How to Solve the Differential Equation (x^2 - 1)y' + 2xy = x?

[josek6]

Solve showing full working
(x^2 -1)y' + 2xy = x

You must show full working.

 

[lurflurf]

No full workings here, were fresh out.
Try to rearrange and find an integrating factor u such that
u(x^2 -1)y' + u(2xy - x)

is and exact differential

 

[josek6]

Find the integerating factor "u"
So I have to rewrite the equation in the standard form y' + Py =Q
IF = e ^ ∫P.dx
where P = 2x ?

 

[josek6]

Thanks...I solved it my solution is
y(x^2 -1)=(x^2)/2

 

[lurflurf]

Good, but you need the arbitrary constant
y(x^2 -1)=(x^2)/2+C

 

[josek6]

I must commend you guys for all the assistance and I will surely recommend you guys to my friends. Thanks again.

 

[CGZ]

(x^2 - 1 ) y' + 2xy = ((x^2-1)y)' = x;-->
(x^2-1)y = \int{x};-->
(x^2-1)y = 0.5 x^2 + C

y = (0.5 x^2 + C) / ( x^2 - 1 )