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.
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
3 answers
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
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.
Sorry
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.
Sorry
2 answers