3.3.12.1 Examples

3.3.12.1.1 Example 1 Solve

$$y^{\prime }=\frac{-y(2x^2{y}^3+3)}{x(x^2{y}^3+1)}$$

The first step is to identify if this is class G and find $F$. We start by multiplying the RHS by $\frac{x}{y}$ (regardless of what is in the RHS) which gives

$$\begin{array}{rl}{y}^{\prime }& =\frac{x}{y}\left(\frac{-y(2x^2{y}^3+3)}{x(x^2{y}^3+1)}\right)\\ & =\frac{-2x^2{y}^3-3}{x^2{y}^3+1}\\ & =F(x,y)\end{array}$$

Next we check if $F(x,y)$ has $y$ or not in it. If so, then let the RHS above be $F(x,y)$ and now do

$$\begin{array}{rl}{f}_x& =x\frac{\partial F}{\partial x}\\ & =x\left(\frac{2y^{3}x}{{(x^2{y}^3+1)}^2}\right)\\ & =\frac{2y^3{x}^2}{{(x^2{y}^3+1)}^2}\end{array}$$

And let

$$\begin{array}{rl}{f}_y& =y\frac{\partial F}{\partial y}\\ & =y\left(\frac{3x^2{y}^2}{{(x^2{y}^3+1)}^2}\right)\\ & =\frac{3x^2{y}^3}{{(x^2{y}^3+1)}^2}\end{array}$$

Now we check, if $f_y=0$ then this is not Homogeneous type G. Else we now need to determine value of $\alpha$. This is done as follows.

$$\begin{array}{rl}\alpha & =\frac{fx}{f_y}\\ & =\frac{2}{3}\end{array}$$

If $\alpha$ comes out not to have in it $x$ nor $y$ as in this case, then we are done. This ode is Homogeneous type G. But we have to do one more check.  We have to check that $F(x,y)$ found above ends up with no $x$ in it. Hence the solution is

$$\begin{array}{}\text{(1)}& \ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\tau \right))}d\tau =0\end{array}$$

Now let $y{x}^{\alpha }=\tau$ or $y=\frac{\tau }{x^{\alpha }}$and substituting this into $F(x,y)$ gives

$$\begin{array}{rl}F\left(\tau \right)& =\frac{-2x^2{\left(\frac{\tau }{x^{\alpha }}\right)}^3-3}{x^2{\left(\frac{\tau }{x^{\alpha }}\right)}^3+1}\\ & =\frac{-2x^2{\left(\frac{\tau }{x^{\frac{2}{3}}}\right)}^3-3}{x^2{\left(\frac{\tau }{x^{\frac{2}{3}}}\right)}^3+1}\\ & =\frac{-2x^2\left(\frac{{\tau }^3}{x^2}\right)-3}{x^2\left(\frac{{\tau }^3}{x^3}\right)+1}\\ & =\frac{-2{\tau }^3-3}{{\tau }^3+1}\end{array}$$

We see that $F(x,y)$ ends up as $F\left(\tau \right)=\frac{-2{\tau }^3-3}{{\tau }^3+1}$ after the transformation. It has no $x$ left in it. If we end up with $x$ then this method can not be used.

The solution (1) becomes

$$\begin{array}{rl}\ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -\left(\frac{-2{\tau }^3-3}{{\tau }^3+1}\right))}d\tau & =0\\ \ln x-c_1+{\int }^{y{x}^{\frac{2}{3}}}\frac{3{\tau }^3+3}{4{\tau }^4+7\tau }d\tau & =0\end{array}$$

Solving the integral gives

$$\ln x-c_1+\frac{3}{7}\ln \left(y{x}^{\frac{2}{3}}\right)+\frac{3}{28}\ln (4x^2{y}^3+7)=0$$

And this is the final answer. Now if earlier we have $F(x,y)$ not have $y$ in it. In this case we check if $F(x,y)$ has $x$. If not, then $\alpha =0$ and we do the same as above. But if $F(x,y)$ has $x$ and not has $y$ then it is not not Homogeneous type G.

