## gerryliyana 3 years ago solve dy/dx = e^{x+y} (x+y)^{-1} - 1 y ????

1. MikiaseKebede

solve what

2. gerryliyana

$\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1$

3. MikiaseKebede

dame i am not at a level like this one, i am sorry

4. gerryliyana

ok..., no problem @MikiaseKebede nevermind :)

5. MikiaseKebede

6. MikiaseKebede

him

7. gerryliyana

ok

8. MikiaseKebede

he is good

9. nubeer

i think you have to apply the product rule

10. gerryliyana

what did u get it??

11. nubeer

hmm i havent solved it just trying to help you with it

12. nubeer

or there is another easier way u =x+y du/dx = 1 du =dx

13. mayankdevnani
14. mayankdevnani

solve dy/dx = e^{x+y}

15. gerryliyana

@mayankdevnani No, i mean $\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1$

16. mukushla

let $z=x+y$and u have$\frac{\text{d}z}{\text{d}x}=1+\frac{\text{d}y}{\text{d}x}$put this in the original equation$\frac{\text{d}z}{\text{d}x}-1=\frac{e^z}{z}-1$$\frac{\text{d}z}{\text{d}x}=\frac{e^z}{z}$u have a separable differential equation :)

17. gerryliyana

ok @mukushla, then i got $\int\limits z dz = \int\limits e^{z} dx$ $\frac{ z^{2} }{ 2 } = e^{z} x$ $\frac{ (x+y)^{2} }{ 2 } = e^{x+y} x$ Did I miss something?

18. mukushla

emm...there is a little mistake in ur separation$ze^{-z}\text{d}z=\text{d}x$

19. gerryliyana

$\int\limits \frac{ z }{ e^{z} } dz = \int\limits dx$

20. mukushla

exactly

21. gerryliyana

$\frac{ -1 + z }{ e^{z} } = x$ $\frac{ -1 + x + y }{ e^{x+y} } = x$ $\frac{ x+y-1 }{ x } = e^{x+y}$

22. mukushla

integration by parts for z :)

23. gerryliyana

$\int\limits ze^{-z} dz = \int\limits dx$ u = z, $$dv = e^{-z} dz$$, then du = dz, and $$v = -e^{-z}$$ So $-ze^{-z} - \int\limits -e^{-z} dz = \int\limits dx$ $-ze^{-z} - \frac{ e^{-z} }{ \ln e } = x$ then ??

24. hartnn

ln e = 1 re substitute z as x+y

25. gerryliyana

$-ze^{-z} - e^{-z} = x$ $-(x+y)e^{-(x+y)} - e^{(x+y)} = x$ $(-x-y - 1) e^{-(x+y)} = x$ $(1+x+y) =- x e^{(x+y)}$

26. gerryliyana

did i miss something?

27. hartnn

nopes, thats correct. :)

28. gerryliyana

so how to find y?

29. hartnn

i don't think, y can be isolated....

30. hartnn
31. gerryliyana

what it means to "W is the product log function"??

32. hartnn

even idk that... not a standard function for sure....

33. gerryliyana

ok Thank you all so much! i really appreciate it :)

34. hartnn

welcome ^_^