How an Integrating Factor Turns a Linear Differential Equation Into One Integral

Many first-order equations refuse to separate — x and y stay tangled together. Two common types still have reliable methods: homogeneous equations, solved with the substitution $y=vx$, and linear equations, solved with an integrating factor.

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

**An equation $\frac{dy}{dx}=F(x,y)$ is homogeneous when F can be written as a function of $\frac{y}{x}$ alone; substituting $y=vx$, so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$, makes the variables separable.

Test for homogeneity.** $F(\lambda x,\lambda y)=F(x,y)$ for every non-zero $\lambda$.

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

- Put $y=vx$: $v+x\frac{dv}{dx}=1+v$
- So $x\frac{dv}{dx}=1$, giving $dv=\frac{dx}{x}$
- Integrating: $v=\log x+C$, so $y=x\log x+Cx$

Check. If $y=x\log x+Cx$, then $\frac{dy}{dx}=\log x+1+C=\frac{y}{x}+1=\frac{x+y}{x}$.

Worked example 2. Solve $\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$. With $y=vx$: $x\frac{dv}{dx}=\frac{1+v^2}{2v}-v=\frac{1-v^2}{2v}$, so $\frac{2vdv}{1-v^2}=\frac{dx}{x}$. Integrating gives $-\log |1-v^2|=\log |x|+C$, which simplifies to $x^2-y^2=Cx$.

An everyday example. Enlarging a photo on a phone screen keeps every shape's proportions, just as a homogeneous equation keeps its form when x and y are scaled together.

The substance. Homogeneous means every term has the same total degree — $x^2+y^2$ and $2xy$ both have degree 2, but $x^2+y$ does not.

**For $\frac{dy}{dx}+Py=Q$, where P and Q are functions of x, multiply through by the integrating factor $\text{I.F.}=e^{\int Pdx}$, which turns the left side into the derivative of $y\times \text{I.F.}$, so the solution is $y\times \text{I.F.}=\int Q\times \text{I.F.}dx+C$.

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

- $P=\frac{1}{x}$, so $\text{I.F.}=e^{\log x}=x$
- $yx=\int x^{3}dx=\frac{x^4}{4}+C$
- $y=\frac{x^3}{4}+\frac{C}{x}$

Worked example 2 (particular solution). Solve $\frac{dy}{dx}+2y=4$ with $y(0)=0$. Here $\text{I.F.}=e^{2x}$, so $y{e}^{2x}=2e^{2x}+C$ and $y=2+C{e}^{-2x}$. The condition gives $C=-2$, so $y=2-2e^{-2x}$, which is about 1.73 at $x=1$.

An everyday example. A water tank filling at a steady rate while leaking in proportion to its level obeys $\frac{dy}{dt}+ky=Q$ — the level rises and settles at $\frac{Q}{k}$, like the value 2 in the second example.

The substance. Write the equation in standard form first — for $x\frac{dy}{dx}+y=x^3$, divide by x before reading off P.

**When an equation is linear in x rather than y, write it as $\frac{dx}{dy}+Px=Q$ with P and Q functions of y, use $\text{I.F.}=e^{\int Pdy}$, and the solution is $x\times \text{I.F.}=\int Q\times \text{I.F.}dy+C$.

Spotting the form.** $\frac{dy}{dx}=\frac{y}{2x+y^2}$ is not linear in y, but flipping it gives $\frac{dx}{dy}=\frac{2x}{y}+y$, which is linear in x.

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

- $P=-\frac{2}{y}$, so $\text{I.F.}=e^{-2\log y}=\frac{1}{y^2}$
- $\frac{x}{y^2}=\int y\times \frac{1}{y^2}dy=\log y+C$
- $x=y^2\log y+C{y}^2$

Check. Differentiating with respect to y gives $\frac{dx}{dy}=2y\log y+y+2Cy$, which equals $\frac{2x}{y}+y$.

An everyday example. Working out delivery time as a function of distance, instead of distance as a function of time, swaps the roles of the variables — just as flipping $\frac{dy}{dx}$ to $\frac{dx}{dy}$ does.

The substance. Flipping the derivative is worth trying when y appears squared or inside a function but x appears only to the first power — the equation may then be linear in x.