Differential equation Watch

Miss Mary
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#1
Report Thread starter 8 years ago
#1
I've got xy'=y+xe^{\frac{y}{x}}.

I've rearranged to get y=\int \frac{y}{x} dx+\int e^{\frac{y}{x}} dx.

Where do I go from here?
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spex
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#2
Report 8 years ago
#2
rearrange to get y'= y/x + exp(y/x)

Then use the substitution u = y/x and see what happens
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Miss Mary
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#3
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(Original post by spex)
rearrange to get y'= y/x + exp(y/x)

Then use the substitution u = y/x and see what happens
You mean y'=u+e^{u}?

Assuming I did everything correctly (integrating both sides and then subbing u), I've got y=\frac{y^2}{2x^2}+\frac{x}{y}e^  {\frac{y}{x}}. Can't see how to make y the subject
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Clarity Incognito
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#4
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(Original post by Miss Mary)
You mean y'=u+e^{u}?

Assuming I did everything correctly (integrating both sides and then subbing u), I've got y=\frac{y^2}{2x^2}+\frac{x}{y}e^  {\frac{y}{x}}. Can't see how to make y the subject
If you're using the substitution  u = \dfrac{y}{x} .

You have to change so that the whole DE is in terms of u and x only. You don't want to be integrating with y's, x's and u's.
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Miss Mary
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#5
Report Thread starter 8 years ago
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(Original post by Clarity Incognito)
If you're using the substitution .

You have to change so that the whole DE is in terms of u and x only. You don't want to be integrating with y's, x's and u's.
u=\frac{y}{x} \longrightarrow y=ux


So then...
y'=u+e^{u}
\int y' dx=\int u du+\int e^{u} du
ux=\int u du+\int e^{u} du
ux=\frac{u^2}{2}+\frac{e^u}{u}

What now?
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ghostwalker
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#6
Report 8 years ago
#6
(Original post by Miss Mary)
....
u=\frac{y}{x} \longrightarrow y=ux

So, differentiating

y'=u+xu'

Now substitute into y'= y/x + exp(y/x), the y' as well as the y/x parts.
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Clarity Incognito
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#7
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#7
(Original post by Miss Mary)
y'=u+e^{u}
\int y' dx=\int u du+\int e^{u} du
Above two lines do not lead on from each other. With the new substitution, what does y' become, it might be easier to look at it as  \dfrac{dy}{dx}
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