3.3.13.2 Examples

3.3.13.2.1 Example 1

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

Here $f(x,y)=\frac{-(y^2+\frac{2}{x})}{2yx}$. We start by checking if it is isobaric or not. To find $m$ such that $f(tx,t^{m}y)=t^{m-1}f(x,y)$ we do (as given in the introduction)

$$\begin{array}{rl}\text{(2)}& m& =\frac{f+x{f}_x}{f-y{f}_y}& =\frac{\frac{-(y^2+\frac{2}{x})}{2yx}+x\left(\frac{x{y}^2+4}{2x^{3}y}\right)}{\frac{-(y^2+\frac{2}{x})}{2yx}-y(-\frac{x{y}^2-2}{2x^2{y}^2})}\\ & =\frac{\frac{1}{x^{2}y}}{-\frac{2}{x^{2}y}}\\ & =-\frac{1}{2}\end{array}$$

Hence this is isobaric of index $m=-\frac{1}{2}$ because it has a numerical solution as a result.

To verify this result, here $M(x,y)=(-y^2-\frac{2}{x}),N(x,y)=2yx$. Let us start by checking for isobaric (since homogeneous is special case).

$$\begin{array}{rl}M(tx,t^{m}y)& =(-t^{2m}{y}^2+\frac{2}{tx})\\ & =\frac{1}{t}(-t^{2m+1}{y}^2+\frac{2}{x})\\ & =t^{-1}(-t^{2m+1}{y}^2+\frac{2}{x})\end{array}$$

The above is same as $(-y^2-\frac{2}{x})$ when $2m+1=0$ or $m=-\frac{1}{2}$. From the above we also see that $r=-1$. This is by comparing the last result above to $t^{r}M(x,y)$. Now that we found candidate $m$ and $r$, then all what we have to do is check $N(tx,t^{m}y)=t^{r-m-1}N(x,y)$ or not. If it is, then we are done and the ode is isobaric of degree $m$

$$\begin{array}{rl}N(tx,t^{m}y)& =2t^{m}ytx\\ & =2t^{\frac{-1}{2}}ytx\\ & =t^{\frac{1}{2}}\left(2yx\right)\\ & =t^{\frac{1}{2}}N(x,y)\end{array}$$

Now we check if $\frac{1}{2}=r-m+1$. Which it is. Since $r-m+1=-1-(-\frac{1}{2})+1=\frac{1}{2}$. Hence this ode is isobaric. From now on Eq (2) will be used to find $m$.

Hence the substitution $y=v{x}^m$ will make the ode separable. This is the whole point of isobaric ode’s. The hardest part is to find $m$. Substituting $y=v{x}^{\frac{=1}{2}}$ in (1) results in

$$v\frac{dv}{dx}=-\frac{1}{x}$$

This is solved for $v$ easily since separable, and then $y$ is found from $y=v{x}^{\frac{=1}{2}}$.

3.3.13.2.2 Example 2

$$\begin{array}{}\text{(1)}& \frac{dy}{dx}=x\sqrt{x^4+4y}-x^3\end{array}$$

We start by checking if it is isobaric or not. Using

$$\begin{array}{rl}m& =\frac{f+x{f}_x}{f-y{f}_y}\\ & =\frac{(x\sqrt{x^4+4y}-x^3)+x(\sqrt{x^4+4y}+\frac{2x^4}{\sqrt{x^4+4y}}-3x^2)}{(x\sqrt{x^4+4y}-x^3)-x^3-\frac{2xy}{\sqrt{x^4+4y}}}\\ & =\frac{4\frac{x}{\sqrt{x^4+4y}}(2y-x^2\sqrt{x^4+4y}+x^4)}{\frac{x}{\sqrt{x^4+4y}}(2y-2x^2\sqrt{x^4+4y}+x^4)}\\ & =\frac{4\frac{x}{\sqrt{x^4+4y}}}{\frac{x}{\sqrt{x^4+4y}}}\\ & =4\end{array}$$

Therefore this is isobaric of order $4$.  Substituting $y=v{x}^m=v{x}^4$ in (1) results in

$$v^{\prime }=\frac{-4v+\sqrt{1+4v}-1}{x}$$

Which is separable. This is solved easily for $v\left(x\right)$ and then $y$ is found from $y=v{x}^4$.

3.3.13.2.3 Example 3

$$\begin{array}{rl}x(x-y^3)\frac{dy}{dx}& =(3x+y^3)y\\ \text{(1)}& \frac{dy}{dx}& =\frac{(3x+y^3)y}{x(x-y^3)}\end{array}$$

We start by checking if it is isobaric or not. Using

$$\begin{array}{rl}m& =\frac{f+x{f}_x}{f-y{f}_y}\\ & =\frac{\frac{(3x+y^3)y}{x(x-y^3)}+x(\frac{3y}{x(-y^3+x)}-\frac{(y^3+3x)y}{x^2(-y^3+x)}-\frac{(y^3+3x)y}{x{(-y^3+x)}^2})}{\frac{(3x+y^3)y}{x(x-y^3)}-y(\frac{3y^3}{x(-y^3+x)}+\frac{y^3+3x}{x(-y^3+x)}+\frac{3(y^3+3x)y^3}{x{(-y^3+x)}^2})}\\ & =\frac{-4\frac{y^4}{{(x-y^3)}^2}}{-12\frac{y^4}{{(x-y^3)}^2}}\\ & =\frac{1}{3}\end{array}$$

$m=\frac{1}{3}$ makes each term the same weight $\frac{4}{3}$. Hence the substitution $y=v{x}^{\frac{1}{3}}$ will make the ode separable. Substituting this in (1) results in

$$\frac{dv}{dx}=\frac{-4}{3x}\frac{v(v^3+2)}{(v^3-1)}$$

Which is separable. This is solved for $v$, and then $y$ is found from $y=v{x}^{\frac{1}{3}}$.