3.3.12.1.2 Example 2 Solve

$$y^{\prime }=\frac{2x(-x^4-2x^{2}y+y^2)}{y^2+2x^{2}y-x^4}$$

The first step is to identify if this is class G and find $F$. We start by multiplying the RHS by $\frac{x}{y}$ (regardless of what is in the RHS) which gives

$$\begin{array}{rl}{y}^{\prime }& =\frac{x}{y}\left(\frac{2x(-x^4-2x^{2}y+y^2)}{y^2+2x^{2}y-x^4}\right)\\ & =\frac{2x^2(x^4+2x^{2}y-y^2)}{y(x^4-2x^{2}y-y^2)}\\ & =F(x,y)\end{array}$$

Next we check if $F(x,y)$ has $y$ or not in it. If so, then let the RHS above be $F(x,y)$ and now do

$$\begin{array}{rl}{f}_x& =x\frac{\partial F}{\partial x}\\ & =x\left(\frac{4x(x^8-4x^{6}y-6x^4{y}^2-4x^2{y}^3+y^4)}{y{(x^4-2x^{2}y-y^2)}^2}\right)\\ & =\frac{4x^2(x^8-4x^{6}y-6x^4{y}^2-4x^2{y}^3+y^4)}{y{(x^4-2x^{2}y-y^2)}^2}\end{array}$$

And let

$$\begin{array}{rl}{f}_y& =y\frac{\partial F}{\partial y}\\ & =y\left(\frac{-2x^2(x^8-4x^{6}y-6x^4{y}^2-4x^2{y}^2-4x^2{y}^3+y^4)}{y^2{(x^4-2x^{2}y-y^2)}^2}\right)\\ & =\frac{-2x^2(x^8-4x^{6}y-6x^4{y}^2-4x^2{y}^2-4x^2{y}^3+y^4)}{y{(x^4-2x^{2}y-y^2)}^2}\end{array}$$

Now we check, if $f_y=0$ then this is not Homogeneous type G. Else we now need to determine value of $\alpha$. This is done as follows.

$$\begin{array}{rl}\alpha & =\frac{fx}{f_y}\\ & =-2\end{array}$$

If $\alpha$ comes out not to have in it $x$ nor $y$ as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

$$y^{\prime }=\frac{y}{x}F\left(\frac{y}{x^{\alpha }}\right)$$

Hence the solution is

$$\begin{array}{}\text{(1)}& \ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\tau \right))}d\tau =0\end{array}$$

Now let $y=\frac{\tau }{x^{\alpha }}$ and substitute this into $F(x,y)$ which results in

$$\begin{array}{rl}F\left(\tau \right)& =\frac{2x^2(x^4+2x^2\frac{\tau }{x^{\alpha }}-{\left(\frac{\tau }{x^{\alpha }}\right)}^2)}{\frac{\tau }{x^{\alpha }}(x^4-2x^2\frac{\tau }{x^{\alpha }}-{\left(\frac{\tau }{x^{\alpha }}\right)}^2)}\\ & =\frac{2x^2(x^4+2x^2\frac{\tau }{x^{-2}}-{\left(\frac{\tau }{x^{-2}}\right)}^2)}{\frac{\tau }{x^{-2}}(x^4-2x^2\frac{\tau }{x^{-2}}-{\left(\frac{\tau }{x^{-2}}\right)}^2)}\\ & =\frac{2x^2(x^4+2x^4\tau -x^4{\tau }^2)}{\tau x^2(x^4-2x^4\tau -{\tau }^2{x}^4)}\\ & =\frac{2(x^4+2x^4\tau -x^4{\tau }^2)}{\tau (x^4-2x^4\tau -{\tau }^2{x}^4)}\\ & =\frac{2}{\tau }\frac{(1+2\tau -{\tau }^2)}{(1-2\tau -{\tau }^2)}\\ & =\frac{2}{\tau }\frac{({\tau }^2-2\tau -1)}{({\tau }^2+2\tau -1)}\end{array}$$

The solution(1) becomes

$$\begin{array}{rl}\ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\alpha \right))}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{x^2}}\frac{1}{\tau (2-\left(\frac{2}{\tau }\frac{({\tau }^2-2\tau -1)}{({\tau }^2+2\tau -1)}\right))}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{x^2}}\frac{1}{2}\frac{{\tau }^2+2\tau -1}{{\tau }^3+{\tau }^2+\tau +1}d\tau & =0\end{array}$$

Solving the integral gives

$$\ln x-c_1-\frac{1}{2}\ln \left(\frac{x^2+y}{x^2}\right)+\frac{1}{2}\ln \left(\frac{x^4+y^2}{x^4}\right)=0$$

