Differential Equations help??? Watch

username3560526
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Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

What I’ve got so far:
dy/du=-1/u^2
u^2(dy/dx) + u = e^(2x)
-du/dx + u = e^(2x)
-du + u dx = e^(2x) dx
But I’m stuck here because you cannot integrate (u dx).

Many thanks
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B_9710
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#2
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(Original post by Constan4699)
Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

What I’ve got so far:
dy/du=-1/u^2
u^2(dy/dx) + u = e^(2x)
-du/dx + u = e^(2x)
-du + u dx = e^(2x) dx
But I’m stuck here because you cannot integrate (u dx).

Many thanks
It looks as though you have just multiplied by dx which you can't do.
You will need to solve the resulting ODE by using an integrating factor because this equation is not separable.
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sindyscape62
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(Original post by Constan4699)
Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

What I’ve got so far:
dy/du=-1/u^2
u^2(dy/dx) + u = e^(2x)
-du/dx + u = e^(2x)
-du + u dx = e^(2x) dx
But I’m stuck here because you cannot integrate (u dx).

Many thanks
Consider your third line:
 \frac{dy}{dx} - u=-e^{2x}

Have you covered the use of integrating factors? If so do you see how to apply one here?
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_gcx
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Do you know how to do first order nonseperable ODEs?

Look at the coefficient of the u term to determine the integrating factor.

A little prod,

Spoiler:
Show
does e^{\int P \mathrm dx} look familiar?
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username3560526
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#5
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No I’ve never seen the use of integrating factors - probably why I’m unable to progress then. Please help?
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_gcx
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(Original post by Constan4699)
No I’ve never seen the use of integrating factors - probably why I’m unable to progress then. Please help?
Maybe watch a video (KhanAcademy is normally good) on integrating factors before continuing? Probably better than explaining it to you here, on the spot.
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username3560526
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#7
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So from my understanding it seems that the integrating factor is e^(-x) (hope this is correct?). So starting again:

du/dx - u = -e^(2x)
So e^(-x)(du/dx) - ue^(-x) = - e^x
Can anyone give me any pointers from here?
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_gcx
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(Original post by Constan4699)
So from my understanding it seems that the integrating factor is e^(-x) (hope this is correct?). So starting again:

du/dx - u = -e^(2x)
So e^(-x)(du/dx) - ue^(-x) = - e^x
Can anyone give me any pointers from here?
Recall the product rule,

(uv)' = u'v + uv'

Also that \displaystyle \frac{\mathrm d}{\mathrm dx} u = \frac{\mathrm du}{\mathrm dx}. (you should also be familiar with the derivative of e^x) Does this help?
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username3560526
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(Original post by _gcx)
Recall the product rule,

(uv)' = u'v + uv'

Also that \displaystyle \frac{\mathrm d}{\mathrm dx} u = \frac{\mathrm du}{\mathrm dx}. (you should also be familiar with the derivative of e^x) Does this help?
Yep thank you very much I think I’ve got it now
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