unfinished differential equation problem

[kcoe05]

dy/dx= (x+5)/ (x^2 - 2x - 3)

so far i have gotten this far:

(x^2 - 2x - 3)dy = (x+5)dx

????????? = 1/2x^2 + 5x + C

would the left side simply be x^2 y - 2xy - 3y? not exactly sure if that is correct.

Thanks for the help

 

[arthur ohlsten]

dy/dx = [x+5] / [x^2 - 2x -3]
factor
dy/dx = [x+5] / { [x-3][x+1] }

we want the equation in the form of A/[x+1] + B /[x-3]

[x+5] / { [x+1][x-3] } = A /[x+1] + B /[x-3]
multiply both sides by x+1
[x+5][x-3] = A + B [x+1] /[x-3]
let x=-1
4/[-4]=A
A=-1

by same approach
B= 2

dy= -dx/[x+1] +2 dx/[x-3]
y= -ln [x+1] + 2 ln[x-3] + C answer
y = ln [[x-3]/[x+1] ] +C answer

Arthur

 

[BigGlenntheHeavy]

\(\displaystyle \frac{dy}{dx} \ = \ \frac{x+5}{x^{2}-2x-3} \ \implies \ \int dy \ = \ \int\frac{x+5}{x^{2}-2x-3}dx\)

\(\displaystyle \implies \ y \ = \ \int \bigg[\frac{2}{x-3}-\frac{1}{x+1}\bigg]dx, \ Hence \ y \ = \ 2ln|x-3|-ln|x+1|+C\)

\(\displaystyle y \ = \ ln|x-3|^{2}-ln|x+1|+C \ = \ ln\bigg|\frac{(x-3)^{2}}{(x+1)}\bigg|+C\)

\(\displaystyle Check: \ \frac{dy}{dx} \ = \ \frac{x+5}{x^{2}-2x-3}\)

 

[kcoe05]

That makes a lot more sense. Thanks a lot guys.