Question

solve the following differential equations
1- (y-2)dx-(x-y-1)dy=0
2- (x-4y-9)dx+(4x+y-2)dy=0

Answer 1

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

$\implies dy/dx=(y-2)/(x-y-1)$

Let $x=u+h;y=v+k$

$\implies dx=du ; dy=dv$

$dy/dx=[(v+k)-2]/[(u+h)-(v+k)-1)]$

$dv/du=[v+(k-2)]/[u-v+(h-k-1)]$

Now;

$k-2=0$ ;

$h-k-1=0$

Solving these equations;

$k=2; h=3$

$\implies x=u+3; y=v+2$

$\implies dv/du= v/(u-v)$

Let; $v/u=z \implies v=uz$

$\implies dv/du=z+u(dz/du)$

$\implies z+u(dz/du)= z/(1-z)$

$\implies \int (z^{-2}-z^{-1})dz=\int du/u$

$\implies ln(uz)=-(1/z)+c$

Substituting the values of $u,v, z, h, k;$ we get;

$(y-2)=ke^{-[(x-3)/(y-2)]}$ (Answer)

2) $(x-4y-9)dx+(4x+y-2)dy=0$

$\implies dy/dx=(4y-x+9)/(4x+y-2)$

Let $x=u+h;y=v+k$

$\implies dx=du ; dy=dv$

$dy/dx=[4(v+k)-(u+h)+9]/$$[4(u+h)+(v+k)-2)]$

$dv/du=[4v-u+(4k-h+9)]/$$[4u+v+(4h+k-2)]$

Now;

$4k-h+9=0$ ;

$4h+k-2=0$

Solving these equations;

$k=-2; h=1$

$\implies x=u+1; y=v-2$

$\implies dv/du= (4v-u)/(4u+v)$

$v/u=z \implies v=uz$

$\implies dv/du=z+u(dz/du)$

$\implies z+u(dz/du)= (4z-1)/(z+4)$

$\implies \int [(z+4)/(z^{2}+1)]dz=-\int du/u$

$\implies 1/2(ln(u^2(z^2+1))=-tan^{-1}z+c$

Substituting the values of $u,v, z, h, k$ ; we get;

$(x-1)^2+(y+2)^2=ke^{-8tan^{-1}[(y+2)/(x-1)]}$ (Answer)