2.62   ODE No. 62

1. Problem in Latex
2. Mathematica input
3. Maple input

$$y^{\prime }(x)-\frac{y(x)-x^2\sqrt{x^2-y(x{)}^2}}{xy(x)\sqrt{x^2-y(x{)}^2}+x}=0$$ ✓ Mathematica : cpu = 4.17572 (sec), leaf count = 36

$$\text{Solve}[2{\tan }^{-1}\left(\frac{y(x)}{\sqrt{x^2-y(x{)}^2}}\right)+x^2+y(x{)}^2=2c_1,y(x)]$$

✓ Maple : cpu = 0.424 (sec), leaf count = 34

$$\{\frac{{\left(y\left(x\right)\right)}^2}{2}+\arctan \left(y\left(x\right)\frac{1}{\sqrt{x^2-{\left(y\left(x\right)\right)}^2}}\right)+\frac{x^2}{2}-\mathit{\_}\mathit{C}\mathit{1}=0\}$$

Hand solution

$$\begin{array}{}\text{(1)}& y^{\prime }=\frac{y-x^2\sqrt{x^2-y^2}}{xy\sqrt{x^2-y^2}+x}\end{array}$$ Let $y=ux$ then $y^{\prime }=u+x{u}^{\prime }$ therefore$$\begin{array}{rl}u+x{u}^{\prime }& =\frac{y-x^2\sqrt{x^2-y^2}}{xy\sqrt{x^2-y^2}+x}\\ & =\frac{ux-x^2\sqrt{x^2-{\left(ux\right)}^2}}{x\left(ux\right)\sqrt{x^2-{\left(ux\right)}^2}+x}\\ & =\frac{ux-x^3\sqrt{1-u^2}}{x^{3}u\sqrt{1-u^2}+x}\\ & =\frac{u-x^2\sqrt{1-u^2}}{x^{2}u\sqrt{1-u^2}+1}\end{array}$$

Hence$$\begin{array}{rl}u(x^{2}u\sqrt{1-u^2}+1)+x{u}^{\prime }(x^{2}u\sqrt{1-u^2}+1)& =u-x^2\sqrt{1-u^2}\\ x^2{u}^2\sqrt{1-u^2}+u+u^{\prime }(x^{3}u\sqrt{1-u^2}+x)& =u-x^2\sqrt{1-u^2}\\ x^2{u}^2\sqrt{1-u^2}+u^{\prime }(x^{3}u\sqrt{1-u^2}+x)& =-x^2\sqrt{1-u^2}\\ x{u}^2\sqrt{1-u^2}+u^{\prime }(x^{2}u\sqrt{1-u^2}+1)& =-x\sqrt{1-u^2}\\ x{u}^2+u^{\prime }(x^{2}u+\frac{1}{\sqrt{1-u^2}})& =-x\\ x(1+u^2)+u^{\prime }(x^{2}u+\frac{1}{\sqrt{1-u^2}})& =0\end{array}$$

Hence$$\begin{array}{}\text{(2)}& x(1+u^2)dx+(x^{2}u+\frac{1}{\sqrt{1-u^2}})du=0\end{array}$$ Let $M=x(1+u^2),N=(x^{2}u+\frac{1}{\sqrt{1-u^2}})$. $$\begin{array}{rl}\frac{\partial M}{\partial u}& =2xu\\ \frac{\partial N}{\partial x}& =2xu\end{array}$$

Therefore (2) is exact.  Let $$x(1+u^2)dx+(x^{2}u+\frac{1}{\sqrt{1-u^2}})du=dU$$ Since $dU=\frac{\partial U}{\partial x}dx+\frac{\partial U}{\partial u}du$. Comparing with the above, we see that $$\begin{array}{}\text{(3)}& \frac{\partial U}{\partial x}& =x(1+u^2)\text{(4)}& \frac{\partial U}{\partial u}& =x^{2}u+\frac{1}{\sqrt{1-u^2}}\end{array}$$

From (3)$$\begin{array}{rl}U& =\int x(1+u^2)dx\\ \text{(5)}& & =\frac{x^2}{2}(1+u^2)+f\left(u\right)\end{array}$$

From (4)$$\begin{array}{rl}\frac{d}{\partial u}(\frac{x^2}{2}(1+u^2)+f\left(u\right))& =x^{2}u+\frac{1}{\sqrt{1-u^2}}\\ x^{2}u+f^{\prime }\left(u\right)& =x^{2}u+\frac{1}{\sqrt{1-u^2}}\\ f^{\prime }\left(u\right)& =\frac{1}{\sqrt{1-u^2}}\end{array}$$

Therefore $$f\left(u\right)=\arcsin \left(u\right)$$ From (5) we find$$U(x,u)=\frac{x^2}{2}(1+u^2)+\arcsin \left(u\right)$$ Since $dU=0$ then$$\begin{array}{rl}\frac{x^2}{2}(1+u^2)+\arcsin \left(u\right)& =C\\ \frac{x^2}{2}(1+u^2)+\arcsin \left(u\right)-C& =0\end{array}$$

Since $y=ux$ then the above can be written as$$\begin{array}{rl}\frac{x^2}{2}(1+{\left(\frac{y}{x}\right)}^2)+\arcsin \left(\frac{y}{x}\right)-C& =0\\ \frac{x^2}{2}\left(\frac{x^2+y^2}{x^2}\right)+\arcsin \left(\frac{y}{x}\right)-C& =0\\ \frac{1}{2}(x^2+y^2)+\arcsin \left(\frac{y}{x}\right)-C& =0\\ \arcsin \left(\frac{y}{x}\right)& =C-\frac{1}{2}(x^2+y^2)\end{array}$$

Hence$$\begin{array}{rl}\frac{y}{x}& =\sin (C-\frac{1}{2}(x^2+y^2))\\ y\left(x\right)& =x\sin (C-\frac{1}{2}(x^2+y^2))\end{array}$$