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

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- May 17th 2006, 10:45 PM #1

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- May 18th 2006, 01:54 AM #2

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Originally Posted by**bobby77**

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 #3

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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.