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I have seen one method which correctly evaluates $P=\frac{2x^2-1}{x(1-x^2)}$.

But I have seen a method that says if $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}$ is a function of just x ,i.e, it is equal to f(x) then its

Integrating Factor (IF)=$e^{\int f(x)dx}$

Here M=$2x^2y-y-ax^3$ and N=$x(1-x^2)$

So, $\frac{\partial M}{\partial y}=2x^2-1$ and $\frac{\partial N}{\partial x}=1-3x^2$

Hence, $$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\;\;\frac{5x^2-2}{x(1-x^2)}$$

Here's the link to the first method: https://haygot.s3.amazonaws.com/questions/1374381_1134731_ans_55dd44fc75824f228cd5c8286c3144f1.jpg

Please tell why the answer is not matching.

Nivedita Chowdhury
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1 Answers1

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The integrating factor need not to be unique for the non-exact equation $M(x,y)dx+N(x,y)dy=0$. The integrating factor $\mu(x)$ obtained by the method you used may lead to a solution which misses some solution of given ODE or conatin an extra solution (kind of weed) which is not the solution of given ODE. Though in this case, the final solution will be same irrespective of the different $\mu(x)$ obtained in the two methods as it is free from variable $y$.

Nitin Uniyal
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  • So how we know when to use which method? Because in certain cases the method I used correctly gives the IF. – Nivedita Chowdhury Aug 09 '22 at 05:36
  • You can certainly use this method but with the caution of loosing or gaining (extra) solution; for which you need to verify at end. – Nitin Uniyal Aug 09 '22 at 05:40
  • Pls tell what you mean by verifying at the end and how do i do it. I am a beginner idk much – Nivedita Chowdhury Aug 09 '22 at 05:46
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    Consider $ydx-xdy=0$ with $\mu(y)=1/y^2; y\ne 0$ as integrating factor. On multiplication of it with ODE and solving you get $x/y=c$ as solution which doesn't include $y(x)=0$ (the lost solution). You can verify at the end that $y(x)=0$ is indeed the solution of original ODE. – Nitin Uniyal Aug 09 '22 at 06:27