Separation of variables to solve DE

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SUMMARY

The discussion focuses on solving the differential equation y' + (2/x)y = 3/x^2 using the method of separation of variables and integrating factors. The correct approach involves recognizing the need for an integrating factor, specifically e^{\int{\frac{2}{x}dx} = x^2, which transforms the equation into x^2y' + 2xy = 3. The general solution is confirmed as y = 3/x + C/x^2. Participants emphasize the importance of treating both sides of the equation consistently during integration.

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  • Understanding of first-order linear differential equations
  • Familiarity with integrating factors
  • Knowledge of separation of variables technique
  • Basic calculus, specifically integration techniques
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beetle2
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Homework Statement



y' + (2/x)y = 3/x^2

Homework Equations



separation of variables

The Attempt at a Solution



First I turned it into

dy/dx + (2/x)y = 3/x^2 dx


then multiplied both sides by dx

dy + (2/x)y = 3/x^2 dx

I then tried to divide both sides by 2/x and got

dy + y = 3/2x

Do I just integrate both sides now?
 
Last edited:
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Separation of variables won't help (you also made an error when multiplying through the equation with dx). It looks like you'll have to solve this using an integrating factor.
 
You also made an error when dividing by (2/x). Make sure that what you do to one term goes for all terms.
 
The integrating factor is e^{\int{\frac{2}{x}dx} = x^2

We than use this and multiply both sides by x^2

which gives


x^2y'+2xy = 3

or

yx^2= 3x+C

divide both sides by x^2

y = \frac{3}{x}+\frac{C}{x^2}
 
Is this the general solution?
 
yes, it's the general solution.
 
Also can be solved as a non-homogeneous Euler equation.
 
boneill3 said:
which gives


x^2y'+2xy = 3

or

yx^2= 3x+C


I see you integrated on the right side wrt x to get 3x+C

but how did x^2y'+2xy become yx^2? did we integrate the LHS wrt x or y?
 
Given a function f(x,y), its total differential with respect to x is \frac{df}{dx} = \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial y}\frac{dy}{dx}.
 
  • #10
beetle2 said:
I see you integrated on the right side wrt x to get 3x+C

but how did x^2y'+2xy become yx^2? did we integrate the LHS wrt x or y?
You have to treat both sides of the equation the same way. If you integrate the RHS wrt x, you have to integrate the LHS wrt x as well.
 

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