1. ## Diff Equation

Hello,

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

Who can help? Thanks.

2. Originally Posted by Bert
Hello,

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

Who can help? Thanks.
Try this:

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

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

which is of variables seperable type.

RonL

3. Yes then $z^{'}(x)=y^{'}+1$ is the same as $z^{'}(x)=y^{'}$ ??

Greets.

4. Originally Posted by Bert
Yes then $z^{'}(x)=y^{'}+1$ is the same as $z^{'}(x)=y^{'}$ ??

Greets.
No $z'=y'+1$, so the LHS of the DE is $z'$

RonL

5. what do you mean with LHS ? Greets.

6. After using CaptainBlack's suggestion you get that $\frac{dz}{dx}=4e^{-z}e^x\sin{(x)}$. And as he also said this is separable. Can you take it from here?

7. Originally Posted by CaptainBlack
Try this:

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

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

which is of variables seperable type.

RonL
I am not am expert in differencial equation (I do not even know how to spell the word) how did you know how to do that? Which class of linear-equations it falls into? Or it that something you can up with yourself?

8. Originally Posted by ThePerfectHacker
I am not am expert in differencial equation (I do not even know how to spell the word) how did you know how to do that?
I didn't. I looked at the problem for a while trying to see a way forward
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.

Which class of linear-equations it falls into? Or it that something you can up with yourself?
Variables separable is one of the simplest classes of ODE, I always check
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 ).

RonL

9. Originally Posted by Bert
what do you mean with LHS ? Greets.
LHS - Left Hand Side
RHS - Right Hand Side

RonL

10. thanks I lot but I' can't see were you stay wit the +1 on the RHS can you drop him of?

Greets.

11. Originally Posted by Bert
thanks I lot but I' can't see were you stay wit the +1 on the RHS can you drop him of?

Greets.
The change of variables gives a new DE for z as a function of x.

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

RonL

12. Of course I see the equation $y^{'}+1=4e^{-y}sin(x)$

And not $y'=4e^{-y}sin(x)$ 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.

13. I have calculated and I get $0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-e^{x}=e^y$

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

Thanks.

14. Originally Posted by Bert
I have calculated and I get $0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-e^{x}=e^y$

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

Thanks.
First thing to note is that you have lost the arbitary constant somewhere.
(that is the $c$ in the second of the equations in your post).

RonL

15. it must so $
0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-ce^{x}=e^y
$

is this oké?