Answer to Question #92857 in Differential Equations for Subhasis Padhy

Question

Solve the following differential equations.

(i) dx/y²(x-y) =dy/x²(x-y) =dz/z(x²+y²)

(ii) (yz+z²)dx - xzdy + xydz =0

(iii) (z+z³)cosx dx - (z+z³)dy + (1-z²) (y-sinx)dz =0

Answer 1

I)$\frac{dx}{y^2(x-y)}=\frac{dy}{x^2(x-y)}\Rightarrow x^2dx=y^2dy\Rightarrow x^3-y^3=C^3$ . We have $\frac{dx}{y^2(x-y)}=\frac{dz}{z(x^2+y^2)}\Rightarrow \frac{(x^2+y^2)}{y^2(x-y)}dx=\frac{dz}{z}.$ Let $x=C\cosh^\frac{2}{3}t$ and $y=C\sinh^\frac{2}{3}t$ , then $dx=\frac{2C\sinh tdt}{3\cosh^\frac{1}{3}t}$ and $\frac{(x^2+y^2)}{y^2(x-y)}dx=$ $=\frac{C^2(\cosh^\frac{4}{3}t+\sinh^\frac{4}{3}t)}{C^2\sinh^\frac{4}{3}t\cdot C(\cosh^\frac{2}{3}t-\sinh^\frac{2}{3}t)}\cdot\frac{2C\sinh tdt}{3\cosh^\frac{1}{3}t}=\frac{2}{3}\frac{\cosh^\frac{4}{3}t+\sinh^\frac{4}{3}t}{\sinh^\frac{1}{3}t\cosh^\frac{1}{3}t(\cosh^\frac{2}{3}t-\sinh^\frac{2}{3}t)}dt$ . Divide the numerator and denominator by $\cosh^\frac{4}{3}t$ : $\frac{2}{3}\frac{\cosh^\frac{4}{3}t+\sinh^\frac{4}{3}t}{\sinh^\frac{1}{3}t\cosh^\frac{1}{3}t(\cosh^\frac{2}{3}t-\sinh^\frac{2}{3}t)}dt=\frac{2}{3}\frac{1+\tanh^\frac{4}{3}t}{\tanh^\frac{1}{3}t(1-\tanh^\frac{2}{3}t)}dt$ Make change of the variable: $\tanh^\frac{2}{3}t=u$ . So $t=artanh(u^\frac{3}{2})$ and $dt=\frac{1}{1-u^3}\cdot\frac{3}{2}\sqrt{u}du$ . Then$\frac{2}{3}\frac{1+\tanh^\frac{4}{3}t}{\tanh^\frac{1}{3}t(1-\tanh^\frac{2}{3}t)}dt=\frac{2}{3}\frac{1+u^2}{\sqrt{u}(1-u)}\cdot\frac{1}{1-u^3}\cdot\frac{3}{2}\sqrt{u}du=\frac{1+u^2}{(1-u)(1-u^3)}du$ Write the partial fraction decomposition of $\frac{1+u^2}{(1-u)(1-u^3)}$ : $\frac{1+u^2}{(1-u)(1-u^3)}=\frac{A}{1-u}+\frac{B}{(1-u)^2}+\frac{Du+E}{u^2+u+1}=$ $=\frac{(A-Au+B)(u^2+u+1)+(Du+E)(1-u^2)}{(1-u)^2(u^2+u+1)}$ $(A-Au+B)(u^2+u+1)+(Du+E)(1-u^2)=(-A+D)u^3+(B-2D+E)u^2+(B+D-2E)u+(A+B+E)$ Equality of the polynoms $(-A+D)u^3+(B-2D+E)u^2+(B+D-2E)u+(A+B+E)$ and $1+u^2$ gives us system of equations: $\begin{cases} -A+D=0\\ B-2D+E=1\\ B+D-2E=0\\ A+B+E=1 \end{cases}$ , where $A=D=0, B=\frac{2}{3}$ and $E=\frac{1}{3}$ . So $\frac{1+u^2}{(1-u)(1-u^3)}=\frac{2}{3}\frac{1}{(1-u)^2}+\frac{1}{3}\frac{1}{u^2+u+1}$ .$\int\bigl(\frac{2}{3}\frac{1}{(1-u)^2}+\frac{1}{3}\frac{1}{u^2+u+1}\bigr)du=\frac{2}{3}\int\frac{1}{(1-u)^2}du+\frac{1}{3}\int\frac{1}{u^2+u+1}du=\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}u+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-u)}+F$ .Since $\frac{1+u^2}{(1-u)(1-u^3)}du=\frac{dz}{z}$ , we have $\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}u+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-u)}+F = \ln |z|$ or $e^Fe^{\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}u+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-u)}} = |z|$ . That is $z=e^Fe^{\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}u+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-u)}}$ or $z=-e^Fe^{\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}u+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-u)}}$ . Union of these 2 cases is $z=Ge^{\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}u+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-u)}}$ . Remembering that $u=\tanh^\frac{2}{3}t$ , we can write $z=Ge^{\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}\tanh^\frac{2}{3}t+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-\tanh^\frac{2}{3}t)}}$ Answer: $x=C\cosh^\frac{2}{3}t$ , $y=C\sinh^\frac{2}{3}t$ and $z=Ge^{\frac{2\sqrt{3}}{9}\arctan\bigl(\frac{2}{\sqrt{3}}\tanh^\frac{2}{3}t+\frac{1}{\sqrt{3}}\bigr)+\frac{2}{3(1-\tanh^\frac{2}{3}t)}}$

