- #1
newtomath
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The problem is :
(y*e^(2xy) +x) + [ b* x*e^(2xy) ] y' =0. Find b so the equation is exact and solve.
I found b=1 and worked the problem to (1/2)e^(2xy) + (x^2/2)e^(2xy) + h(y); where I found h(y) to be simply c.
The answer in the text states that e^(2xy) + x^2 =c
Would I be incorrect to multiply [ (1/2)e^(2xy) + (x^2/2)e^(2xy) = -c ] by 2 to arrive at the answer since c is an arbitrary constant?
Thanks
(y*e^(2xy) +x) + [ b* x*e^(2xy) ] y' =0. Find b so the equation is exact and solve.
I found b=1 and worked the problem to (1/2)e^(2xy) + (x^2/2)e^(2xy) + h(y); where I found h(y) to be simply c.
The answer in the text states that e^(2xy) + x^2 =c
Would I be incorrect to multiply [ (1/2)e^(2xy) + (x^2/2)e^(2xy) = -c ] by 2 to arrive at the answer since c is an arbitrary constant?
Thanks