Solve (x+y-1)dy - (x-y+2)dx = 0
Answers
Answer:
1) (y-2)dx-(x-y-1)dy=0(y−2)dx−(x−y−1)dy=0
\implies dy/dx=(y-2)/(x-y-1)⟹dy/dx=(y−2)/(x−y−1)
Let x=u+h;y=v+kx=u+h;y=v+k
\implies dx=du ; dy=dv⟹dx=du;dy=dv
dy/dx=[(v+k)-2]/[(u+h)-(v+k)-1)]dy/dx=[(v+k)−2]/[(u+h)−(v+k)−1)]
dv/du=[v+(k-2)]/[u-v+(h-k-1)]dv/du=[v+(k−2)]/[u−v+(h−k−1)]
Now;
k-2=0k−2=0 ;
h-k-1=0h−k−1=0
Solving these equations;
k=2; h=3k=2;h=3
\implies x=u+3; y=v+2⟹x=u+3;y=v+2
\implies dv/du= v/(u-v)⟹dv/du=v/(u−v)
Let; v/u=z \implies v=uzv/u=z⟹v=uz
\implies dv/du=z+u(dz/du)⟹dv/du=z+u(dz/du)
\implies z+u(dz/du)= z/(1-z)⟹z+u(dz/du)=z/(1−z)
\implies \int (z^{-2}-z^{-1})dz=\int du/u⟹∫(z
−2
−z
−1
)dz=∫du/u
\implies ln(uz)=-(1/z)+c⟹ln(uz)=−(1/z)+c
Substituting the values of u,v, z, h, k;u,v,z,h,k; we get;
(y-2)=ke^{-[(x-3)/(y-2)]}(y−2)=ke
−[(x−3)/(y−2)]
(Answer)
2) (x-4y-9)dx+(4x+y-2)dy=0(x−4y−9)dx+(4x+y−2)dy=0
\implies dy/dx=(4y-x+9)/(4x+y-2)⟹dy/dx=(4y−x+9)/(4x+y−2)
Let x=u+h;y=v+kx=u+h;y=v+k
\implies dx=du ; dy=dv⟹dx=du;dy=dv
dy/dx=[4(v+k)-(u+h)+9]/dy/dx=[4(v+k)−(u+h)+9]/[4(u+h)+(v+k)-2)][4(u+h)+(v+k)−2)]
dv/du=[4v-u+(4k-h+9)]/dv/du=[4v−u+(4k−h+9)]/[4u+v+(4h+k-2)][4u+v+(4h+k−2)]
Now;
4k-h+9=04k−h+9=0 ;
4h+k-2=04h+k−2=0
Solving these equations;
k=-2; h=1k=−2;h=1