Solution formula for linear first-order equations

Builds on Linear first-order equation Integrating factor Initial value problem

Statement

Let $I$ be an open interval, let $p, q : I \to \mathbb{R}$ be continuous, let $x_0 \in I$ and let $y_0 \in \mathbb{R}$ . Let $P(x) = \int_{x_0}^{x} p(s)\, ds$ for $x \in I$ . Then the initial value problem $y' + p(x)\, y = q(x)$ , $y(x_0) = y_0$ has exactly one solution on $I$ , and it is $y(x) = e^{-P(x)} \left( y_0 + \int_{x_0}^{x} e^{P(s)}\, q(s)\, ds \right), \qquad x \in I.$

Remarks

Proof idea: $\mu = e^{P}$ is an integrating factor, so a solution satisfies $(\mu y)' = \mu q$ . Integrate from $x_0$ to $x$ and use $\mu(x_0) = 1$ . The solution exists on all of $I$ . A linear equation has no solution that blows up inside the interval of continuity of $p$ and $q$ . Example: for $y' + y = 1$ , $y(0) = 0$ , the formula gives $y(x) = 1 - e^{-x}$ .

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Differential equation concept
Function concept
Interval concept
Ordinary differential equation definition
Linear first-order equation definition
Solution of an ordinary differential equation definition
Integrating factor definition
Initial value problem definition
Solution formula for linear first-order equations statement

Tag	`06B9W45XV0`
URI	`m2://06B9W45XV0`
Kind	statement, theorem
Writing	W0 stub
MSC	34A30
Status	This copy of the site has no build data. Run `m2 status 06B9W45XV0`.
Uses	Linear first-order equation, Integrating factor, Initial value problem

Solution formula for linear first-order equations

Statement

Remarks

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