a) (ye2xy+x)dx+(bxe2xy)dy=dU, then ye2xy+x=∂x∂U and bxe2xy=∂y∂U.
We have ∂x∂y∂2U=∂y∂x∂2U, that is ∂x∂(bxe2xy)=∂y∂(ye2xy+x)
∂x∂(bxe2xy)=be2xy+2bxye2xy, ∂y∂(ye2xy+x)=e2xy+2xye2xy , so be2xy+2bxye2xy=e2xy+2xye2xy, and we obtain b=1.
Hence ∂y∂U=xe2xy, so U=2e2xy+f(x).
Then ∂x∂U=ye2xy+f′(x). Since ∂x∂U=ye2xy+x, we have f′(x)=x.
Let f(x)=2x2, then U=2e2xy+x2
So the given equation we can rewrite as d(2e2xy+x2)=0, general solution of which is 2e2xy+x2=2C or e2xy+x2=C
b)General solution we will look in form y=y1+u=u+sinx
We have y′=u′+cosx, then
u′+cosx=2cosx2cos2x−sin2x+(u+sinx)2
After expanding brackets and moving cosx to the right side, we obtain
u′=2cosxu2+2usinx or dxdu=2cosxu2+2usinx
So 2cosxdu=u2dx+2usinxdx or 2cosxdu=u2dx−2ud(cosx)
Since d(UV)=VdU+UdV for every differentiable U and V, we have 2d(ucosx)=u2dx
Let ucosx=v, then 2dv=cos2xv2dx.
So we have v=0 or v22dv=cos2xdx or −2d(v1)=d(tanx) . We obtain the general solution v=0 or tanx+v2=C.
Reurning to u, we obtain u=0 or tanx+ucosx2=C. And, returning to y, we obtain y=sinx or tanx+(y−sinx)cosx2=C. Expessing y, we obtain y=sinx or y=sinx+Ccosx−sinx2
Answer: a)b=1, e2xy+x2=C, b)y=sinx or y=sinx+Ccosx−sinx2
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