Linear differential equations

The form $\frac{dy}{dx} + P(x)\,y = Q(x)$ is a first-order linear differential equation in $y$. Linear here means $y$ and $\dfrac{dy}{dx}$ appear to first power only , no $y^2$, no $\sin y$. The functions $P$ and $Q$ depend only on $x$.

This is among the most useful DE forms in applied work , it models RC and RL circuits, mixing problems with constant flow, motion under linear drag, and many growth problems with constant input rates.

The integrating-factor idea

Suppose we multiply both sides by a function $\mu(x)$ , the integrating factor: $\mu(x)\frac{dy}{dx} + \mu(x) P(x)\,y = \mu(x) Q(x).$ The left side resembles the derivative of a product $\dfrac{d}{dx}(\mu y) = \mu y' + \mu' y$, provided $\mu' = \mu P$. Solving this for $\mu$: $\dfrac{d\mu}{\mu} = P\,dx$, so $\mu = e^{\int P(x)\,dx}$.

So the integrating factor is $\mu(x) = e^{\int P(x)\,dx}.$ After multiplying, the equation becomes $\dfrac{d}{dx}(\mu y) = \mu Q$, which integrates directly: $\mu(x) y = \int \mu(x) Q(x)\,dx + C, \qquad y = \frac{1}{\mu(x)}\left[\int \mu(x) Q(x)\,dx + C\right].$

The procedure

1. Put the equation in standard form $\dfrac{dy}{dx} + P(x) y = Q(x)$ , coefficient of $\dfrac{dy}{dx}$ must be $1$.
2. Compute $P(x)$ and the integrating factor $\mu(x) = e^{\int P\,dx}$.
3. Multiply both sides by $\mu$.
4. Recognise the left side as $\dfrac{d}{dx}[\mu y]$.
5. Integrate both sides; solve for $y$.

When the roles swap: linear in $x$

Sometimes a DE is not linear in $y$ but is linear in $x$. The form $\frac{dx}{dy} + P(y)\,x = Q(y)$ is solved identically , treating $x$ as the dependent variable and $y$ as the independent variable. Integrating factor: $\mu(y) = e^{\int P(y)\,dy}$.

You should always test both forms when the obvious arrangement fails.

Worked examples

Example 1. Solve $\dfrac{dy}{dx} + y = e^x$.

Here $P = 1$, $Q = e^x$. $\mu = e^x$. Multiply: $e^x y' + e^x y = e^{2x}$, i.e. $(e^x y)' = e^{2x}$. Integrate: $e^x y = \dfrac{e^{2x}}{2} + C$, so $y = \dfrac{e^x}{2} + C e^{-x}$.

Example 2. Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$.

$P = 1/x$, $\mu = e^{\int dx/x} = e^{\ln|x|} = |x|$, take $\mu = x$ (positive branch). Multiply: $xy' + y = x^3$, i.e. $(xy)' = x^3$. Integrate: $xy = \dfrac{x^4}{4} + C$, so $y = \dfrac{x^3}{4} + \dfrac{C}{x}$.

Example 3. Solve $\dfrac{dy}{dx} + y \tan x = \sec x$.

$P = \tan x$, $\int \tan x\,dx = \ln|\sec x|$, so $\mu = \sec x$. Multiply: $\sec x \cdot y' + y \sec x \tan x = \sec^2 x$, i.e. $(\sec x \cdot y)' = \sec^2 x$. Integrate: $\sec x \cdot y = \tan x + C$, so $y = \sin x + C \cos x$.

Example 4. Solve $(x + y)\dfrac{dy}{dx} = 1$.

Rewrite: $\dfrac{dy}{dx} = \dfrac{1}{x + y}$, then flip: $\dfrac{dx}{dy} = x + y$, i.e. $\dfrac{dx}{dy} - x = y$. Linear in $x$. $P = -1$, $\mu = e^{-y}$. Multiply: $(e^{-y} x)' = y e^{-y}$. Integrate (by parts): $\int y e^{-y}\,dy = -y e^{-y} - e^{-y} + C$. So $e^{-y} x = -y e^{-y} - e^{-y} + C$, giving $x = -y - 1 + C e^y$.

Example 5. Solve $\dfrac{dy}{dx} + 2y = e^{3x}$, $y(0) = 1$.

$P = 2$, $\mu = e^{2x}$. Multiply: $(e^{2x} y)' = e^{5x}$. Integrate: $e^{2x} y = \dfrac{e^{5x}}{5} + C$, so $y = \dfrac{e^{3x}}{5} + C e^{-2x}$. Apply IC: $1 = \dfrac{1}{5} + C$, so $C = \dfrac{4}{5}$. Particular: $y = \dfrac{e^{3x} + 4e^{-2x}}{5}$.

Example 6. Solve $x\dfrac{dy}{dx} - y = x^2$.

Divide by $x$: $\dfrac{dy}{dx} - \dfrac{y}{x} = x$. $P = -1/x$, $\mu = e^{-\ln x} = 1/x$. Multiply: $\dfrac{1}{x}\dfrac{dy}{dx} - \dfrac{y}{x^2} = 1$, i.e. $\left(\dfrac{y}{x}\right)' = 1$. Integrate: $\dfrac{y}{x} = x + C$, so $y = x^2 + Cx$.

Try it yourself

1. $\dfrac{dy}{dx} + y = x$
2. $\dfrac{dy}{dx} + 2xy = x$
3. $\dfrac{dy}{dx} - \dfrac{y}{x} = 2x^2$
4. $\dfrac{dy}{dx} + y \cot x = \sin 2x$
5. $(1 + x^2)\dfrac{dy}{dx} + 2xy = 4x^2$
6. $\dfrac{dy}{dx} + y = \cos x$
7. $\dfrac{dy}{dx} - y \tan x = e^x \sec x$
8. $x\dfrac{dy}{dx} + y = x \cos x$, $y(\pi/2) = 0$.
9. $\dfrac{dx}{dy} + x = y$ (linear in $x$)
10. $(x + 1)\dfrac{dy}{dx} - y = e^x (x + 1)^2$
11. $\dfrac{dy}{dx} - 2y = x^2 e^{2x}$
12. $\dfrac{dy}{dx} = \dfrac{y - x}{x}$ (rearrange to standard form)
13. $\dfrac{dy}{dx} - 2y \cot 2x = 1 - 2x \cot 2x - 2 \csc 2x$
14. Verify that for $\dfrac{dy}{dx} + P y = Q P$, the answer is $y = Q + C e^{-\int P\,dx}$ if $Q$ is constant.

Pitfalls and tricks

• Always normalise to coefficient $1$ on $\dfrac{dy}{dx}$ before computing $P$.
• $\mu = e^{\int P\,dx}$. No constant of integration here , any antiderivative works (the constant just rescales $\mu$).
• The product rule check. After multiplying by $\mu$, the left side must equal $(\mu y)'$. Verify by expanding.
• Linear in $x$, not $y$? When the natural form has nonlinearity in $y$ but linearity in $x$, swap roles and proceed.
• Don't forget $+C$ when integrating $\mu Q$. It generates the family of solutions.