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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx } = (x + 4y  1)^{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 can u help me?

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353368437237:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm sorry @ksaimouli , but i cannot see clearly

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0do u want to solve for dy/dx

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0then how do u find x ?

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0nothing but 1 dx/dxso y =1 dy/dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok.., is it use separable equation or linear?

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0it is linear no need to find y until they mentioned y=something u will find it in optimization

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0hold on is this integration

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0You have to differentiate it? Right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have to find the solution of differential equations...,

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0Meaning you have to differentiate that equation?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1dy/dx= (x + 4y  1)^{2} u = x+4y1 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ du }{ dx } = 4u^{2} + 1\] \[\int\limits \frac{ du }{ 4u^{2}+1 } = \int\limits dx\] \[\int\limits \frac{ du }{ (2u)^{2}+1 } = \int\limits dx\] \[\tan^{1} (2u) = x\] \[\tan^{1} (x+4y1) = x\] \[\tan x = x +4y1\] \[4y = \tan x  x + 1\] \[y = \frac{ \tan x  x + 1 }{ 4 }\] am i right @hartnn ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i'm sorry, i made a mistake...,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\tan^{1} (2(x+4y1)) = x\] \[\tan^{1} (2x+8y2)) = x\] \[\tan x = 2x + 8y  2\] \[y = \frac{ \tan x  2x + 2 }{ 8 }\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1the coefficient is divided, not multiplied.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1tan2x = x+4y1 + c don't forget +c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok \[\tan 2x = x + 4y 1 + C\] \[y = \frac{ \tan 2x  x + 1  C }{ 4 }\] is it right ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yes, seems correct now :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i got it.., Thanks a lot. I appreciate all your help :)
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