dy/dx= (2x-y)/(x+y) all subject to y(1) = 1 I am not sure where to take this. I divided everything by x and made the subsequent y/x = V and then worked from there. please can you help
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Originally Posted by manalive04 dy/dx= (2x-y)/(x+y) all subject to y(1) = 1 I am not sure where to take this. I divided everything by x and made the subsequent y/x = V and then worked from there. please can you help That sounds like a very good idea! What did you get?
i got an improper fraction which i am not sure what to do with x(dv/dx) = (2 - 2v - v^2)/1+v then took everything to the other side ==> (1+v)/(2-2v-v^2)dv = dx/x then to intergrate both sides but how do you do the LHS? it is topheavy?
Originally Posted by manalive04 i got an improper fraction which i am not sure what to do with x(dv/dx) = (2 - 2v - v^2)/1+v then took everything to the other side ==> (1+v)/(2-2v-v^2)dv = dx/x then to intergrate both sides but how do you do the LHS? it is topheavy? can you integrate the LHS now?
Originally Posted by skeeter can you integrate the LHS now? can you plese show me how im sorry im not confident here
hint:
Originally Posted by skeeter hint: what is u' im sorry i dont understand
Originally Posted by manalive04 what is u' im sorry i dont understand u is just a variable in this case. is a special case of integral and it equals it is saying that differentiating (the denominator) gives -2-2v which is the numerator. The poster said that u = 2-2v-v^2 and thus u' = -2-2v
Originally Posted by e^(i*pi) u is just a variable in this case. is a special case of integral and it equals it is saying that differentiating (the denominator) gives -2-2v which is the numerator. The poster said that u = 2-2v-v^2 and thus u' = -2-2v so what does this imply??? im so confused can u please make this clearer im sorry
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