3.3.12.1.3 Example 3 Solve

$$y^{\prime }=-\frac{1}{2}\frac{3y^2-x}{y(y^2-3x)}$$

The first step is to identify if this is class G and find $F$. We start by multiplying the RHS by $\frac{x}{y}$ (regardless of what is in the RHS) which gives

$$\begin{array}{rl}{y}^{\prime }& =\frac{x}{y}(-\frac{1}{2}\frac{3y^2-x}{y(y^2-3x)})\\ & =-\frac{1}{2}\frac{3x{y}^2-x^2}{y^4-3x{y}^2}\\ & =F(x,y)\end{array}$$

Next we check if $F(x,y)$ has $y$ or not in it. If so, then let the RHS above be $F(x,y)$ and now do

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

And let

$$\begin{array}{rl}{f}_y& =y\frac{\partial F}{\partial y}\\ & =\frac{3x{y}^4-3x^2{y}^2+3x^3}{y^2{(-y^2+3x)}^2}\end{array}$$

Now we check, if $f_y=0$ then this is not Homogeneous type G. Else we now need to determine value of $\alpha$. This is done as follows.

$$\begin{array}{rl}\alpha & =\frac{fx}{f_y}\\ & =-\frac{1}{2}\end{array}$$

If $\alpha$ comes out not to have in it $x$ nor $y$ as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

$$y^{\prime }=\frac{y}{x}F\left(\frac{y}{x^{\alpha }}\right)$$

Hence the solution is

$$\begin{array}{}\text{(1)}& \ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\tau \right))}d\tau =0\end{array}$$

Now let $y=\frac{\tau }{x^{\alpha }}$ and substitute this into $F(x,y)$ which results in

$$\begin{array}{rl}F\left(\tau \right)& =-\frac{1}{2}\frac{3x{y}^2-x^2}{y^4-3x{y}^2}\\ & =-\frac{1}{2}\frac{3x{\left(\frac{\tau }{x^{\alpha }}\right)}^2-x^2}{{\left(\frac{\tau }{x^{\alpha }}\right)}^4-3x{\left(\frac{\tau }{x^{\alpha }}\right)}^2}\\ & =-\frac{1}{2}\frac{3x{\left(\frac{\tau }{x^{-\frac{1}{2}}}\right)}^2-x^2}{{\left(\frac{\tau }{x^{-\frac{1}{2}}}\right)}^4-3x{\left(\frac{\tau }{x^{-\frac{1}{2}}}\right)}^2}\\ & =-\frac{1}{2}\frac{3x^2{\tau }^2-x^2}{{\tau }^4{x}^2-3x{\tau }^{2}x}\\ & =-\frac{1}{2}\frac{3{\tau }^2-1}{{\tau }^4-3{\tau }^2}\end{array}$$

The solution(1) becomes

$$\begin{array}{rl}\ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\alpha \right))}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{\sqrt{x}}}\frac{1}{\tau (\frac{1}{2}-(-\frac{1}{2}\frac{3{\tau }^2-1}{{\tau }^4-3{\tau }^2}))}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{\sqrt{x}}}2\tau \frac{{\tau }^2-3}{{\tau }^4-1}d\tau & =0\end{array}$$

Solving the integral gives

$$\ln x-c_1-\frac{1}{2}\ln (\frac{y}{\sqrt{x}}-1)-\ln (\frac{y}{\sqrt{x}}+1)+2\ln (\frac{y^2}{x}+1)=0$$

3.3.12.1.4 Example 4 Solve

$$y^{\prime }=-\frac{1}{2}\frac{y(1+\sqrt{x^2{y}^4+1})}{x}$$

The first step is to identify if this is class G and find $F$. We start by multiplying the RHS by $\frac{x}{y}$ (regardless of what is in the RHS) which gives

$$\begin{array}{rl}{y}^{\prime }& =\frac{x}{y}(-\frac{1}{2}\frac{y(1+\sqrt{x^2{y}^4+1})}{x})\\ & =-\frac{1}{2}(1+\sqrt{x^2{y}^4+1})\\ & =F(x,y)\end{array}$$

Next we check if $F(x,y)$ has $y$ or not in it. If so, then let the RHS above be $F(x,y)$ and now do

$$\begin{array}{rl}{f}_x& =x\frac{\partial F}{\partial x}\\ & =-\frac{1}{2}\frac{x^2{y}^4}{\sqrt{x^2{y}^4+1}}\end{array}$$

