u=sin2x cos3y

Question

solve and find partial equations

Answer 1

Solve:

u = sin 2x cos 3y

cos 3y = u / sin2x

cos 3y = u cosec2x

Let t = 3y

The genaral solution of cos(x) = k is x = 2nπ ± α where α = cos^-1(k), where n is an integer.

t = 2nπ ± cos^-1(u cosec2x) where n Є z.

3y = 2nπ ± cos^-1(u cosec2x)

y = 1/3 { 2nπ ± cos^-1(u cosec2x) }

Therefore solution is sin2x ≠ 0,  y = 1/3 { 2nπ + cos^-1(u cosec2x) }, sin2x ≠ 0,  y = 1/3 { 2nπ - cos^-1(u cosec2x) }

Partial differential equations:

u = sin 2x cos 3y

du/dx = d/dx (sin2x cos3y)

          = cos3y d/dx (sin2x)

          = cos3y (cos2x) d/dx (2x)

          = cos3y (cos2x) (2)

          = 2 cos2x cos3y

d²u/dydx = d/dy (du/dx)

              = d/dy (2 cos2x cos3y)

              = 2 cos2x d/dy (cos3y)

              = 2 cos2x (-sin3y) d/dy(3y)

              = 2 cos2x (-sin3y) (3)

              = -6 sin3y cos2x

du/dy = d/dy (sin2x cos3y)

         = sin2x d/dy (cos3y)

         = sin2x (-sin3y) d/dy (3y)

         = sin2x (-sin3y) (3)

         = -3 sin2x sin3y

d²u/dxdy = d/dx (du/dy)

              = d/dy (-3 sin2x sin3y)

              = -3 sin3y d/dx (sin2x)

              = -3 sin3y (cos2x) d/dy(2x)

              = -3 sin3y (cos2x) (2)

              = -6 sin3y cos2x

Partial equations are du/dx = 2 cos2x cos3y,

du/dy = -3 sin2x sin3y and

d²u/dydx = d²u/dxdy = -6 sin3y cos2x.