# Differential Equations help???Watch

#1
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
0
1 year ago
#2
(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.
0
1 year ago
#3
(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

Have you covered the use of integrating factors? If so do you see how to apply one here?
0
1 year ago
#4
Do you know how to do first order nonseperable ODEs?

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

A little prod,

Spoiler:
Show
does look familiar?
0
#5
No I’ve never seen the use of integrating factors - probably why I’m unable to progress then. Please help?
0
1 year ago
#6
(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.
0
#7
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?
0
1 year ago
#8
(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,

Also that . (you should also be familiar with the derivative of e^x) Does this help?
0
#9
(Original post by _gcx)
Recall the product rule,

Also that . (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
1
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