Answer to Question #204912 in Differential Equations for shun

Question

Equation in which M(x,y) and N(x,y) are Linear but not Homogeneous

Find the general solution of the following:

1.) (2x+3y-1)dx-4(x+1)dy=0

2.) (x-4y-3)dx-(x-6y-5)dy=0

Answer 1

Let us find the general solution of the following:

1) $(2x+3y-1)dx-4(x+1)dy=0$

Let $x=u-1,\ y =v+1.$ Then $(2(u-1)+3(v+1)-1)du-4udv=0.$

It follows that $(2u+3v)du-4udv=0.$

Let $u=tv,$ then $du=tdv+vdt.$ Then $(2tv+3v)(tdv+vdt)-4tvdv=0.$

It follows that $(2t+3)(tdv+vdt)-4tdv=0,$ and hence $(2t^2+3t)dv+(2t+3)vdt-4tdv=0$ which is equivalent to $(2t^2-t)dv+(2t+3)vdt=0.$ Taking into account that $v=0$, and thus $y=1$ is not a solution, we have that $\frac{dv}{v}=-\frac{2t+3}{2t^2-t}dt=\frac{-2t-3}{(2t-1)t}dt=(\frac{3}{t}-\frac{8}{2t-1})dt,$ and thus $\int\frac{dv}{v}=\int(\frac{3}{t}-\frac{8}{2t-1})dt.$ It follows that $\ln v =3\ln t -4\ln|2t-1|+ln|C|,$ and hence $v=\frac{Ct^3}{(2t-1)^4}.$ Since $t=\frac{u}{v}=\frac{x+1}{y-1},$ we have that $y-1=\frac{C(\frac{x+1}{y-1})^3}{(2(\frac{x+1}{y-1})-1)^4}=\frac{C(x+1)^3(y-1)}{(2x-y+3)^4}$. Since $x=-1$ is a solution, the general solution we can write in the form:

$(x+1)^3=C(2x-y+3)^4.$

2) $(x-4y-3)dx-(x-6y-5)dy=0$

Let $x=u-1,\ y=v-1.$ Then $(u-1-4(v-1)-3)du-(u-1-6(v-1)-5)dv=0.$

It follows that $(u-4v)du-(u-6v)dv=0.$

Let $u=tv,$ then $du=tdv+vdt.$ Then $(tv-4v)(tdv+vdt)-(tv-6v)dv=0.$ Taking into account that $v=0$, and thus $y=-1$ is not a solution, we have that $(t-4)(tdv+vdt)-(t-6)dv=0.$ It follows that $(t^2-4t)dv+(t-4)vdt-(t-6)dv=0,$

and hence $(t^2-5t+6)dv+(t-4)vdt=0.$ We have that $\frac{dv}{v}=-\frac{t-4}{t^2-5t+6}dt=\frac{-t+4}{(t-2)(t-3)}dt=(\frac{1}{t-3}-\frac{2}{t-2})dt,$ and thus $\int\frac{dv}{v}=\int(\frac{1}{t-3}-\frac{2}{t-2})dt.$ Then $\ln |v|=\ln|t-3|-2\ln|t-2|+\ln|C|,$ and thus $v=\frac{C(t-3)}{(t-2)^2}.$ Taking into account that $t=\frac{u}{v}=\frac{x+1}{y+1},$ we conclude that $y+1=\frac{C(\frac{x+1}{y+1}-3)}{(\frac{x+1}{y+1}-2)^2}=\frac{C(y+1)(x-3y-2)}{(x-2y-1)^2}.$ It follows that he general solution we can write in the form: $(x-2y-1)^2=C(x-3y-2).$