Question

(x-4y-9)dx+(4x+y-2)

Answer 1

Let us solve the differential equation $(x-4y-9)dx+(4x+y-2)dy=0.$

Firstly, consider the system $\begin{cases}x-4y-9=0 \\ 4x+y-2=0\end{cases}$ which is equivalent to $\begin{cases}x-4y-9=0 \\ 16x+4y-8=0\end{cases}$ and to $\begin{cases}x-4y-9=0 \\ 17x-17=0\end{cases}.$ It follows that $\begin{cases} x=1 \\ y=-2\end{cases}.$

Let us use the transformation $x=u+1,\ y=v-2.$ Then we get $(u-4v)du+(4u+v)dv=0.$

Also let $u=tv,$ and hence $du=tdv+vdt.$ It follows that $(tv-4v)(tdv+vdt)+(4tv+v)dv=0.$

Since for $v=0$ we have that $y=-2$ is not a solution of the differential equation, let us divide both sides by $v:$ $(t-4)(tdv+vdt)+(4t+1)dv=0.$ The last equation is equivalent to $(t^2-4t)dv+(t-4)vdt+(4t+1)dv=0,$ and hence to $(t^2+1)dv=-(t-4)vdt.$

It follows that $-\frac{dv}{v}=\frac{(t-4)dt}{t^2+1},$ and consequently $-\int\frac{dv}{v}=\int\frac{(t-4)dt}{t^2+1}.$ We get that $-\ln|v|=\frac{1}{2}\int\frac{2tdt}{t^2+1}-4\int\frac{dt}{t^2+1}=\frac{1}{2}\ln(t^2+1)-4\arctan t+\ln |C|,$ and hence

$8\arctan t=\ln(t^2+1)+2\ln|v|+\ln |C_1|=\ln|(t^2+1)v^2C_1|.$

It follows that $(t^2+1)v^2C_1=e^{8\arctan t}.$

Since $t=\frac{u}{v}=\frac{x-1}{y+2}$ and $v=y+2,$ we have that

$((\frac{x-1}{y+2})^2+1)(y+2)^2C_1=e^{8\arctan \frac{x-1}{y+2}}.$

We conclude that the general solution is the following:

$((x-1)^2+(y+2)^2)C_1=e^{8\arctan \frac{x-1}{y+2}}.$