dy/dx= e^(5x-5y)
a. y = 5ln (e^5x + C)
b. y = 1/5(ln (e^5x + C) )
c. y = ln (e^5x + C)
d. y = 5e^5x + C
2007-03-24 19:05:35 · 3 answers · asked by namereg b 1 in Science & Mathematics ➔ Mathematics
I think its (b), because when taking the derivative of b, the 5 that comes out of the 5x cancels with 1/5 to produce the constant of 1 thats currently before the e^(5x-5y).
2007-03-24 19:12:39 · answer #1 · answered by hello_be_happy 2 · 0⤊ 0⤋
dy/dx = (e^5x)*(e^-5y)
e^5y dy = e^5x dx
â«e^5y dy = â«e^5x dx
(1/5)e^5y = (1/5)e^5x + K
e^5y = e^5x + C
5y = ln(e^5x + C)
y = (1/5)ln(e^5x + C)
2007-03-24 19:20:30 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
the 'b' option seems to beto some extent correct tho ther is a slight changein it.....
=> dy/dx= e^(5x-5y)
=> dy/dx=e^5x/e^5y
=> dy.e^5y=dxe^5x
e^5/5=e^5x+c
taking log on both sides,,
we get
5y=1/5(log5{e^5x/5+C)
2007-03-24 19:18:38 · answer #3 · answered by jedi Knight 2 · 0⤊ 0⤋