Separable Differential Equations

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SUMMARY

The discussion focuses on the validity of manipulating separable differential equations, specifically in the form \(\frac{dy}{dx} = h(x)g(y)\). Participants confirm that dividing by \(g(y)\) is justified, as it allows the equation to be rearranged and integrated correctly. The integration leads to the conclusion \(G(y) = H(x) + c\), where \(G(y)\) and \(H(x)\) are antiderivatives of \(\frac{1}{g(y)}\) and \(h(x)\), respectively. This method is widely accepted as an intuitive approach to solving separable differential equations.

PREREQUISITES
  • Understanding of differential equations, specifically separable forms.
  • Familiarity with integration techniques and antiderivatives.
  • Knowledge of the chain rule in calculus.
  • Basic algebraic manipulation of fractions and limits.
NEXT STEPS
  • Study the derivation and application of the chain rule in calculus.
  • Explore advanced integration techniques, including substitution and integration by parts.
  • Learn about different types of differential equations beyond separable ones.
  • Investigate the implications of manipulating differential equations in various mathematical contexts.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and differential equations, as well as educators seeking to clarify the concepts of separable differential equations.

Bashyboy
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I have read that, if you given a differential equation [itex]\frac{dy}{dx} = f(x,y)[/itex], and can write it in the form [itex]\frac{dy}{dx} = h(x)g(y)[/itex], then you can proceed with the following steps:

[itex]\frac{dy}{g(y)} = h(x)dx[/itex]

integrating

[itex]G(y) = H(x) + c[/itex]

Why are these steps vaild? I thought that one was not supposed to regard [itex]\frac{dy}{dx}[/itex]. I have heard that you can regard it as a fraction, because, before taking the limit, you can manipulate the fraction [itex]\frac{\Delta y}{\Delta x}[/itex].

Could someone please help?
 
Last edited:
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The justification is that dividing by g(y) results in

[tex]\frac{\mathrm{d}y}{\mathrm{d}x}=h(x)g(y)\implies h(x)=\frac{1}{g(y)}\frac{ \mathrm{d} y}{\mathrm{d}x}[/tex]

which then, supposing (chain rule) that for some function G(y)
[tex]\frac{\mathrm{d}}{\mathrm{d}y}G(y)=\frac{1}{g(y)}\implies\frac{\mathrm{d}}{\mathrm{d}x}G(y)=\frac{1}{g(y)}\frac{\mathrm{d}y}{\mathrm{d}x},[/tex]

can be integrated to

[tex]\int h(x) \mathrm{d} x = \int \frac{1}{g(y)}\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x = \int (\frac{\mathrm{d}}{\mathrm{d}x}G(y)) \mathrm{d}x= G(y) = \int (\frac{\mathrm{d}}{\mathrm{d}y}G(y))\mathrm{d}y=\int \frac{1}{g(y)} \mathrm{d}y,[/tex]

the result you get if you "divide by dy". The steps you made are commonly used as they are often considered more intuitive.
 
Last edited:
You can think of the steps that seem incorrect as mnemonics to obtain a correct result. The justification is the one given above.
 

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