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

2. 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 $(x-4)^4$ are of the form:

$
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

3. 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.