How to find du and dv?

Question

Find $du$ and $dv$ if $u+v=x+y$ and $\frac{\sin(u)}{\sin(v)}=\frac{x}{y}$.
How to solve this?
Found almost an answer:
But how do we get $du=...$ from the second?

Answer 1

You know they treated $du, dv$ as variables and solved the Linear equation

• So, Is it OK to treat differentials as variables?: YES! The rest you can do either by using Cramer's Rule or else as follow: $$u + v = x + y \implies du + dv = dx + dy $$ $$\color{red}{du} = dx + dy - dv \text{ & } \color{blue}{dv} = dx + dy - du$$

$$y\sin u - x\sin v =0 \implies dyS_u + yC_u\color{red}{du} - dxS_v -xC_v\color{blue}{dv} = 0 $$ $$dyS_u + yC_u\color{red}{du} - dxS_v -xC_v\color{blue}{(dx + dy - du)} = 0 $$ $$\color{red}{du}(yC_u+xC_v) = \left(dx(S_v+xC_y) -dy(S_u-xC_v)\right)$$ $$\implies du = \frac{\left[dx(\sin v+ x\cos y) - (\sin u - x\cos v)dy\right]}{(x\cos v+y\cos u)}$$

Answer 2

To expand on my comments, (2) rearranges to $-y\cos u\mathrm{d}u+x\cos v\mathrm{d}v=-\sin v\mathrm{d}x+\sin u\mathrm{d}y$. Then the numbered equations can be restated as$$\left(\begin{array}{cc} 1 & 1\\ -y\cos u & x\cos v \end{array}\right)\left(\begin{array}{c} \mathrm{d}u\\ \mathrm{d}v \end{array}\right)=\left(\begin{array}{cc} 1 & 1\\ -\sin v & \sin u \end{array}\right)\left(\begin{array}{c} \mathrm{d}x\\ \mathrm{d}y \end{array}\right).$$Hence$$\left(\begin{array}{c} \mathrm{d}u\\ \mathrm{d}v \end{array}\right)=\frac{1}{x\cos v+y\cos u}\left(\begin{array}{cc} x\cos v & -1\\ y\cos u & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 1\\ -\sin v & \sin u \end{array}\right)\left(\begin{array}{c} \mathrm{d}x\\ \mathrm{d}y \end{array}\right),$$which is your final pair of equations.

Answer 3

Let $\;u=u(x,y),\;v=v(x,y),\;$ where $\;x\;$ and $\;y\;$ are independent variables.

Then \begin{cases} u_x+v_x=1\\ y\cos u\, u_x=\sin v+x\cos v\, v_x\\ u_y+v_y=1\\ \sin u +y\cos u\,u_y =x\cos v\, v_y, \end{cases} \begin{cases} v_x=1-u_x,\;(y\cos u+x\cos v)\,u_x=\sin v+x\cos v\\ u_x=1-v_x,\;(y\cos u+x\cos v)\,v_x=-\sin v+y\cos u\\ v_y=1-u_y,\;(y\cos u+x\cos v)\,u_y=-\sin u+x\cos v\\ u_y=1-v_y,\;(y\cos u+x\cos v)\,v_y=\sin u+y\cos u. \end{cases}

Finally, $$\text du=u_x\,\text dx +u_y\,\text dy = \dfrac{\sin v+x\cos v}{y\cos u+x\cos v}\,\text dx +\dfrac{-\sin u+x\cos v}{y\cos u+x\cos v}\,\text dy,$$ $$\text dv=v_x\,\text dx +v_y\,\text dy = \dfrac{-\sin v+y\cos u}{y\cos u+x\cos v}\,\text dx +\dfrac{\sin u+y\cos u}{y\cos u+x\cos v}\,\text dy.$$