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solve dy/dx= (x + 4y - 1)^{2}

Differential Equations
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\[\frac{ dy }{ dx } = (x + 4y - 1)^{2}\]
@jim_thompson5910 can u help me?
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Other answers:

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

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