3.3.13.2.4 Example 4

$$\begin{array}{}\text{(1)}& y^{\prime }=\frac{y}{x}\ln (xy-1)\end{array}$$

We start by checking if it is isobaric or not. Using

$$\begin{array}{rl}m& =\frac{f+x{f}_x}{f-y{f}_y}\\ & =\frac{\frac{y}{x}\ln (xy-1)+x(\frac{-y\ln (xy-1)}{x^2}+\frac{y^2}{x(xy-1)})}{\frac{y}{x}\ln (xy-1)-y(\frac{\ln (xy-1)}{x}+\frac{y}{xy-1})}\\ & =\frac{\frac{y^2}{xy-1}}{-\frac{y^2}{xy-1}}\\ & =-1\end{array}$$

Hence the substitution $y=\frac{v}{x}$ will make the ode separable. Substituting this in (1) results in

$$v^{\prime }=\frac{v\ln \left(v\right)}{x}$$

Which is separable. This is solved for $v$, and then $y$ is found from $y=\frac{v}{x}$.

3.3.13.2.5 Example 5

$$\begin{array}{}\text{(1)}& {\left(y^{\prime }\right)}^2=y{(y-2y^{\prime }x)}^3\end{array}$$

One way to handle this is to first solve for $y^{\prime }$ and then apply the above method. This will result in $m=-1$.

3.3.13.2.6 Example 6

$$\begin{array}{rl}(x-y)y^{\prime }-x-y& =0\\ \text{(1)}& y^{\prime }& =\frac{x+y}{x-y}& =f(x,y)\end{array}$$

We start by checking if it homogenous or not. Using

$$\begin{array}{rl}m& =\frac{f+x{f}_x}{f-y{f}_y}\\ & =\frac{\frac{x+y}{x-y}+x(\frac{1}{x-y}-\frac{x+y}{{(x-y)}^2})}{\frac{x+y}{x-y}-y\left((\frac{1}{x-y}+\frac{x+y}{{(x-y)}^2})\right)}\\ & =\frac{x(\frac{1}{x-y}-\frac{x+y}{{(x-y)}^2})}{-y\left((\frac{1}{x-y}+\frac{x+y}{{(x-y)}^2})\right)}\\ & =1\end{array}$$

Since $m=1$ then this is homogeneous ode (special case of isobaric). Hence the substitution $v=\frac{y}{x}$ makes the ode (1) separable.

3.3.13.2.7 Example 7

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

We start by checking if it homogenous or not. Using

$$\begin{array}{rl}m& =\frac{f+x{f}_x}{f-y{f}_y}\\ & =\frac{\frac{y+2\sqrt{xy}}{x}+x(\frac{y}{x\sqrt{xy}}-\frac{y+2\sqrt{xy}}{x^2})}{\frac{y+2\sqrt{xy}}{x}-y\left(\frac{1+\frac{x}{\sqrt{xy}}}{x}\right)}\\ & =1\end{array}$$

Since $m=1$ then this is homogeneous ode (special case of isobaric). Hence the substitution $v=\frac{y}{x}$ makes the ode (1) separable.

3.3.13.2.8 Example 8

$$\begin{array}{}\text{(1)}& y^{\prime }=\frac{-y(y^2+3x^2+2x)}{x^2+y^2}\end{array}$$

We start by checking if it homogenous or not. Using

$$\begin{array}{rl}m& =\frac{f+x{f}_x}{f-y{f}_y}\\ & =\frac{\frac{-y(y^2+3x^2+2x)}{x^2+y^2}+x\frac{d}{dx}\left(\frac{-y(y^2+3x^2+2x)}{x^2+y^2}\right)}{\frac{-y(y^2+3x^2+2x)}{x^2+y^2}-y\frac{d}{dy}\left(\frac{-y(y^2+3x^2+2x)}{x^2+y^2}\right)}\\ & =\frac{\frac{-y(y^2+3x^2+2x)}{x^2+y^2}+x(-2\frac{y(-x^2+2x{y}^2+y^2)}{{(x^2+y^2)}^2})}{\frac{-y(y^2+3x^2+2x)}{x^2+y^2}-y(-\frac{3x^4+2x^3-2x{y}^2+y^4}{{(x^2+y^2)}^2})}\\ & =\frac{3x^4+8x^2{y}^2+4x{y}^2+y^4}{4x^2{y}^2+4x{y}^2}\end{array}$$

Since this does not simplify to numerical value, it is not homogenous ode. This turns out to be homogenous type D. See earlier note on this. There is a slight difference in definition between homogenous ode and homogenous type D. In Maple terms, homogenous ode is called homogenous ode type A. A homogenous type D is one in which the substitution $y=ux$ makes the ode separable or quadrature.

3.3.13.2.9 Example 9

$$\begin{array}{}\text{(1)}& y^{\prime }=\frac{{(-108y^2+12\sqrt{-108y^3{x}^3+81y^4})}^{\frac{2}{3}}+12xy}{6{(-108y^2+12\sqrt{-108y^3{x}^3+81y^4})}^{\frac{1}{3}}}\end{array}$$

We start by checking if it homogenous or not. Using

$$m=\frac{f+x{f}_x}{f-y{f}_y}$$

Which simplifies to

$$m=3$$

Hence the substitution $y=v{x}^m$ will make the ode separable. Substituting $y=v{x}^3$ in (1) results in separable ode. But for this case, we have to assume $x>0$ in order to simplify it. The resulting ode is too long to write now, but verified to be separable using the computer.