## gerryliyana 3 years ago solve dy/dx= (x + 4y - 1)^{2}

1. gerryliyana

$\frac{ dy }{ dx } = (x + 4y - 1)^{2}$

2. gerryliyana

@jim_thompson5910 can u help me?

3. ksaimouli

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4. gerryliyana

i'm sorry @ksaimouli , but i cannot see clearly

5. ksaimouli

do u want to solve for dy/dx

6. ksaimouli

use chain rule

7. gerryliyana

how to find y ??

8. ksaimouli

then how do u find x ?

9. ksaimouli

nothing but 1 dx/dxso y =1 dy/dx

10. gerryliyana

ok.., is it use separable equation or linear?

11. ksaimouli

it is linear no need to find y until they mentioned y=something u will find it in optimization

12. ksaimouli

hold on is this integration

13. gerryliyana

ok i'll try., :)

14. saifoo.khan

You have to differentiate it? Right?

15. gerryliyana

i have to find the solution of differential equations...,

16. saifoo.khan

Meaning you have to differentiate that equation?

17. gerryliyana

yes

18. hartnn

dy/dx= (x + 4y - 1)^{2} u = x+4y-1 du/dx = 1 + 4dy/dx so, dy/dx = (1/4)*(du/dx -1) (1/4)*(du/dx -1) = u^2 du/dx = 4u^2 +1 now variables are easily seperable.

19. gerryliyana

@hartnn

20. gerryliyana

$\frac{ du }{ dx } = 4u^{2} + 1$ $\int\limits \frac{ du }{ 4u^{2}+1 } = \int\limits dx$ $\int\limits \frac{ du }{ (2u)^{2}+1 } = \int\limits dx$ $\tan^{-1} (2u) = x$ $\tan^{-1} (x+4y-1) = x$ $\tan x = x +4y-1$ $4y = \tan x - x + 1$ $y = \frac{ \tan x - x + 1 }{ 4 }$ am i right @hartnn ??

21. gerryliyana

oh i'm sorry, i made a mistake...,

22. hartnn

tan^-1 2u /2 this ?

23. gerryliyana

$\tan^{-1} (2(x+4y-1)) = x$ $\tan^{-1} (2x+8y-2)) = x$ $\tan x = 2x + 8y - 2$ $y = \frac{ \tan x - 2x + 2 }{ 8 }$

24. hartnn

the co-efficient is divided, not multiplied.

25. hartnn

(tann^-1.... )/2

26. hartnn

tan2x = x+4y-1 + c don't forget +c

27. gerryliyana

ok $\tan 2x = x + 4y -1 + C$ $y = \frac{ \tan 2x - x + 1 - C }{ 4 }$ is it right ?

28. hartnn

yes, seems correct now :)

29. gerryliyana

yes i got it.., Thanks a lot. I appreciate all your help :)

30. hartnn

welcome ^_^