1. ## Differntial Equations Problem

A problem that my friend gave me. I never took diff-e-q but my guess would be to integrate with respect to x then substitue x(0) to find the constant, but not sure if that's right, doesn't seem like diff-e-q material that way.

Separate variables and use partial fractions to solve the initial value problem $\frac{dt}{dx}=17x(x-15);x(0)=17$

2. Originally Posted by eXist
A problem that my friend gave me. I never took diff-e-q but my guess would to integrate with respect to x then substitue x(0) to find the constant, but not sure if that's right.

Separate variables and use partial fractions to solve the initial value problem $\frac{dt}{dx}=17x(x-15);x(0)=17$
I think you mean ...

$\frac{dx}{dt}=17x(x-15);x(0)=17$

separate variables ...

$\frac{dx}{x(x-15)} = 17 \, dt$

partial fraction decomposition ...

$\frac{1}{x(x-15)} =\frac{A}{x} + \frac{B}{x-15}$

$1 = A(x-15) + Bx$

let $x = 15$ ... $B = \frac{1}{15}$

let $x = 0$ ... $A = -\frac{1}{15}$

$\frac{1}{x(x-15)} = \frac{1}{15}\left(\frac{1}{x-15} - \frac{1}{x}\right)$

$\frac{1}{15}\int \left(\frac{1}{x-15} - \frac{1}{x}\right) \, dx = \int 17 \, dt$

finish from here?

3. Ya I think I have the rest . Thanks a lot, you're the man skeets.

4. Originally Posted by skeeter
partial fraction decomposition ...

$\frac{1}{x(x-15)} =\frac{A}{x} + \frac{B}{x-15}$

$1 = A(x-15) + Bx$

let $x = 15$ ... $B = \frac{1}{15}$

let $x = 0$ ... $A = -\frac{1}{15}$

$\frac{1}{x(x-15)} = \frac{1}{15}\left(\frac{1}{x-15} - \frac{1}{x}\right)$

$\frac{1}{15}\int \left(\frac{1}{x-15} - \frac{1}{x}\right) \, dx = \int 17 \, dt$
Can you explain the partial fraction decomposition steps you've taken here?