MHB Lauren's question at Yahoo Answers regarding inexact ODE

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The discussion addresses Lauren's question about solving a linear first-order ordinary differential equation (ODE) given by 2y(y^2-x)dy=dx. The equation is reformulated into the standard form, revealing that it is inexact due to the differing partial derivatives of M and N. An integrating factor, μ(y)=e^(y^2), is determined to make the equation exact. After applying this integrating factor, the solution is derived, leading to the implicit solution (x+1-y^2)e^(y^2)=C. The discussion effectively demonstrates the steps required to solve the inexact ODE using integrating factors.
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Here is the question:

Help me with my Linear Differential Equation of Order One: The Integrating Factor?

Help me with my Linear Differential Equation of Order One: The Integrating Factor

2y(y^2-x)dy=dx

I have posted a link there to this topic so the OP can see my work.
 
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Re: Lauren's question from Yahoo! Answers regarding inexact ODE

Hello Lauren,

We are given to solve:

$$2y\left(y^2-x \right)\,dy=dx$$

I would first write the ODE in the form:

$$M(x,y)\,dx+N(x,y)\,dy=0$$

and so we have:

$$(1)\,dx+\left(2xy-2y^3 \right)\,dy=0$$

We see the equation is inexact, since:

$$\frac{\delta M}{\delta y}=0\ne\frac{\delta N}{\delta x}=2y$$

Thus, we observe that:

$$\frac{\frac{\delta N}{\delta x}-\frac{\delta M}{\delta y}}{M}=2y$$

Since this is a function of $y$ alone, we take as our integrating factor:

$$\mu(y)=e^{\int 2y\,dy}=e^{y^2}$$

And so we write the new ODE as:

$$e^{y^2}\,dx+\left(2xy-2y^3 \right)e^{y^2}\,dy=0$$

We see now that:

$$\frac{\delta M}{\delta y}=2ye^{y^2}=\frac{\delta N}{\delta x}=2ye^{y^2}$$

Hence, our new equation is exact. Since it is exact, we must have:

$$\frac{\delta F}{\delta x}=e^{y^2}$$

Integrating with respect to $x$, we obtain:

$$F(x,y)=xe^{y^2}+g(y)$$

Now, to determine $g(y)$, we may take the partial derivative with respect to $y$, substituting using $$\frac{\delta F}{\delta y}=\left(2xy-2y^3 \right)e^{y^2}$$:

$$\left(2xy-2y^3 \right)e^{y^2}=2xye^{y^2}+g'(y)$$

$$g'(y)=-2y^3e^{y^2}$$

Next, we want to integrate this to determine $g(y)$, and we may use integration by parts:

$$u=y^2\,\therefore\,du=2y\,dy$$

$$dv=2ye^{y^2}\,dy\,\therefore\,v=e^{y^2}$$

and we may write:

$$g(y)=-\left(y^2e^y-\int 2ye^{y^2}\,dy \right)=e^{y^2}-y^2e^{y^2}=e^{y^2}\left(1-y^2 \right)$$

Thus, the solution to the original ODE is given implicitly by:

$$xe^{y^2}+e^{y^2}\left(1-y^2 \right)=C$$

$$\left(x+1-y^2 \right)e^{y^2}=C$$
 
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