Differential Equation, Separable, I believe

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SUMMARY

The discussion focuses on solving the differential equation (3y²)x(dy/dx) - x + 1 = 0 with the initial condition y(e) = 1. The solution process involves separating variables to obtain (3y²)dy = ((x-1)/x)dx, followed by integration leading to y³ = x - ln(x) + C. The unique solution is determined by substituting the initial condition, resulting in C = 2 - e.

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  • Understanding of differential equations, specifically separable equations.
  • Familiarity with integration techniques.
  • Knowledge of initial value problems.
  • Basic algebraic manipulation skills.
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  • Study the method of separation of variables in differential equations.
  • Learn about integrating factors for solving first-order differential equations.
  • Explore initial value problems and their significance in differential equations.
  • Review the properties of logarithmic functions in the context of integration.
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Students studying calculus, particularly those focusing on differential equations, as well as educators looking for examples of solving separable equations.

Marylander
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Haven't done one of these in awhile and I was looking for a place to make sure I was doing it right. Hopefully one of you can take the time to look it over?

Homework Statement



Find the unique solution of the differential equation (3y^2)x(dy/dx)-x+1=0 for which y(e)=1

Homework Equations



None.

The Attempt at a Solution



(3y^2)dy=((x-1)/x)dx

Integrate

y^3=x-lnx+C

Substitute in for unique solution

1^3=e-ln(e)+C
2-e=C
 
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Welcome to PF!

Hi Marylander ! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
Marylander said:
Find the unique solution of the differential equation (3y^2)x(dy/dx)-x+1=0 for which y(e)=1

(3y^2)dy=((x-1)/x)dx

Integrate

y^3=x-lnx+C

Substitute in for unique solution

1^3=e-ln(e)+C
2-e=C

Looks good! :biggrin:
 
Didn't think to. Too used to not having it.

Good, thanks for checking.
 

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