II)Dividing the equation by $xy^2$ , we have $\frac{(yz+z^2)dx}{xy^2}+\frac{ydz-zdy}{y^2}=0$ . Since $\frac{ydz-zdy}{y^2}=d\bigl(\frac{z}{y}\bigr)$ , we can rewrite the equation: $\bigl(\frac{z}{y}+\bigl(\frac{z}{y}\bigr)^2\bigr)\frac{dx}{x}+d\bigl(\frac{z}{y}\bigr)=0$ . After making change of the variables $\frac{z}{y}=t$ , we have $(t+t^2)\frac{dx}{x}+dt=0$ or $\frac{dx}{x}=-\frac{dt}{t+t^2}=\frac{dt}{t+1}-\frac{dt}{t}$ . Solution of this equation is $\ln |x|+C=\ln |t+1|-\ln |t|=\ln\bigl |\frac{t+1}{t}\bigr|$ . Rewrite this equaliy: $e^c|x|=\bigl|\frac{t+1}{t}\bigr|$ . That is $e^cx=\frac{t+1}{t}$ or $-e^cx=\frac{t+1}{t}$ . Union of these 2 cases is $Dx=\frac{t+1}{t}$ . Remembering that $t=\frac{z}{y}$ , we have $Dx=\frac{y+z}{z}$

Answer: $Dx=\frac{y+z}{z}$

III)Rewrite the equation: $(z+z^3)(\cos xdx-dy)+(1-z^2)(y-\sin x)dz=0$ . Note that $\cos xdx-dy=d(\sin x-y)$ .

So after making change of the variables $\sin x-y=t$ , we have $(z+z^3)dt-(1-z^2)tdz=0$ or $\frac{dt}{t}=\frac{(1-z^2)dz}{z+z^3}$

$\frac{1-z^2}{z+z^3}=\frac{A}{z}+\frac{Bz+C}{z^2+1}=\frac{A(z^2+1)+z(Bz+C)}{z^3+z}=\frac{(A+B)z^2+Cz+A}{z^3+z}$ , so $\begin{cases} A+B=-1\\ C=0\\ A=1 \end{cases}$ , where $A=1, B=-2, C=0$

We have $\frac{1-z^2}{z+z^3}=\frac{1}{z}-\frac{2z}{z^2+1}$ . Since $\frac{dt}{t}=\frac{dz}{z}-\frac{2zdz}{z^2+1}$ , we have $\ln |t| = \ln |z| - \ln |z^2+1| +C$ or $|t|=e^C\bigl|\frac{z}{z^2+1}\bigr|$ . That is $t=e^C\frac{z}{z^2+1}$ or $t=-e^C\frac{z}{z^2+1}$ . Union of these 2 cases is $t=D\frac{z}{z^2+1}$ . Remembering that $t=\sin x-y$ , we have $\sin x-y=D\frac{z}{z^2+1}$

Answer: $\sin x-y=D\frac{z}{z^2+1}$