3.3.18.2 Integrating factor that depends on $y$ only

Let

$$\begin{array}{rl}\text{(1)}& \mu M(x,y)+\mu N(x,y)\frac{dy}{dx}& =d\varphi (x,y)& =\frac{\partial \varphi }{\partial x}\frac{dx}{dx}+\frac{\partial \varphi }{\partial y}\frac{dy}{dx}\\ \text{(2)}& & =\frac{\partial \varphi }{\partial x}+\frac{\partial \varphi }{\partial y}\frac{dy}{dx}\end{array}$$

Comparing (1),(2) then

$$\begin{array}{rl}\frac{\partial \varphi }{\partial x}& =\mu M\\ \frac{\partial \varphi }{\partial y}& =\mu N\end{array}$$

The compatibility condition is $\frac{{\partial }^2\varphi }{\partial y\partial x}=\frac{{\partial }^2\varphi }{\partial x\partial y}$ then this implies

$$\begin{array}{rl}\frac{\partial }{\partial y}\left(\frac{\partial \varphi }{\partial x}\right)& =\frac{\partial }{\partial x}\left(\frac{\partial \varphi }{\partial y}\right)\\ \frac{\partial \mu M}{\partial y}& =\frac{\partial \mu N}{\partial x}\\ {\mu }_{y}M+\mu M_y& ={\mu }_{x}N+\mu N_x\\ {\mu }_{y}M& ={\mu }_{x}N+\mu N_x-\mu M_y\\ {\mu }_{y}M& ={\mu }_{x}N+\mu (N_x-M_y)\\ {\mu }_y& =\frac{{\mu }_{x}N}{M}+\frac{1}{M}\mu (N_x-M_y)\end{array}$$

Assuming $\mu \equiv \mu \left(y\right)$ then ${\mu }_x=0$ and the above simplifies to

$$\begin{array}{rl}{\mu }_y& =\frac{1}{M}\mu (N_x-M_y)\\ \frac{d\mu }{dy}\frac{1}{\mu }& =\frac{1}{M}(N_x-M_y)\end{array}$$

Let $\frac{1}{M}(N_x-M_y)=B$. If $B\equiv B\left(y\right)$ which depends only on $y$ then we can solve the above.

$$\begin{array}{rl}\frac{d\mu }{dy}\frac{1}{\mu }& =B\left(y\right)\\ \mu & =e^{\int Bdy}\end{array}$$

Let $\bar{M}=\mu M,\bar{N}=\mu N$ then the ode

$$\bar{M}(x,y)+\bar{N}(x,y)y^{\prime }=0$$

is now exact.

3.3.18.2.1 Example 1 Solve

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{\sqrt{(x^2-1)(y^2-1)}}{(x^2-1)}\\ dy& =\frac{\sqrt{(x^2-1)(y^2-1)}}{(x^2-1)}dx\\ -\frac{\sqrt{(x^2-1)(y^2-1)}}{(x^2-1)}dx+dy& =0\end{array}$$

Comparing to

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

Shows that $M=-\frac{\sqrt{(x^2-1)(y^2-1)}}{(x^2-1)},N=1$. We see that $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$. Hence not exact. Lets try

$$\begin{array}{rl}B& =\frac{1}{M}(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\\ & =\frac{(1-x^2)}{\sqrt{(x^2-1)(y^2-1)}}(0-\frac{-y}{\sqrt{(x^2-1)(y^2-1)}})\\ & =\frac{(1-x^2)}{\sqrt{(x^2-1)(y^2-1)}}\frac{y}{\sqrt{(x^2-1)(y^2-1)}}\\ & =\frac{(1-x^2)y}{(x^2-1)(y^2-1)}\\ & =\frac{-y}{(y^2-1)}\end{array}$$

Since $B$ does not depend on $x$ then we can use this for an integrating factor.

$$\begin{array}{rl}\mu & =e^{\int Bdy}\\ & =e^{-\int \frac{y}{(y^2-1)}dy}\\ & =\frac{1}{\sqrt{y-1}\sqrt{y+1}}\end{array}$$

Hence the ode now becomes

$$\begin{array}{rl}\mu Mdx+\mu Ndy& =0\\ \text{(A1)}& \overline{M}dx+\overline{N}dy& =0\end{array}$$

Where

$$\begin{array}{rl}\overline{M}& =\mu M\\ & =\frac{1}{\sqrt{y-1}\sqrt{y+1}}\frac{\sqrt{(x^2-1)(y^2-1)}}{(x^2-1)}\\ & =\frac{\sqrt{(x^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}(x^2-1)}\end{array}$$

And

$$\begin{array}{rl}\overline{N}& =\mu N\\ & =\frac{1}{\sqrt{y-1}\sqrt{y+1}}\end{array}$$

Now ode (A1) is exact. Now we follow the main method for solving an exact ode on the above. Let

$$\begin{array}{}\text{(1)}& \frac{\partial \varphi }{\partial x}& =\overline{M}\text{(2)}& \frac{\partial \varphi }{\partial y}& =\overline{N}\end{array}$$

Since $M$ has both $y$ and $x$, in it and $N$ has only $y$ in it, then in this case we start differently than before. We start with (2) and not (1) as it makes things simpler when integrating.

