Question

solve the differential equation:
dy/dx = x+1/(x^2+2x-3)^2

Answer 1

You isolate the x's with the dx and the dy on it's own on their own side. Doing this you multiply by dx to both right hand side and left hand side. Then you have dy=x+1/(x^2+2x-3)^2 dx. And to solve for this you integrate both sides.

Answer 2

well its just integration with respect to x

Answer 3

\[\int\limits_{}^{}dy=\int\limits_{}^{}(x+1)/(x^2+2x-3)^2dx\]
\[y=\int\limits\limits_{}^{}(x+1)/(x^2+2x-3)^2dx\]
Simplify the integrand on the left hand side and integrate :-)

Answer 4

*integrand in the right hand side sorry LOL

Answer 5

i dont know how to integrate this one... y= (x+1)(x^2+2x-3)^-2 ? do i multiply it out first and then integrate?

Answer 6

Well actually I didn't even check to see if you can simplify it but let me try to work it out :-) Then I'll post.

Answer 7

ok great thank you

Answer 8

it is an inspection/substitution method

do you see if we differentiate x^2+2x-3 we get double this: x+1

Answer 9

Yes you have \[(x+1)/(x-1)^2(x+3)^2\] in the integrand

Answer 10

so if we were to differentiate a function f(x^2+2x-3) we would get 2f'(x^2+2x-3)(x+1)

Answer 11

alternatively use the substitution u = x^2+2x-3

Answer 12

so we have\[\frac{du}{dx} = 2(x+1) \text{ , } dx = \frac{du}{2(x+1)}\]

\[\int\limits{\frac{x+1}{(x^2+2x-3)^2}}dx = \int\limits{\frac{1}{2u^2}du}\]

Answer 13

how did you get the 2 for 2u^2?

Answer 14

where did the 1/2 come from?

Answer 15

u was chosen to be \[u = x^2+2x-3\] so du=\[du=(2x+2)dx\] Solving for \[dx=du/(2x+2)=du/2(x+1) \] so when you put in back into the integrand you have the 1/2 to account for the 2 in the du equation

Answer 16

krystabell is spot on :)

Answer 17

ok thank you guys!