Solve the initial value problem dy/dx = (x – 4)^4 where y(1) = 1.

Results 1 to 3 of 3

- May 17th 2006, 11:45 PM #1

- Joined
- Sep 2005
- Posts
- 136

- May 18th 2006, 02:54 AM #2

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 4

Originally Posted by**bobby77**

whose derivative is are of the form:

So to solve this you need to do the integral, then plug in the initial condition

to determin the value of constant that you require.

RonL

- May 18th 2006, 04:10 AM #3

- Joined
- Apr 2005
- Posts
- 1,631

Originally Posted by**bobby77**

That means when x=1, then y=1.

dy/dx = (x-4)^4

y = (1/5)(x-4)^5 +C

So, when x=1, and at the same time, y=1,

1 = (1/5)(1 -4)^5 +C

1 = (1/5)(-3)^5 +C

1 = (1/5)(-243) +C

Clear the fraction, multiply both sides by 5,

5 = -243 +5C

5 +243 = 5C

C = 248/5

Therefore,

y = f(x) = (1/5)(x-4)^5 +248/5

y = f(x) = (1/5)[(x-4)^5 +248] ------------answer.