Find f(e^-1).

Question

If f is the solution of x*f'(x)-f(x)=x such that f(-1)=1, find f(e^-1).

 

Answer: -2e^-1

 

 

Answer 1

Rewrite the equation as f'(x) = 1 + f(x)/x, and you see that it is homogeneous (i.e., f'(x) can be written in terms of f(x)/x).

 

For homogeneous equations, use the change of variable f(x)=u(x) x, f'(x) = u'(x) x + u(x). You get

x(u'(x) x + u(x)) - (u(x) x) = x 

⇒ u'(x)=1/x

⇒ u(x)= ln(abs(x)) +c

⇒ f(x) = x ln(abs(x)) + cx

Impose the initial condition to find c:

f(-1) = -ln(1)-c = 1

⇒c=-1

So the unique solution is

f(x) = x ln(abs(x)) - x

with

f(e^-1) = e^-1 ln(e^-1) - e^-1 = -2e^-1.

Answer 2

Divide both sides by x² and use the quotient rule with (u/v)' = (u'v-uv')/v² where u=f(x) and v=x.

 

x·f'(x) - f(x) = x

 

(f'(x)·x - f(x)·1)/x² = x^-1

 

(f(x)/x)' = x^-1

 

f(x)/x = ln |x| + C

 

f(x) = x ln |x| +Cx

 

Now given f(-1)=1 you get

 

-1·ln |-1| + C·(-1) = 1 ⇒ C = -1

 

Thus f(x) = x ln |x| - x

 

Hence f(e^-1) = e^-1·ln |e^-1| - e^-1 = -2e^-1