Linear and Bernoulli equations

A first-order linear ODE is any equation that can be written in the standard form

where: $P(x)$ and $Q(x)$ are given functions of $x$ only — crucially, $y$ appears just to the first power and never multiplied by itself or by $dy/dx$. Standard form matters because the method that follows only works once the coefficient of $dy/dx$ is exactly $1$; if the original equation has some other coefficient in front, you must divide through first before reading off $P(x)$. The central device is the integrating factor

where: the integral $\int P(x)\,dx$ can be any one antiderivative of $P$ (the arbitrary constant of integration is deliberately dropped here, since it only rescales the whole IF by a nonzero constant that cancels later). Multiplying the standard-form equation by $\mathrm{IF}$ makes the left side collapse into a single derivative,

after which one integration finishes the job:

Bernoulli's equation looks like a small variation but is genuinely nonlinear:

(When $n=0$ or $n=1$ the right side is already linear in $y$, so no substitution is needed at all — a common exam trap is not noticing this shortcut.) Dividing through by $y^n$ and substituting $v = y^{1-n}$ turns it, after a short chain-rule step, into a linear equation for $v(x)$, which the integrating-factor method above then solves; back-substituting $y = v^{1/(1-n)}$ recovers the answer.

Solve $\dfrac{dy}{dx} + \dfrac{2}{x}y = x^2$ for $x>0$, given $y(1)=1$, and then extend the same equation family to a Bernoulli version.

Step 1. Read off $P(x) = 2/x$ and $Q(x) = x^2$; the equation is already in standard form.

Step 2. Compute the integrating factor: $\mathrm{IF} = e^{\int (2/x)\,dx} = e^{2\ln x} = x^2$.

Step 3. Multiply through by $\mathrm{IF}=x^2$ and recognize the product-rule collapse: $x^2\dfrac{dy}{dx} + 2xy = x^4 \implies \dfrac{d}{dx}(x^2 y) = x^4$.

Step 4. Integrate both sides: $x^2 y = \displaystyle\int x^4\,dx = \dfrac{x^5}{5} + C$, so $y = \dfrac{x^3}{5} + \dfrac{C}{x^2}$.

Step 5. Apply $y(1)=1$: $1 = \dfrac{1}{5} + C \implies C = \dfrac{4}{5}$, giving the particular solution

Competitive stretch

Now solve the related Bernoulli equation $\dfrac{dy}{dx} - y = xy^3$, where $P=-1$, $Q=x$, $n=3$. Dividing by $y^3$ gives $y^{-3}y' - y^{-2} = x$. Substituting $v = y^{1-n} = y^{-2}$, so $v' = -2y^{-3}y'$, transforms the equation into the linear form $v' + 2v = -2x$. Its integrating factor is $e^{2x}$, giving $\frac{d}{dx}(ve^{2x}) = -2xe^{2x}$; integrating the right side by parts yields $ve^{2x} = -xe^{2x} + \tfrac{1}{2}e^{2x} + C_1$, so $v = -x + \tfrac{1}{2} + C_1e^{-2x}$. Back-substituting $v=y^{-2}$ gives the implicit general solution

This is the general solution of the Bernoulli equation, expressed implicitly in $y$.

• Reading off $P(x)$ before the equation is in standard form. Always divide through so the coefficient of $dy/dx$ is exactly $1$ first; skipping this gives the wrong $P(x)$ and a useless integrating factor.
• Forgetting the integrating factor multiplies every term. It must be applied to $Q(x)$ as well as the left side — a common arithmetic slip that silently drops a term.
• Mis-forming the Bernoulli substitution. Using $v=y^n$ instead of $v=y^{1-n}$, or losing the factor $(1-n)$ when computing $dv/dx$, wrecks the reduction to a linear equation.
• Missing the $n=0$ or $n=1$ shortcut. These special cases are already linear and need no substitution at all — trying to substitute anyway just adds unnecessary work.
• Dropping the final constant $C$. It is fine (and conventional) to omit the constant when computing $\int P\,dx$ for the integrating factor itself, but the constant of integration in the general solution must never be dropped.

Ask the tutor: stuck computing an integrating factor with a tricky $\int P(x)\,dx$, or want to double-check a Bernoulli substitution on a new example? Ask the AI tutor to work through it with you step by step.