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gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ dy }{ dx } = (x + 4y  1)^{2}\]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
@jim_thompson5910 can u help me?
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1353368437237:dw
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
i'm sorry @ksaimouli , but i cannot see clearly
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
do u want to solve for dy/dx
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
use chain rule
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
how to find y ??
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
then how do u find x ?
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
nothing but 1 dx/dxso y =1 dy/dx
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
ok.., is it use separable equation or linear?
 one year ago

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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
hold on is this integration
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
ok i'll try., :)
 one year ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
You have to differentiate it? Right?
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
i have to find the solution of differential equations...,
 one year ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
Meaning you have to differentiate that equation?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
dy/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.
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
@hartnn
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ du }{ dx } = 4u^{2} + 1\] \[\int\limits \frac{ du }{ 4u^{2}+1 } = \int\limits dx\] \[\int\limits \frac{ du }{ (2u)^{2}+1 } = \int\limits dx\] \[\tan^{1} (2u) = x\] \[\tan^{1} (x+4y1) = x\] \[\tan x = x +4y1\] \[4y = \tan x  x + 1\] \[y = \frac{ \tan x  x + 1 }{ 4 }\] am i right @hartnn ??
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
oh i'm sorry, i made a mistake...,
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
tan^1 2u /2 this ?
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
\[\tan^{1} (2(x+4y1)) = x\] \[\tan^{1} (2x+8y2)) = x\] \[\tan x = 2x + 8y  2\] \[y = \frac{ \tan x  2x + 2 }{ 8 }\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
the coefficient is divided, not multiplied.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
(tann^1.... )/2
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
tan2x = x+4y1 + c don't forget +c
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
ok \[\tan 2x = x + 4y 1 + C\] \[y = \frac{ \tan 2x  x + 1  C }{ 4 }\] is it right ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yes, seems correct now :)
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
yes i got it.., Thanks a lot. I appreciate all your help :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago
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