# general solution of a differential equation

• Dec 2nd 2009, 12:59 PM
georgiahelo
general solution of a differential equation
Find the general solution of the differential equation

dy/dx = y^4 ( (x+1) / (x-2)(x^2 + 5)) and any other solutions if they exist.
• Dec 2nd 2009, 07:20 PM
mr fantastic
Quote:

Originally Posted by georgiahelo
Find the general solution of the differential equation

dy/dx = y^4 ( (x+1) / (x-2)(x^2 + 5)) and any other solutions if they exist.

The DE is seperable. (To integrate the x stuff I suggest using partial fractions).

If you need more help please show all your working and say where you get stuck.
• Dec 5th 2009, 06:20 AM
georgiahelo
This is where I get stuck, I need a step by step on how to solve the integral of

(x+1)/(x-2)(x^2+5)

It confuses me as to what I do with the top, and the x^2 on the bottom when dealing with partial fractions.can you help?
• Dec 5th 2009, 07:30 AM
TheEmptySet
Quote:

Originally Posted by georgiahelo
This is where I get stuck, I need a step by step on how to solve the integral of

(x+1)/(x-2)(x^2+5)

It confuses me as to what I do with the top, and the x^2 on the bottom when dealing with partial fractions.can you help?

Since you have one linear factor and one irreducible quadratic factor with neither factor repeated, the partial fraction decomposition will have the form.

$\frac{A}{x-2}+\frac{Bx+C}{x^2+5}=\frac{x+1}{(x-2)(x^+5)}$

Now multiply both sides by the LCD to get

$A(x^2+5)+(Bx+C)(x-2)=x+1$

expand out the left hand side and collect the powers of x to get

$(A+B)x^2+(-2B+C)x+(5A-2C)=x+1$

Now we want these two polynomials to be equation so we set the coeffients on each power of x equal to get the following system of equations

$A+B=0;-2B+C=1;5A-2C=1$

Solving this system gives

$A=\frac{1}{3},B=-\frac{1}{3},C=\frac{1}{3}$