And let

$$\begin{array}{rl}{f}_y& =y\frac{\partial F}{\partial y}\\ & =\frac{-x^2{y}^4}{\sqrt{x^2{y}^4+1}}\end{array}$$

Now we check, if $f_y=0$ then this is not Homogeneous type G. Else we now need to determine value of $\alpha$. This is done as follows.

$$\begin{array}{rl}\alpha & =\frac{fx}{f_y}\\ & =\frac{1}{2}\end{array}$$

If $\alpha$ comes out not to have in it $x$ nor $y$ as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

$$y^{\prime }=\frac{y}{x}F\left(\frac{y}{x^{\alpha }}\right)$$

Hence the solution is

$$\begin{array}{}\text{(1)}& \ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\tau \right))}d\tau =0\end{array}$$

Now let $y=\frac{\tau }{x^{\alpha }}$ and substitute this into $F(x,y)$ which results in

$$\begin{array}{rl}F\left(\tau \right)& =-\frac{1}{2}(1+\sqrt{x^2{y}^4+1})\\ & =-\frac{1}{2}(1+\sqrt{x^2{\left(\frac{\tau }{x^{\alpha }}\right)}^4+1})\\ & =-\frac{1}{2}(1+\sqrt{x^2{\left(\frac{\tau }{x^{\frac{1}{2}}}\right)}^4+1})\\ & =-\frac{1}{2}(1+\sqrt{x^2\frac{{\tau }^4}{x^2}+1})\\ & =-\frac{1}{2}(1+\sqrt{{\tau }^4+1})\end{array}$$

The solution(1) becomes

$$\begin{array}{rl}\ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\alpha \right))}d\tau & =0\\ \ln x-c_1+{\int }^{y\sqrt{x}}\frac{1}{\tau (-\frac{1}{2}-(-\frac{1}{2}(1+\sqrt{{\tau }^4+1})))}d\tau & =0\\ \ln x-c_1+{\int }^{y\sqrt{x}}\frac{2}{\tau \sqrt{{\tau }^4+1}}d\tau & =0\end{array}$$

Solving the integral gives

$$\ln x-c_1-\mathrm{arctanh}\left(\frac{1}{\sqrt{x^2{y}^4+1}}\right)=0$$

3.3.12.1.5 Example 5 Solve

$$y^{\prime }=x(1+\frac{2y}{x}+\frac{y^2}{x^4})$$

The first step is to identify if this is class G and find $F$. We start by multiplying the RHS by $\frac{x}{y}$ (regardless of what is in the RHS) which gives

$$\begin{array}{rl}{y}^{\prime }& =\frac{x}{y}\left(x(1+\frac{2y}{x}+\frac{y^2}{x^4})\right)\\ & =\frac{x^2}{y}+2x+\frac{y}{x^2}\\ & =\frac{{(x^2+y)}^2}{x^{2}y}\\ & =F(x,y)\end{array}$$

Next we check if $F(x,y)$ has $y$ or not in it. If so, then let the RHS above be $F(x,y)$ and now do

$$\begin{array}{rl}{f}_x& =x\frac{\partial F}{\partial x}\\ & =\frac{2x^4-2y^2}{x^{2}y}\end{array}$$

And let

$$\begin{array}{rl}{f}_y& =y\frac{\partial F}{\partial y}\\ & =\frac{-x^4+y^2}{x^{2}y}\end{array}$$

Now we check, if $f_y=0$ then this is not Homogeneous type G. Else we now need to determine value of $\alpha$. This is done as follows.

$$\begin{array}{rl}\alpha & =\frac{fx}{f_y}\\ & =-2\end{array}$$

If $\alpha$ comes out not to have in it $x$ nor $y$ as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

$$y^{\prime }=\frac{y}{x}F\left(\frac{y}{x^{\alpha }}\right)$$

Hence the solution is

$$\begin{array}{}\text{(1)}& \ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\tau \right))}d\tau =0\end{array}$$

