Special case when solving D.E.'s

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Discussion Overview

The discussion revolves around the conditions for linear independence of terms in the complementary function and particular integral when solving differential equations. Participants explore the implications of having common terms in these components and the necessity of modifying terms to maintain linear independence.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions why common terms cannot exist in both the complementary function and particular integral, using a specific example involving exponential functions.
  • Another participant suggests testing the proposed solution by substituting it back into the differential equation to solve for constants.
  • A participant notes that due to the nature of differentiation, certain products will lead to internal cancellations, affecting the solution's form.
  • There is a suggestion that if terms are not linearly independent, certain constants would vanish, which raises questions about practical implications.
  • One participant asks for clarification on the derivation of a specific equation related to the differentiation of a product.
  • A later reply explains the origin of the equation by defining a function in terms of an exponential and applying the differential operator.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of linear independence in the context of the differential equation, and the discussion remains unresolved regarding the implications of having common terms.

Contextual Notes

Some assumptions about the nature of linear independence and the behavior of exponential functions in differential equations are not fully explored, leaving certain mathematical steps and implications unresolved.

unseensoul
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Why can't there be common terms on both complementary function and particular integral when solving differential equations?

For instance,

dy/dx + 3y = exp(-x) + exp(-3x)

y(CF) = Aexp(−3x)

y(PI) = Cexp(−x) + Dxexp(−3x)

The term Dexp(-3x) in the P.I. has to be multiplied by x to be linearly independent of Aexp(-3x) in the C.F.. Why? What would happen if it wasn't?
 
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Try it and see! Take y(x)= C exp(-x)+ D exp(-3x) and put it into the equation. Try to solve for C and D to make it true.
 
Note that due to the characteristics of the differentiation of a product, ALL products of the form Y(x)=f(x)e^(-3x) will experience some "internal cancellation".

In your case, we will get on LHS:
\frac{dY}{dx}+3Y=(f^{,}(x)e^{-3x}-3f(x)e^{-3x})+3f(x)e^{-3x}=f^{,}(x)e^{-3x}
Thus, since this must equal identically RHS, i.e, e^{-3x}, we get the subsidiary equation on f:
f^{,}(x)=1, which is readily solved.
 
HallsofIvy said:
Try it and see! Take y(x)= C exp(-x)+ D exp(-3x) and put it into the equation. Try to solve for C and D to make it true.

D would vanish... That's why they have to be linearly independent? I can't think of it in practical situations.
 
arildno said:
\frac{dY}{dx}+3Y=(f^{,}(x)e^{-3x}-3f(x)e^{-3x})+3f(x)e^{-3x}=f^{,}(x)e^{-3x}

Where did you get this from?
 
From defining Y(x) as f(x)e^(-3x), and applying the LHS operator on it!
 

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