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## Exact ODEs

An ODE of the form \begin{align} M(x,y)+N(x,y)\frac{dy}{dx}&=0 \;\;\;\; (12)\\ \text{ or, } M(x,y)\, dx+N(x,y)\, dy&=0. \nonumber \end{align} is called exact if it can be written as $\frac{d}{dx}\left(f(x,y)\right) =0$ for some function $$f(x,y)$$ in which $$y$$ is a function of $$x$$. Then the solution is $f(x,y)=c.$ Note that comparing $\frac{d}{dx}\left(f(x,y)\right) =f_x(x,y)+f_y(x,y)\frac{dy}{dx}=0$ with (12) we get $\begin{equation*} f_x(x,y)= M(x,y),\, f_y(x,y)=N(x,y). \end{equation*}$ Thus we have $M_y(x,y)=f_{yx}(x,y)=f_{xy}(x,y)=N_x(x,y).$
Theorem. Let $$M,N,M_y$$ and $$N_x$$ be continuous functions on $$R=(a,b)\times (c,d)$$. Then the ODE $M(x,y)\, dx+N(x,y)\, dy=0$ is exact in $$R$$ if and only if $M_y(x,y)=N_x(x,y)$ at each point in $$R$$.

Example. $2xy\, dx+(x^2+3y^2)\, dy=0$ is exact since $M_y=\frac{\partial}{\partial y}\left(2xy \right)=2x=\frac{\partial}{\partial x}\left(x^2+3y^2 \right)=N_x.$
Example. $2xy^2\, dx+(x^2+3y^2)\, dy=0$ is non-exact since $M_y=\frac{\partial}{\partial y}\left(2xy^2 \right)=4xy\neq 2x=\frac{\partial}{\partial x}\left(x^2+3y^2 \right)=N_x.$

Steps to solve:

1. Find a function $$f(x,y)$$ such that \begin{align*} f_x(x,y)&= M(x,y)\\ f_y(x,y)&=N(x,y). \end{align*}

2. The solution is $$f(x,y)=c.$$

Example. Solve the ODE $2xy+(x^2+3y^2)\, \frac{dy}{dx}=0.$ Solution. Note that $M_y=\frac{\partial}{\partial y}\left(2xy \right) =2x=\frac{\partial}{\partial x}\left(x^2+3y^2 \right)=N_x.$ So the given ODE is exact and there is a function $$f(x,y)$$ such that \begin{align*} f_x(x,y)&= 2xy & (13)\\ f_y(x,y)&=x^2+3y^2.& (14) \end{align*} Integrating (13) partially with respect to $$x$$, we get $\begin{eqnarray} \int f_x(x,y)\, dx &=& \displaystyle\int 2xy \, dx \nonumber \\ \implies f(x,y) &=& x^2y+h(y). \;\;\;\; (15) \end{eqnarray}$ Differentiating (15) partially with respect to $$y$$, we get $$$f_y(x,y)=x^2+h'(y).\;\;\;\; (16)$$$ Comparing (14) and (16), we get $$x^2+h'(y)=x^2+3y^2 \implies h'(y)=3y^2$$. Thus $h(y)=\int 3y^2\, dy=y^3.$ From (15) we get $f(x,y)=x^2y+h(y)=x^2y+y^3.$ Thus the solution is $x^2y+y^3=c.$

Non-exact to Exact:
Sometimes a non-exact ODE can made exact when multiplied by an integrating factor $$\mu(x,y)$$. For example, $(2x^2+y^2)\, dx+xy\, dy=0$ is non-exact, but \begin{align*} 2x\, [ (2x^2+y^2)\, dx+xy\, dy ]&=0\\ \text{i.e., } (4x^3+2xy^2)\, dx+2x^2y\, dy&=0 \end{align*} is exact where $$\mu(x,y)=2x$$.

Finding such integrating factor $$\mu(x,y)$$ is complicated. So we will concentrate on the IF that is a function of $$x$$, i.e., $$\mu(x,y)=\mu(x)$$ and $$\mu_y=0$$. Suppose the following ODE is exact. $\mu(x) M(x,y)+\mu(x) N(x,y)\frac{dy}{dx}=0$ \begin{align*} (\mu M)_y&=(\mu N)_x\\ \mu_yM+\mu M_y&=\mu_xN+\mu N_x\\ \mu M_y&=\mu_xN+\mu N_x \text{ since } \mu_y=0\\ \mu_xN&=\mu M_y-\mu N_x\\ \frac{d\mu}{dx}&=\mu\frac{M_y-N_x}{N}\\ \int \frac{d\mu}{\mu}&=\int \frac{M_y-N_x}{N}\, dx\\ \ln \mu&=\int \frac{M_y-N_x}{N}\, dx\\ \mu&=e^{\int \frac{M_y-N_x}{N}\, dx}\\ \end{align*}
Example. Find an integrating factor that makes the following non-exact ODE exact when multiplied by it. $(2x^2+y^2)-xy\, \frac{dy}{dx}=0$
Solution. Here $$M=2x^2+y^2$$ and $$N=-xy$$. So $\frac{M_y-N_x}{N}= \frac{(2y)-(-y)}{-xy}= -\frac{3}{x}.$ \begin{align*} \frac{d\mu}{dx}&=\mu\frac{M_y-N_x}{N}=-\mu\frac{3}{x}\\ \int \frac{d\mu}{\mu}&=-3\int \frac{dx}{x}\\ \ln \mu&=-3\ln x=\ln(x)^{-3}=\ln\left(\frac{1}{x^3} \right)\\ \mu&=\frac{1}{x^3}\\ \end{align*} So the IF is $$\mu=\frac{1}{x^3}$$. We can verify that the following ODE is exact. \begin{align*} \frac{1}{x^3}[(2x^2+y^2)-xy\, \frac{dy}{dx}]&=0\\ \left(\frac{2}{x}+\frac{y^2}{x^3}\right)-\frac{y}{x^2}\, \frac{dy}{dx}&=0 \end{align*} Checking: $M_y=\frac{\partial}{\partial y}\left(\frac{2}{x}+\frac{y^2}{x^3}\right)=\frac{2y}{x^3} =\frac{\partial}{\partial x}\left(-\frac{y}{x^2}\right)=N_x.$ The solution is $$2\ln x-\displaystyle\frac{y^2}{2x^2}=c$$. (verify)

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