Differential Equations Problem

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SUMMARY

The discussion focuses on solving the differential equation x y' + 3y = x² using the correct integrating factor. The initial attempt incorrectly identified the integrating factor as e^log(x) or x, leading to an incorrect solution. The correct integrating factor is x³, which simplifies the equation to d/dx(x³y) = x⁴. This adjustment allows for the proper integration and solution of the differential equation.

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tomeatworld
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Homework Statement


x y' + 3y = x2

Homework Equations


The Attempt at a Solution


So I tried using an integrating factor (as I couldn't separate variables).
So I've said that the function p(x)= 1/x and q(x)=x. So the integrating factor is elog(x) or x. Putting this in:
x y = [tex]\int x^{2}[/tex] so
x y = [tex]\frac{1}{3} x^{3}[/tex] + c
and finally: y(x) = [tex]\frac{1}{3} x^{2} + \frac{c}{x}[/tex] but this isn't the right answer. Where have I gone wrong?
 
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You don't have the right integrating factor. Starting from y' + (3/x)y = x, your integrating factor should be x3. Multiplying by the integrating factor gives x3y' + 3x2y = x4.

This equation can be written as d/dx(x3y) = x4. Can you take it from there?
 
ahh I see! yeah, that's great! thanks a load!
 

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