# Again, Saperable equation!

• Feb 19th 2010, 02:42 PM
Ted
Again, Saperable equation!
Hello.
Am stuck at this problem in saperable equations' section in my book.

Problem: Solve the differential equation $\frac{dy}{dx}=x+y$ by making the change of variable $u=x+y$.

My solution:
I did not see any similar problems in my class' lectures and in the book's section.
All what I can do it is rewrite it as:
$?=u$.
I put "?" because I do not know how to express $\frac{dy}{dx}$ in terms of u. i.e, expressing it using the suggested substiution.
Any help??
• Feb 19th 2010, 03:06 PM
Krizalid
we have $\frac{du}{dx}=1+\frac{dy}{dx}$ so the ODE becomes $\frac{du}{dx}-1=u,$ which is separable.
• Feb 19th 2010, 03:22 PM
Ted
Quote:

Originally Posted by Krizalid
we have $\frac{du}{dx}=1+\frac{dy}{dx}$ so the ODE becomes $\frac{du}{dx}-1=u,$ which is separable.

Sorry, but I did not follow you.
How did you got:
$\frac{du}{dx}=1+\frac{dy}{dx}$
?
• Feb 19th 2010, 03:35 PM
o_O
Take the derivative of both sides with respect to $x$ ..

$u = x + y \ \Rightarrow \ \frac{du}{dx} = 1 + \frac{dy}{dx}$
• Feb 19th 2010, 03:41 PM
Ted
My bad =(. I did not notice that.
$u=\pm Ae^x - 1$, where $A=e^C$ , $C$ is a constant. Right?
• Feb 20th 2010, 10:39 AM
Krizalid
no, actually i'm getting $\ln \left| x+y+1 \right|=x+k.$
• Feb 20th 2010, 10:44 AM
o_O
Those two are equivalent though.
• Feb 20th 2010, 11:51 AM
shayan603
true they are equivilent
• Feb 20th 2010, 04:46 PM
Prove It
Quote:

Originally Posted by Ted
Hello.
Am stuck at this problem in saperable equations' section in my book.

Problem: Solve the differential equation $\frac{dy}{dx}=x+y$ by making the change of variable $u=x+y$.

My solution:
I did not see any similar problems in my class' lectures and in the book's section.
All what I can do it is rewrite it as:
$?=u$.
I put "?" because I do not know how to express $\frac{dy}{dx}$ in terms of u. i.e, expressing it using the suggested substiution.
Any help??

There's no need for change of variable. Why not use the integrating factor method?

$\frac{dy}{dx} = x + y$

$\frac{dy}{dx} - y = x$

The integrating factor is $e^{\int{-1\,dx}} = e^{-x}$.

Multiplying through gives

$e^{-x}\frac{dy}{dx} - e^{-x}y = x\,e^{-x}$

$\frac{d}{dx}(e^{-x}y) = x\,e^{-x}$

$e^{-x}y = \int{x\,e^{-x}\,dx}$

Now letting $u = x$ so that $du = 1$ and $dv = e^{-x}$ so that $v = -e^{-x}$...

$e^{-x}y = -x\,e^{-x} - \int{-e^{-x}\,dx}$

$e^{-x}y = -x\,e^{-x} - e^{-x} + C$

$y = -x - 1 + Ce^{x}$
• Feb 20th 2010, 04:53 PM
o_O
Quote:

Originally Posted by Prove It
There's no need for change of variable. Why not use the integrating factor method?

Because the question said so.

Quote:

Originally Posted by Ted
Problem: Solve the differential equation $\frac{dy}{dx} = x + y$ by making the change of variable $u=x+y$.

• Feb 20th 2010, 05:09 PM
Prove It
Quote:

Originally Posted by o_O
Because the question said so.

That's fair enough, but I still don't see the point in changing variables just to turn a first order linear ODE into another first order linear ODE...
• Feb 20th 2010, 05:54 PM
Krizalid
Quote:

Originally Posted by Prove It
That's fair enough, but I still don't see the point in changing variables just to turn a first order linear ODE into another first order linear ODE...

let me tell ya something:

the same as computing very easy integrals by using trig. substitutions.

us: what's the point?

them: because we were told we needed to do it on that way.

so people are learning ways on solving ODEs, of course we can find faster ways, but the same happens when computing an easy integral when using trig. sub., since for us makes no sense.
• Feb 20th 2010, 07:08 PM
Prove It
Quote:

Originally Posted by Krizalid
let me tell ya something:

the same as computing very easy integrals by using trig. substitutions.

us: what's the point?

them: because we were told we needed to do it on that way.

so people are learning ways on solving ODEs, of course we can find faster ways, but the same happens when computing an easy integral when using trig. sub., since for us makes no sense.

I'm just playing devil's advocate here.

If the point is to learn how to solve ODEs using a change of variables, then why not give them some examples and exercises that CAN'T be solved in quicker and easier ways?
• Feb 21st 2010, 04:42 AM
Ted
Ohhh.
What happened here? @@
OK ProveIt, thanks for the full solution.
Sure, I noticed its a linear equation.
but i want to see how to solve it using the change of variable.
since am new to the ODE, I wondered when I see "by change of variables", because I never see this method.
Thanks all.
• Feb 21st 2010, 01:08 PM
Prove It
Quote:

Originally Posted by Ted
Ohhh.
What happened here? @@
OK ProveIt, thanks for the full solution.
Sure, I noticed its a linear equation.
but i want to see how to solve it using the change of variable.
since am new to the ODE, I wondered when I see "by change of variables", because I never see this method.
Thanks all.

Have a read of the Differential Equations tutorial in the Differential Equations sub-forum...