# Calculus Integration problem. NO clue

• Mar 11th 2008, 08:15 PM
Naveed103
Calculus Integration problem. NO clue
Question is as follows :

Let f be the function satisfying f ' (x) = -3xf(x) for all real numbers x, with f(1) = 4 and lim as x -> infinity of f(x) = 0.

a.) Evaluate the integral of -3xf(x) dx with bounds 1 and infinity.

b.) Use Euler's method, starting at x = 1 with a step size of .5 to approximate f(2)

c.) Write an expression for y = f(x) by solving the differential equation dy/dx = -3xy with the initial condition f(1) = 4

I dont really need help with b, once I figure out the equation I can do that one. But A and C have me completely stumped.
• Mar 11th 2008, 08:45 PM
mr fantastic
Quote:

Originally Posted by Naveed103
Question is as follows :

Let f be the function satisfying f ' (x) = -3xf(x) for all real numbers x, with f(1) = 4 and lim as x -> infinity of f(x) = 0.

a.) Evaluate the integral of -3xf(x) dx with bounds 1 and infinity.

b.) Use Euler's method, starting at x = 1 with a step size of .5 to approximate f(2)

c.) Write an expression for y = f(x) by solving the differential equation dy/dx = -3xy with the initial condition f(1) = 4

I dont really need help with b, once I figure out the equation I can do that one. But A and C have me completely stumped.

a.) Integrate both sides of f ' (x) = -3xf(x): $f(x) = \int -3x f(x) \, dx$.

Therefore $\int_{1}^{\infty} -3x f(x) \, dx = \lim_{a \rightarrow +\infty} f(a) - f(1) = 0 - 4 = -4$.

c.) dy/dx = -3xy is seperable: $\frac{1}{y}\, dy = -3x \, dx$ etc. etc.
• Mar 11th 2008, 09:10 PM
Naveed103
thanks a lot for your help on a. So if I integrate c, I get y/4 = -3/2 x^2 + c. f(1) = 4, and c = 10, so the final solution is y/4 = -3/2 x^2 + 5/2 or y = -6x^2 + 10?
• Mar 11th 2008, 09:23 PM
mr fantastic
Quote:

Originally Posted by Naveed103
thanks a lot for your help on a. So if I integrate c, I get y/4 = -3/2 x^2 + c. f(1) = 4, and c = 10, so the final solution is y/4 = -3/2 x^2 + 5/2 or y = -6x^2 + 10?

First of all, you can check a solution by substituting it back into the DE. Does your answer work?

I cannot see how you could possibly figure that the integral of 1/y is y/4, which I guess preempts the answer to my question above .......
• Mar 11th 2008, 09:46 PM
Naveed103
oh I'm sorry. The 1/y I wrote on my paper looks like 1/4, ><. I've got bad handwriting and it's late. so if you integrate 1/y you get lny = -3/2 x^2 + c. If I solve for c I get 2.88, which seems like a really odd number. I'm sorry if these questions are stupid, I'm not very good at this stuff.