anonymous
  • anonymous
solve dy/dx= (x + 4y - 1)^{2}
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ dy }{ dx } = (x + 4y - 1)^{2}\]
anonymous
  • anonymous
@jim_thompson5910 can u help me?
ksaimouli
  • ksaimouli
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anonymous
  • anonymous
i'm sorry @ksaimouli , but i cannot see clearly
ksaimouli
  • ksaimouli
do u want to solve for dy/dx
ksaimouli
  • ksaimouli
use chain rule
anonymous
  • anonymous
how to find y ??
ksaimouli
  • ksaimouli
then how do u find x ?
ksaimouli
  • ksaimouli
nothing but 1 dx/dxso y =1 dy/dx
anonymous
  • anonymous
ok.., is it use separable equation or linear?
ksaimouli
  • ksaimouli
it is linear no need to find y until they mentioned y=something u will find it in optimization
ksaimouli
  • ksaimouli
hold on is this integration
anonymous
  • anonymous
ok i'll try., :)
saifoo.khan
  • saifoo.khan
You have to differentiate it? Right?
anonymous
  • anonymous
i have to find the solution of differential equations...,
saifoo.khan
  • saifoo.khan
Meaning you have to differentiate that equation?
anonymous
  • anonymous
yes
hartnn
  • 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.
anonymous
  • anonymous
@hartnn
anonymous
  • anonymous
\[\frac{ du }{ dx } = 4u^{2} + 1\] \[\int\limits \frac{ du }{ 4u^{2}+1 } = \int\limits dx\] \[\int\limits \frac{ du }{ (2u)^{2}+1 } = \int\limits dx\] \[\tan^{-1} (2u) = x\] \[\tan^{-1} (x+4y-1) = x\] \[\tan x = x +4y-1\] \[4y = \tan x - x + 1\] \[y = \frac{ \tan x - x + 1 }{ 4 }\] am i right @hartnn ??
anonymous
  • anonymous
oh i'm sorry, i made a mistake...,
hartnn
  • hartnn
tan^-1 2u /2 this ?
anonymous
  • anonymous
\[\tan^{-1} (2(x+4y-1)) = x\] \[\tan^{-1} (2x+8y-2)) = x\] \[\tan x = 2x + 8y - 2\] \[y = \frac{ \tan x - 2x + 2 }{ 8 }\]
hartnn
  • hartnn
the co-efficient is divided, not multiplied.
hartnn
  • hartnn
(tann^-1.... )/2
hartnn
  • hartnn
tan2x = x+4y-1 + c don't forget +c
anonymous
  • anonymous
ok \[\tan 2x = x + 4y -1 + C\] \[y = \frac{ \tan 2x - x + 1 - C }{ 4 }\] is it right ?
hartnn
  • hartnn
yes, seems correct now :)
anonymous
  • anonymous
yes i got it.., Thanks a lot. I appreciate all your help :)
hartnn
  • hartnn
welcome ^_^

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