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 identified as \(e^{5x}\), which, when multiplied by both sides of the equation, transforms the left-hand side into the derivative of \(y \cdot e^{5x}\). This leads to the equation \(\frac{d}{dx}(y \cdot e^{5x}) = e^{7x}\), clarifying the relationship between the terms in the equation.

PREREQUISITES
  • Understanding of first-order linear differential equations
  • Familiarity with the integrating factor method
  • Basic knowledge of derivatives and exponential functions
  • Experience with K.A. Stroud's "Engineering Mathematics" concepts
NEXT STEPS
  • Study the derivation of integrating factors in differential equations
  • Practice solving first-order linear differential equations using integrating factors
  • Explore applications of linear differential equations in engineering contexts
  • Review the section on derivatives in K.A. Stroud's "Engineering Mathematics" for deeper understanding
USEFUL FOR

Students and self-learners in mathematics, particularly those studying differential equations, as well as educators seeking to clarify the integrating factor method in linear equations.

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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