Integrating (2) w.r.t. $y$ gives

$$\begin{array}{rl}\varphi & =\int \overline{N}dy+f\left(x\right)\\ & =\int \frac{1}{\sqrt{y-1}\sqrt{y+1}}dy+f\left(x\right)\end{array}$$

But $\int \frac{1}{\sqrt{y-1}\sqrt{y+1}}dy=\frac{\sqrt{(y-1)(y+1)}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}=\frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}$, hence the above becomes

$$\begin{array}{}\text{(3)}& \varphi =\frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}+f\left(x\right)\end{array}$$

Taking derivative of (3) w.r.t. $x$ gives

$$\begin{array}{rl}\frac{\partial \varphi }{\partial x}& =\frac{d}{dx}\left(\frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}\right)+f^{\prime }\left(x\right)\\ \text{(4)}& \frac{\partial \varphi }{\partial x}& =f^{\prime }\left(x\right)\end{array}$$

But $\frac{\partial \varphi }{\partial x}=\overline{M}$. Hence the above becomes

$$\begin{array}{rl}\overline{M}& =f^{\prime }\left(x\right)\\ \frac{\sqrt{(x^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}(x^2-1)}& =f^{\prime }\left(x\right)\end{array}$$

To solve for $f\left(x\right)$ we now integrate the above w.r.t. $x$ which gives

$${\int }^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau =f\left(x\right)$$

No need to add constant of integration, as that will be absorbed anyway. Substituting the above back into (3) gives

$$\varphi =\frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}+{\int }^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau$$

$\varphi =c$, hence the solution is

$$\begin{array}{}\text{(4A)}& \frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}+{\int }^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau +c=0\end{array}$$

Lets now see what happens if after Eq (2), we started with $M$ and not $N$ as we always do. Integrating (1) w.r.t. $x$ gives

$$\begin{array}{rl}\varphi & =\int \overline{M}dx+f\left(y\right)\\ & =\int \frac{\sqrt{(x^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}(x^2-1)}dx+f\left(y\right)\\ \text{(5)}& & ={\int }^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau +f\left(y\right)\end{array}$$

Taking derivative w.r.t. $y$ the above becomes

$$\begin{array}{rl}\frac{\partial \varphi }{\partial y}& =\frac{d}{dy}{\int }^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau +f^{\prime }\left(y\right)\\ & ={\int }^x\frac{\partial }{\partial y}\left(\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}\right)d\tau +f^{\prime }\left(y\right)\\ & =0+f^{\prime }\left(y\right)\\ & =f^{\prime }\left(y\right)\end{array}$$

But $\frac{\partial \varphi }{\partial y}=\overline{N}$, hence the above becomes

$$\frac{1}{\sqrt{y-1}\sqrt{y+1}}=f^{\prime }\left(y\right)$$

Integrating w.r.t. $y$ gives

$$\begin{array}{rl}f\left(y\right)& =\int \frac{1}{\sqrt{y-1}\sqrt{y+1}}dy+c\\ f\left(y\right)& =\frac{\sqrt{(y-1)(y+1)}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}+c\end{array}$$

Substituting this into (5) gives the solution as (after combining constants)

$$c_1={\int }^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau +\frac{\sqrt{(y-1)(y+1)}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}$$

Which is same answer as (4A).  So starting with $M$ or $N$ gives same result. But if $N$ depends on $x,y$ and $M$ depends on only one of these, it can be simpler to pick $M$. Same for the other way around. If $N$ depends on only one, and $M$ depends on both $x,y$, then it will be easier to start with $N$. But in both cases, same result should be obtained.

3.3.18.2.2 Example 2 This is same example as above but with initial conditions $y\left(x_0\right)=y_0$ to show how to handle IC when unable to do the integration.

$$\begin{array}{rl}-\frac{\sqrt{(x^2-1)(y^2-1)}}{(x^2-1)}dx+dy& =0\\ y\left(x_0\right)& =y_0\end{array}$$

The solution found in above example is

$$\frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}+{\int }^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau +c=0$$

At $y\left(x_0\right)=y_0$ the above becomes

$$\frac{\sqrt{y_0^2-1}\ln (y_0+\sqrt{y_0^2-1})}{\sqrt{y_0-1}\sqrt{y_0+1}}+{\int }_{x_0}^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y_0-1}\sqrt{y_0+1}({\tau }^2-1)}d\tau +c=0$$

Substituting this value of $c$ in the solution gives

$$\frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}+{\int }_{x0}^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}d\tau =\frac{\sqrt{y_0^2-1}\ln (y_0+\sqrt{y_0^2-1})}{\sqrt{y_0-1}\sqrt{y_0+1}}+{\int }_{x_0}^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y_0-1}\sqrt{y_0+1}({\tau }^2-1)}d\tau$$

Or

$$(\frac{\sqrt{y^2-1}\ln (y+\sqrt{y^2-1})}{\sqrt{y-1}\sqrt{y+1}}-\frac{\sqrt{y_0^2-1}\ln (y_0+\sqrt{y_0^2-1})}{\sqrt{y_0-1}\sqrt{y_0+1}})+{\int }_{x0}^x\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y-1}\sqrt{y+1}({\tau }^2-1)}-\frac{\sqrt{({\tau }^2-1)(y^2-1)}}{\sqrt{y_0-1}\sqrt{y_0+1}({\tau }^2-1)}d\tau =0$$