Linear equations - use of integrating factor

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SUMMARY

The discussion focuses on solving the linear differential equation \(\frac{dy}{dx} + 5y = e^{2x}\) using the integrating factor method as outlined in K.A. Stroud's "Engineering Mathematics." The integrating factor is \(e^{5x}\), which, when multiplied by the equation, transforms the left-hand side into the derivative of \(y \cdot e^{5x}\). The participant clarifies their misunderstanding regarding the application of the product rule, realizing that \(e^{5x} \frac{dy}{dx}\) is indeed part of the derivative of \(y \cdot e^{5x}\).

PREREQUISITES
  • Understanding of first-order linear differential equations
  • Familiarity with the integrating factor method
  • Knowledge of the product rule in calculus
  • Basic proficiency in manipulating exponential functions
NEXT STEPS
  • Study the application of integrating factors in solving linear differential equations
  • Learn more about the product rule and its implications in calculus
  • Explore examples of first-order linear differential equations in K.A. Stroud's "Engineering Mathematics"
  • Practice solving differential equations using various integrating factors
USEFUL FOR

Students and self-learners in mathematics, particularly those studying differential equations, as well as educators looking for clarification on integrating factors and product rule applications.

nufc365
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Hi, I am doing self study and have hit a snag in the road. Can someone please clarfiy this for me. I am reading K.A. Stroud's Engineering Mathematics which so far has been great.Consider the equation [itex]\frac{dy}{dx}[/itex] + 5y = e2x

In this case, we begin multiplying both sides by e5x. This gives

e5x[itex]\frac{dy}{dx}[/itex] + y5e5x = e2x.e5x = e7x

We now find that the left-hand-side is the derivative of y.e5x.

[itex]\frac{d}{dx}[/itex](y.e5x) = e7x____________________

Can someone please clarify how the left hand side is the derivative of y.e5x.

I get that y5e5x is the derivative of y.e5x, but what about the first part of the left-hand-side. What happens to the e5x[itex]\frac{dy}{dx}[/itex]
 
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The product rule is

(uv)'=u'v+uv'

here u=y v=e^(5x)

You have taken y'=0
 
Thanks a lot lurflurf. It seems so obvious now!

I was treating y as a constant. doh
 

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