How to find psi

Question

Find solution in form psi(x,y) = 0

(5xye^(xy) - 6y + 5e^(xy) - 7sin(x)cox(y)) dy/dx + (-10x + 5y^2e^(xy) - 7sin(y)cos(x))

Could someone explain the process to finding the solution? Thank you.

Answer 1

Your differential equation

[-10x + 5y^2e^(x y) - 7sin(y)cos(x)] d x + [5xye^(x y) - 6y + 5e^(x y) - 7sin(x)cox(y)] d y = 0

is an exact one

PROOF FOR EXACTENESS

Let M (x, y) = -10x + 5y^2e^(x y) - 7sin(y)cos(x)

and N( x, y ) = 5xye^(x y) - 6y + 5e^(x y) - 7sin(x)cox(y) Then

∂ M / ∂ y = 10 y e^{x y} + 5xy²e^xy − 7 cosx cosy

∂ N / ∂ x = 10 y e^{x y} + 5xy²e^xy − 7 cosx cosy

Since ∂ M / ∂ y = ∂ N / ∂ x the differential equation is an exact one . That means there exist a function

Ψ(x ,y ) = c whose total differential is the given differential equation.

In other words

d Ψ / d x = -10x + 5y^2e^(x y) - 7sin(y)cos(x)

d Ψ = [-10x + 5y^2e^(x y) - 7sin(y)cos(x) ] dx

Ψ = ∫ [-10x + 5y^2e^(x y) - 7sin(y)cos(x) ] dx

Ψ = ∫ [-10x] d x + ∫ [ 5y^2e^(x y) ] dx - ∫ [ 7sin(y)cos(x) ] dx + F(y)

Ψ = - 5 x² + 5ye^xy - 7sin(y) sin (x) + F(y)

Now we must find F(y).

But we know that ( ∂/∂y) (- 5 x² + 5ye^xy - 7sin(y) sin (x) + F(y) ) = N (x ,y) That is

( ∂/∂y) [- 5 x² + 5ye^xy - 7sin(y) sin (x) + F(y) ] = 5xye^(x y) - 6y + 5e^(x y) - 7sin(x)cox(y)

5xye^(x y) + 5e^(x y)- 7sin(x)cox(y) + F' (y ) = 5xye^(x y) - 6y + 5e^(x y) - 7sin(x)cox(y)

Or equivalently F' (y ) = - 6y Then F( y) = -3y²

Finally

Ψ(x,y) = - 5 x² + 5ye^xy - 7sin(y) sin (x) + F(y)

Ψ(x,y) = - 5 x² + 5ye^xy - 7sin(y) sin (x) - 3y²