Hello,

Who can you solve this $\displaystyle y^{'}+1=4e^{-y}\sin x $ the problem is that I don't know who to bring in a standaard form

Who can help? Thanks.

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- May 8th 2006, 06:05 AMBertDiff Equation
Hello,

Who can you solve this $\displaystyle y^{'}+1=4e^{-y}\sin x $ the problem is that I don't know who to bring in a standaard form

Who can help? Thanks. - May 8th 2006, 06:53 AMCaptainBlackQuote:

Originally Posted by**Bert**

let $\displaystyle z(x)=y(x)+x$, then the equation becomes:

$\displaystyle

z'=4e^{-z+x}\sin(x)=4 e^{-z}\ e^x \sin(x)

$,

which is of variables seperable type.

RonL - May 8th 2006, 07:19 AMBert
Yes then $\displaystyle z^{'}(x)=y^{'}+1$ is the same as $\displaystyle z^{'}(x)=y^{'}$ ??

Greets. - May 8th 2006, 07:35 AMCaptainBlackQuote:

Originally Posted by**Bert**

RonL - May 8th 2006, 12:25 PMBert
what do you mean with LHS ? Greets.

- May 8th 2006, 12:43 PMJameson
After using CaptainBlack's suggestion you get that $\displaystyle \frac{dz}{dx}=4e^{-z}e^x\sin{(x)}$. And as he also said this is separable. Can you take it from here?

- May 8th 2006, 02:41 PMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

- May 8th 2006, 07:20 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

short of brute force (that is getting out my copy of Piaggio or Kreysig).

After about 20 seconds it became obvious that the LHS (left hand side)

could be converted to a derivative by that change of variable, and "Bob

was my uncle".

I expect that I have seen this all before, and it is just subconscious memories

surfacing.

Quote:

Which class of linear-equations it falls into? Or it that something you can up with yourself?

if something can be converted to one of these first (Its the only class,

other than linear constant coefficients, I can solve without having to

remind myself what to do next :confused: ).

RonL - May 8th 2006, 07:20 PMCaptainBlackQuote:

Originally Posted by**Bert**

RHS - Right Hand Side

RonL - May 8th 2006, 11:48 PMBert
thanks I lot but I' can't see were you stay wit the +1 on the RHS can you drop him of?

Greets. - May 9th 2006, 12:32 AMCaptainBlackQuote:

Originally Posted by**Bert**

Solve this, then change back to y, using y(x)=z(x)-x.

RonL - May 9th 2006, 12:55 AMBert
Of course I see the equation $\displaystyle y^{'}+1=4e^{-y}sin(x)$

And not $\displaystyle y'=4e^{-y}sin(x)$ I try to do your substitution on this equation and of course it can't

So thank you verry well you solove the problem Greets. - May 9th 2006, 05:24 AMBert
I have calculated and I get $\displaystyle 0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-e^{x}=e^y$

Is this the same as $\displaystyle e^y=2(sin(x)-cos(x))+ce^{-x}$

Thanks. - May 9th 2006, 07:05 AMCaptainBlackQuote:

Originally Posted by**Bert**

(that is the $\displaystyle c$ in the second of the equations in your post).

RonL - May 9th 2006, 07:10 AMBert
it must so $\displaystyle

0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-ce^{x}=e^y

$

is this oké?