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

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SUMMARY

The differential equation (x^2 - 1)y' + 2xy = x can be solved by finding an integrating factor u. The equation is rearranged into the standard form y' + Py = Q, with P = 2x. The integrating factor is calculated as IF = e ^ ∫P.dx. The final solution is y = (0.5 x^2 + C) / (x^2 - 1), where C is the arbitrary constant.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with integrating factors in differential equations
  • Knowledge of standard forms of differential equations
  • Basic calculus, specifically integration techniques
NEXT STEPS
  • Study the method of integrating factors in more detail
  • Learn about exact differential equations and their solutions
  • Explore the application of arbitrary constants in differential equations
  • Practice solving various first-order differential equations
USEFUL FOR

Students studying differential equations, educators teaching calculus, and anyone looking to enhance their problem-solving skills in mathematical analysis.

josek6
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Solve showing full working
(x^2 -1)y' + 2xy = x

You must show full working.
 
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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
 
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 ?
 
Thanks...I solved it my solution is
y(x^2 -1)=(x^2)/2
 
Good, but you need the arbitrary constant
y(x^2 -1)=(x^2)/2+C
 
I must commend you guys for all the assistance and I will surely recommend you guys to my friends. Thanks again.
 
(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 )
 

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