Posted by sammy on Monday, May 19, 2008 at 3:25am.
consider the differential equation dy/dx= (y - 1)/ x squared where x not = 0
a) find the particular solution y= f(x) to the differential equation with the initial condition f(2)=0
(b)for the particular solution y = F(x) described in part (a) find lim F(x) X goes to infinity
calculus - drwls, Monday, May 19, 2008 at 6:32am
dy/dx = (y - 1)/ x^2
Use the integration method of separation of variables.
dy/(y-1) = dx/x^2
ln (y-1) = -1/x + C
You say that the initial condition is
f(x) = y = 0 at x =2
Are you sure the initial condition is not f(0) = 2 ? You cannot have a logarithm of a negative number.
calculus - alame, Tuesday, March 16, 2010 at 6:27pm
Start like above, but once you gget to the logarithm part, it is in absolute values, so it is ln(1) which is 0. Solve from there
calculus - Andy, Saturday, March 19, 2011 at 2:56pm
The two statements above is wrong.
you first move around the
dy/dx = (y-1)/x^2 into
dy/(y-1) = dx/x^2
Then you integral where it turns to
ln( (y-1)/C ) = -1/x
Next you move the e from the ln to the other side
(y-1)/C = e^(-1/x)
After that you times C to both sides and move the -1 afterwords
y-1 = Ce^(-1/x) --> y = Ce^(-1/x)+1
I not sure how you do part (b) though.
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