $e^x\frac{dy}{dx}+e^{x}y=x{e}^x$eˣ dy/dx + eˣy = x eˣ, and the left side is $\frac{d}{dx}\left(e^{x}y\right)$d/dx (eˣy). Integrating $x{e}^x$x eˣ by parts gives $x{e}^x-e^x+c$x eˣ − eˣ + c, so $e^{x}y=x{e}^x-e^x+c$eˣy = x eˣ − eˣ + c. Divide by $e^x$eˣ: $y=x-1+c{e}^{-x}$y = x − 1 + c e^(−x).

If y = 1 when x = 0, then 1 = −1 + c, so c = 2 and $y=x-1+2e^{-x}$y = x − 1 + 2e^(−x). Check at x = 1: $y=2e^{-1}$y = 2e^(−1), about 0.736, and $\frac{dy}{dx}=1-2e^{-1}$dy/dx = 1 − 2e^(−1), about 0.264. Their sum is 1, which is x.

As x grows, $c{e}^{-x}$c e^(−x) shrinks toward 0 whatever c is, so every solution comes closer and closer to the line y = x − 1.

The equation $\frac{dy}{dx}+y=x$dy/dx + y = x, rearranged as $\frac{dy}{dx}=x-y$dy/dx = x − y: each dash has gradient x − y at its point. The gold curve is the solution through the probe, at (0, 0) the curve $y=x-1+e^{-x}$y = x − 1 + e^(−x). Drag the probe to (0, 1) and it picks $y=x-1+2e^{-x}$y = x − 1 + 2e^(−x), which starts with gradient −1 and runs alongside the line y = x − 1. Each position of the probe sets the constant c and picks one curve of the family.

The usual mistakes

Building the factor from Q. The integrating factor is $e^{\int Pdx}$e^(∫P dx), made from the coefficient of y. Q only appears on the right after the multiplication.

Leaving out the exponential. $\int Pdx$∫P dx on its own does not make the left side a product rule; $e^{\int Pdx}$e^(∫P dx) does.

Multiplying only the left side by I. Every term is multiplied, so the right side becomes I Q.

Adding the constant after dividing by I. For the first example that gives $y=\frac{x^2}{4}+c$y = x²/4 + c, and then $\frac{dy}{dx}+\frac{2y}{x}=x+\frac{2c}{x}$dy/dx + 2y/x = x + 2c/x, which is not x unless c = 0. The constant is added at the integration, so it is divided by $x^2$x² as well.

A hospital’s technetium generator

In the application below, the number of technetium atoms y satisfies a linear equation of this form, with the time h in hours in place of x. P is the constant b, the technetium’s decay constant, and Q is $aN{e}^{-ah}$a N e^(−a h), the rate at which it is made. The model method tries a multiple of $e^{-ah}$e^(−a h) for y.

The integrating factor reaches the same answer. Here $I=e^{bh}$I = e^(b h), and multiplying through makes the left side the derivative of $y{e}^{bh}$y e^(b h), so $y{e}^{bh}=K{e}^{\left(b-a\right)h}+C$y e^(b h) = K e^((b − a)h) + C, where $K=a\frac{N}{b-a}$K = a N/(b − a). Dividing by $e^{bh}$e^(b h) gives $y=K{e}^{-ah}+C{e}^{-bh}$y = K e^(−a h) + C e^(−b h), the model method’s solution.