Now let $y=\frac{\tau }{x^{\alpha }}$ and substitute this into $F(x,y)$ which results in

$$\begin{array}{rl}F\left(\tau \right)& =\frac{{(x^2+y)}^2}{x^{2}y}\\ & =\frac{{(x^2+\frac{\tau }{x^{\alpha }})}^2}{x^2\frac{\tau }{x^{\alpha }}}\\ & =\frac{{(x^2+\frac{\tau }{x^{-2}})}^2}{x^2\frac{\tau }{x^{-2}}}\\ & =\frac{{(x^2+\tau x^2)}^2}{x^4\tau }\\ & =\frac{x^4+{\tau }^2{x}^4+2\tau x^4}{x^4\tau }\\ & =\frac{1+{\tau }^2+2\tau }{\tau }\end{array}$$

The solution(1) becomes

$$\begin{array}{rl}\ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\alpha \right))}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{x^2}}\frac{1}{\tau (2-\left(\frac{1+{\tau }^2+2\tau }{\tau }\right))}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{x^2}}-\frac{1}{{\tau }^2+1}d\tau & =0\\ \ln x-c_1-{\int }^{\frac{y}{x^2}}\frac{1}{{\tau }^2+1}d\tau & =0\end{array}$$

Solving the integral gives

$$\begin{array}{rl}\ln x-c_1-\arctan \left(\frac{y}{x^2}\right)& =0\\ y& =-\tan (c_1-\ln x)x^2\end{array}$$

3.3.12.1.6 Example 6 Solve

$${\left(y^{\prime }\right)}^2=4y-x^2$$

Hence

$$y^{\prime }=\pm \sqrt{4y-x^2}$$

For the first ode, the first step is to identify if this is class G and find $F$. We start by multiplying the RHS by $\frac{x}{y}$ (regardless of what is in the RHS) which gives

$$\begin{array}{rl}{y}^{\prime }& =\frac{x}{y}\sqrt{4y-x^2}\\ & =F(x,y)\end{array}$$

Next we check if $F(x,y)$ has $y$ or not in it. If so, then let the RHS above be $F(x,y)$ and now do

$$\begin{array}{rl}{f}_x& =x\frac{\partial F}{\partial x}\\ & =-\frac{2x(x^2-2y)}{y\sqrt{4y-x^2}}\end{array}$$

And let

$$\begin{array}{rl}{f}_y& =y\frac{\partial F}{\partial y}\\ & =\frac{x(x^2-2y)}{y\sqrt{4y-x^2}}\end{array}$$

Now we check, if $f_y=0$ then this is not Homogeneous type G. Else we now need to determine value of $\alpha$. This is done as follows.

$$\begin{array}{rl}\alpha & =\frac{fx}{f_y}\\ & =-2\end{array}$$

If $\alpha$ comes out not to have in it $x$ nor $y$ as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

$$y^{\prime }=\frac{y}{x}F\left(\frac{y}{x^{\alpha }}\right)$$

Hence the solution is

$$\begin{array}{}\text{(1)}& \ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\tau \right))}d\tau =0\end{array}$$

Now let $y=\frac{\tau }{x^{\alpha }}$ and substitute this into $F(x,y)$ which results in

$$\begin{array}{rl}F\left(\tau \right)& =\frac{x}{y}\sqrt{4y-x^2}\\ & =\frac{x}{\frac{\tau }{x^{\alpha }}}\sqrt{4\frac{\tau }{x^{\alpha }}-x^2}\\ & =\frac{x}{\tau x^2}\sqrt{4\tau x^2-x^2}\end{array}$$

Since the requirement is that $F\left(\tau \right)$ ends up free of $x$, then the only way to use this method and simplify the above to eliminate $x$ is to assume $x>0$. Now the above simplifies to

$$F\left(\tau \right)=\frac{1}{\tau }\sqrt{4\tau -1}$$

The solution(1) becomes

$$\begin{array}{rl}\ln x-c_1+{\int }^{y{x}^{\alpha }}\frac{1}{\tau (-\alpha -F\left(\alpha \right))}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{x^2}}\frac{1}{\tau (2-\frac{1}{\tau }\sqrt{4\tau -1})}d\tau & =0\\ \ln x-c_1+{\int }^{\frac{y}{x^2}}\frac{1}{2\tau -\sqrt{4\tau -1}}d\tau & =0\end{array}$$

Solving the integral gives long complicated expression which is verified correct. So better to keep the solution implicit as the above. Now we solve the second ode $y^{\prime }=-\sqrt{4y-x^2}$ in similar way.