Question

solve D.E y4dx+(x-3/4-y3x)dy=0​

Answer 1

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)]