# calculus

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

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1. 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.

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posted by drwls
2. 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

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posted by alame
3. 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

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posted by Andy

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