Why Does dy Change in Solving Linear Differential Equations?

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SUMMARY

The discussion focuses on the transformation of the differential equation \(\frac{dy}{dx} + P(x) y = Q(x)\) when applying an integrating factor \(v(x)\). Participants clarify that \(dy\) represents a differential change in \(y\) and that the term \(+P(x)\) vanishes after multiplication by \(v(x)\). The disappearance of \(d\) in step three is attributed to the integration process, which operates on the entire equation with respect to \(x\

. PREREQUISITES
  • Understanding of basic calculus concepts, specifically differentiation and integration.
  • Familiarity with differential equations and their standard forms.
  • Knowledge of integrating factors and their role in solving linear differential equations.
  • Proficiency in notation related to derivatives, such as \(dy/dx\) and differential operators.
NEXT STEPS
  • Study the method of integrating factors in linear differential equations.
  • Learn how to manipulate and simplify expressions involving differentials.
  • Explore the properties of differential operators and their applications in calculus.
  • Practice solving linear differential equations using various integrating factors.
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Students studying calculus, particularly those focusing on differential equations, as well as educators seeking to clarify concepts related to integrating factors and differential notation.

robertjford80
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Homework Statement



Screenshot2012-05-15at41120AM.png




I don't understand why in step 2 dy turns into d and why +P(x) dissappears
I also don't see a difference between v(x)y and v(x)*y

In step three why does the d disappear. I see that dx is going over to the right side, well, what about the numerator d?
 
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d/dx is the differential operator. dy/dx just means d/dx operating on y. You should probably read a bit more elementary calculus to get used to this notation.

You start from a differential equation
[tex]\frac{dy}{dx} + P(x) y = Q(x)[/tex]
and then you multiply both sides of this equation with a function [itex]v(x)[/itex] (which you have to find first) which has the property that the left hand sides of steps 1 and 2 are equal.

In step 3, you integrate both sides of the equation with respect to x.
 

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