How an Integrating Factor Turns a Hard Equation Into an Easy One

Many first order equations, such as $\frac{dy}{dx}=\frac{x+y}{x}$ or $\frac{dy}{dx}+y=x$, refuse to split into an $x$ side and a $y$ side. Two classic methods handle them: a substitution for homogeneous equations and an integrating factor for linear ones.

This part covers homogeneous equations, linear equations in $y$ and in $x$, and particular solutions.

**A first order equation $\frac{dy}{dx}=F(x,y)$ is homogeneous when $F(\lambda x,\lambda y)=F(x,y)$, so $F$ depends only on $\frac{y}{x}$; substituting $y=vx$ turns it into a separable equation in $v$ and $x$.

The method:**

- Put $y=vx$, so $\frac{dy}{dx}=v+x\frac{dv}{dx}$
- Substitute, separate $v$ and $x$, and integrate
- Replace $v$ by $\frac{y}{x}$

Worked example. Solve $\frac{dy}{dx}=\frac{x+y}{x}$.

Since $\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}$, the equation is homogeneous. With $y=vx$:

$$v+x\frac{dv}{dx}=1+v\Rightarrow x\frac{dv}{dx}=1\Rightarrow dv=\frac{dx}{x}$$

$$v=\log |x|+C\Rightarrow y=x\log |x|+Cx$$

**When $x=vy$ is better.** If the equation is given as $\frac{dx}{dy}=G(x,y)$ with $G$ homogeneous, substitute $x=vy$ instead.

An everyday example. Enlarging a photograph keeps its shape because every length scales by the same factor — a homogeneous equation is unchanged in the same way when $x$ and $y$ are scaled together.

The substance. **Homogeneous here means unchanged under scaling of $x$ and $y$**, which is a different idea from a homogeneous polynomial.

**Multiply $\frac{dy}{dx}+Py=Q$ by the integrating factor $e^{\int Pdx}$, which turns the left side into the derivative of $y\cdot \text{IF}$, and integrate to get $y\cdot \text{IF}=\int Q\cdot \text{IF}dx+C$.

Why it works.** With $\text{IF}=e^{\int Pdx}$, the derivative of $\text{IF}$ is $P\cdot \text{IF}$, so

$$\frac{d}{dx}(y\cdot \text{IF})=\text{IF}\frac{dy}{dx}+P\cdot \text{IF}\cdot y=Q\cdot \text{IF}$$

Worked example 1. Solve $\frac{dy}{dx}+y=x$. Here $P=1$, $Q=x$ and $\text{IF}=e^x$:

$$y{e}^x=\int x{e}^{x}dx=x{e}^x-e^x+C\Rightarrow y=x-1+C{e}^{-x}$$

Worked example 2. Solve $x\frac{dy}{dx}+2y=x^2$ for $x>0$. In standard form, $\frac{dy}{dx}+\frac{2}{x}y=x$:

$$\text{IF}=e^{\int \frac{2}{x}dx}=e^{2\log x}=x^2$$

$$y{x}^2=\int x^{3}dx=\frac{x^4}{4}+C\Rightarrow y=\frac{x^2}{4}+\frac{C}{x^2}$$

An everyday example. A water tank filling from a tap while a leak drains it in proportion to the water inside follows a linear equation of exactly this type.

The substance. Put the equation in standard form first — the coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$.

**When an equation is linear in $x$ with $y$ as the independent variable, use the integrating factor $e^{\int P_{1}dy}$ and integrate with respect to $y$: $x\cdot \text{IF}=\int Q_1\cdot \text{IF}dy+C$.

When to switch.** If $\frac{dy}{dx}$ is not linear in $y$ but $\frac{dx}{dy}$ is linear in $x$, treat $x$ as the unknown.

Worked example. Solve $\frac{dx}{dy}-\frac{x}{y}=2y$ for $y>0$.

$$P_1=-\frac{1}{y},\text{IF}=e^{-\int \frac{dy}{y}}=e^{-\log y}=\frac{1}{y}$$

$$x\cdot \frac{1}{y}=\int 2y\cdot \frac{1}{y}dy=2y+C\Rightarrow x=2y^2+Cy$$

Recognising the form. An equation like $ydx-(x+2y^2)dy=0$ rearranges to $\frac{dx}{dy}-\frac{x}{y}=2y$, the example above.

An everyday example. Reading a train timetable by station instead of by clock time swaps which quantity is treated as the input — just as this form treats $y$ as the independent variable.

The substance. Neither form is harder — the real skill is spotting which variable the equation is linear in.

**Solve the equation completely, substitute the given values of $x$ and $y$ to find the constant, and rewrite the solution with that constant.

Worked example 1 — linear.** Solve $\frac{dy}{dx}+y=x$ with $y(0)=1$. From $y=x-1+C{e}^{-x}$:

$$1=0-1+C\Rightarrow C=2y=x-1+2e^{-x}$$

Check. $\frac{dy}{dx}=1-2e^{-x}$, so $\frac{dy}{dx}+y=1-2e^{-x}+x-1+2e^{-x}=x$.

Worked example 2 — homogeneous. Solve $\frac{dy}{dx}=\frac{x+y}{x}$ with $y=0$ when $x=1$. From $y=x\log |x|+Cx$:

$$0=1\cdot 0+C\Rightarrow C=0y=x\log x\text{ for }x>0$$

An everyday example. A tank's filling law describes every possible starting level, but knowing it held $500$ litres at noon fixes one exact curve — the particular solution.

The substance. Find the constant only after integrating completely — substituting too early loses the constant's effect.