Solving Differential Eq.

Question

a. x(dy/dx) + y = e^x , x > 0

Answer 1

x(dy/dx) + y = e^x , x > 0 seems like an exact differential equation. So let check if it is.

For exact DE:

M(x,y) + N(x,y) y' = 0 iff ∂M/∂y = ∂N/∂x

So there exist a function f satisfying

∂f/∂x = M(x,y)

∂f/∂y = N(x,y).

So subtract e^x on both sides of the equation and change the notation of dy/dx to y':

y - e^x + xy' = 0

∴M(x,y) = y - e^x

N(x,y) = x

∂M/∂y = 1

∂N/∂x = 1. Therefore the DE is exact!

To get the complete solution, first, let's use the ∂f/∂x = M(x,y):

∂f/∂x = y - e^x

Partially integrate it with respect to x (treating y as constant):

f(x,y) = xy - e^x + h(y)

Then get the partial derivative with respect to y (treating x as constant):

∂f/∂y = x + h'(y).

Since ∂f/∂y = N(x,y), we can now have:

x + h'(y) = x

h'(y) = 0

∴h(y) = 0

Therefore the complete solution will be:

C = xy - e^x