Question

x sec^2 y – x^2 cos y) dy = (tan y - 3x^4) dx​

Answer 1

Answer:

step 1: considering a simplified version, xsec2ydy=(tany−3x4)dxxsec2⁡ydy=(tan⁡y−3x4)dx

(tany−3x4)dx−xsec2ydy=0(tan⁡y−3x4)dx−xsec2⁡ydy=0

let M(x,y)=tany−3x4,N(x,y)=−xsec2yM(x,y)=tan⁡y−3x4,N(x,y)=−xsec2⁡y

partial derivatives: My=sec2y,Nx=−sec2yMy=sec2⁡y,Nx=−sec2⁡y

My−NxN=−2xMy−NxN=−2x

let u(x)=e∫−2xdx=1x2u(x)=e∫−2xdx=1x2

considering u(x)(tany−3x4)dx−u(x)xsec2ydy=0

(1x2tany−3x2)dx−1xsec2ydy=0(1x2tan⁡y−3x2)dx−1xsec2⁡ydy=0

(−1x)tany−x3=C(−1x)tan⁡y−x3=C

step 2: let (−1x)tany−x3=u(x,y)(−1x)tan⁡y−x3=u(x,y) then

1x2tanydx−1xsec2ydy−3x2dx=uxdx+uydy

multiple by x2x2 : tanydx−xsec2ydy−3x4dx=x2uxdx+x2uydytan⁡ydx−xsec2⁡ydy−3x4dx=x2uxdx+x2uydy

x2cosydy=x2uxdx+x2uydy−x2cos⁡ydy=x2uxdx+x2uydy

−cosydy=uxdx+uydy−cos⁡ydy=uxdx+uydy

uxdx+uydy+cosydy=0uxdx+uydy+cos

du(x,y)+dsiny=0du(x,y)+dsin

u(x,y)+siny=Cu(x,y)+siny= C

so, the final result should be (−1x)tany−x3=C−siny(−1x)tan⁡y−x3=C−sin⁡y