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Using an Integrating Factor The expression $u(x)$ is an integrating factor for $y^{\prime}+P(x) y=Q(x) .$ Which of the following is equal to $u^{\prime}(x) ?$ Verify your answer.(a) $P(x) u(x) \quad$ (b) $P^{\prime}(x) u(x)$(c) $Q(x) u(x) \quad$ (d) $Q^{\prime}(x) u(x)$

$(a) \quad U^{\prime}(x)=P(x) U(x)$

Calculus 2 / BC

Chapter 6

Differential Equations

Section 4

First-Order Linear Differential Equations

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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for the differential equation. Why Prime plus p affects times why To cure fact. I think the grating factor is given by the function you effects equal to here to the integral off B f x t x. So this term here is just a function of X that we can call, for example, geo fix. Then we will have that Hugh Effects sequel toe A to the Geo fix and did the realty off you off X is going to be there. The reality off this exponential that we know that is G prime off x times be gx now G prime off X is the derivative respect to X off the integral off P of X t X and this is just Pierre fix. And that means that they're the reality off. The integrating factor is going to be be affects times e to the integral be of X t x and this term here it's a integrating factor. So your prime is p affects tire you affects. So the solution is it

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