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Solving dy/dx=x-y

  • Thread starter computerex
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  • #1
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Homework Statement



[tex]\int (x-y) [/tex]

What is [tex]\int y[/tex] ? I don't mean [tex]\int y dy[/tex].

The Attempt at a Solution



[tex]dy/dx = x-y[/tex]
[tex]y + dy = x dx[/tex]
[tex]\int y + \int dy = \int x dx[/tex]
[tex]\int y + y = x^2/2 + c[/tex]

I am stuck at this point because I don't know how to integrate y without respect to anything...If that even makes sense.

EDIT:

Nvm...I am stupid xD

[tex]\int x dx - \int y dx [/tex]
 
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Answers and Replies

  • #2
gabbagabbahey
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EDIT:

Nvm...I am stupid xD

[tex]\int x dx - \int y dx [/tex]
That still doesn't help you though. How exactly do you plan on integrating [itex]\int y(x) dx[/itex] when you don't know what [itex]y(x)[/itex] is?

You can't solve this differential equation just by integrating both sides. Instead, try using the substitution [itex]u=x-y[/itex] to rewrite the DE in terms of [itex]u(x)[/itex] and [itex]u'(x)[/itex].
 
  • #3
D H
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Instead, try using the substitution [itex]u=x-y[/itex] to rewrite the DE in terms of [itex]u(x)[/itex] and [itex]u'(x)[/itex].
That will result in another nonhomogeneous ODE. A tiny bit simpler perhaps, but still nonhomogeneous.

computerex: What have you been taught regarding solving nonhomogeneous differential equations?
 
  • #4
gabbagabbahey
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That will result in another nonhomogeneous ODE. A tiny bit simpler perhaps, but still nonhomogeneous.
Something about separable ODE's appeals to me though:wink:
 
  • #5
D H
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Depends on the OP's background. The original problem can be rewritten as

[tex]\frac{dy}{dx} + y = x[/tex]

The homogeneous and particular solutions can be read off just by inspection.
 

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