• May 17th 2006, 10:45 PM
bobby77
Solve the initial value problem dy/dx = (x – 4)^4 where y(1) = 1.
• May 18th 2006, 01:54 AM
CaptainBlack
Quote:

Originally Posted by bobby77
Solve the initial value problem dy/dx = (x – 4)^4 where y(1) = 1.

The fundamental theorem of calculus tells us that all the functions
whose derivative is $\displaystyle (x-4)^4$ are of the form:

$\displaystyle y(x)=\int_0^x (u-4)^4 du + C$

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, 03:10 AM
ticbol
Quote:

Originally Posted by bobby77
Solve the initial value problem dy/dx = (x – 4)^4 where y(1) = 1.

"...where y(1) = 1"
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.