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Lauren's question at Yahoo! Answers regarding inexact ODE
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Re: Lauren's question from Yahoo! Answers regarding inexact ODE
Hello Lauren,
We are given to solve:
\(\displaystyle 2y\left(y^2-x \right)\,dy=dx\)
I would first write the ODE in the form:
\(\displaystyle M(x,y)\,dx+N(x,y)\,dy=0\)
and so we have:
\(\displaystyle (1)\,dx+\left(2xy-2y^3 \right)\,dy=0\)
We see the equation is inexact, since:
\(\displaystyle \frac{\delta M}{\delta y}=0\ne\frac{\delta N}{\delta x}=2y\)
Thus, we observe that:
\(\displaystyle \frac{\frac{\delta N}{\delta x}-\frac{\delta M}{\delta y}}{M}=2y\)
Since this is a function of $y$ alone, we take as our integrating factor:
\(\displaystyle \mu(y)=e^{\int 2y\,dy}=e^{y^2}\)
And so we write the new ODE as:
\(\displaystyle e^{y^2}\,dx+\left(2xy-2y^3 \right)e^{y^2}\,dy=0\)
We see now that:
\(\displaystyle \frac{\delta M}{\delta y}=2ye^{y^2}=\frac{\delta N}{\delta x}=2ye^{y^2}\)
Hence, our new equation is exact. Since it is exact, we must have:
\(\displaystyle \frac{\delta F}{\delta x}=e^{y^2}\)
Integrating with respect to $x$, we obtain:
\(\displaystyle F(x,y)=xe^{y^2}+g(y)\)
Now, to determine $g(y)$, we may take the partial derivative with respect to $y$, substituting using \(\displaystyle \frac{\delta F}{\delta y}=\left(2xy-2y^3 \right)e^{y^2}\):
\(\displaystyle \left(2xy-2y^3 \right)e^{y^2}=2xye^{y^2}+g'(y)\)
\(\displaystyle g'(y)=-2y^3e^{y^2}\)
Next, we want to integrate this to determine $g(y)$, and we may use integration by parts:
\(\displaystyle u=y^2\,\therefore\,du=2y\,dy\)
\(\displaystyle dv=2ye^{y^2}\,dy\,\therefore\,v=e^{y^2}\)
and we may write:
\(\displaystyle g(y)=-\left(y^2e^y-\int 2ye^{y^2}\,dy \right)=e^{y^2}-y^2e^{y^2}=e^{y^2}\left(1-y^2 \right)\)
Thus, the solution to the original ODE is given implicitly by:
\(\displaystyle xe^{y^2}+e^{y^2}\left(1-y^2 \right)=C\)
\(\displaystyle \left(x+1-y^2 \right)e^{y^2}=C\)
Hello Lauren,
We are given to solve:
\(\displaystyle 2y\left(y^2-x \right)\,dy=dx\)
I would first write the ODE in the form:
\(\displaystyle M(x,y)\,dx+N(x,y)\,dy=0\)
and so we have:
\(\displaystyle (1)\,dx+\left(2xy-2y^3 \right)\,dy=0\)
We see the equation is inexact, since:
\(\displaystyle \frac{\delta M}{\delta y}=0\ne\frac{\delta N}{\delta x}=2y\)
Thus, we observe that:
\(\displaystyle \frac{\frac{\delta N}{\delta x}-\frac{\delta M}{\delta y}}{M}=2y\)
Since this is a function of $y$ alone, we take as our integrating factor:
\(\displaystyle \mu(y)=e^{\int 2y\,dy}=e^{y^2}\)
And so we write the new ODE as:
\(\displaystyle e^{y^2}\,dx+\left(2xy-2y^3 \right)e^{y^2}\,dy=0\)
We see now that:
\(\displaystyle \frac{\delta M}{\delta y}=2ye^{y^2}=\frac{\delta N}{\delta x}=2ye^{y^2}\)
Hence, our new equation is exact. Since it is exact, we must have:
\(\displaystyle \frac{\delta F}{\delta x}=e^{y^2}\)
Integrating with respect to $x$, we obtain:
\(\displaystyle F(x,y)=xe^{y^2}+g(y)\)
Now, to determine $g(y)$, we may take the partial derivative with respect to $y$, substituting using \(\displaystyle \frac{\delta F}{\delta y}=\left(2xy-2y^3 \right)e^{y^2}\):
\(\displaystyle \left(2xy-2y^3 \right)e^{y^2}=2xye^{y^2}+g'(y)\)
\(\displaystyle g'(y)=-2y^3e^{y^2}\)
Next, we want to integrate this to determine $g(y)$, and we may use integration by parts:
\(\displaystyle u=y^2\,\therefore\,du=2y\,dy\)
\(\displaystyle dv=2ye^{y^2}\,dy\,\therefore\,v=e^{y^2}\)
and we may write:
\(\displaystyle g(y)=-\left(y^2e^y-\int 2ye^{y^2}\,dy \right)=e^{y^2}-y^2e^{y^2}=e^{y^2}\left(1-y^2 \right)\)
Thus, the solution to the original ODE is given implicitly by:
\(\displaystyle xe^{y^2}+e^{y^2}\left(1-y^2 \right)=C\)
\(\displaystyle \left(x+1-y^2 \right)e^{y^2}=C\)