Question

What is the general solution of the differential equation `e^(dy/dx) = x`?

Answer 1

`dy/dx=ln x`

Explanation:

If `e^(dy/dx)=x`
then taking the natural log of both sides:
LS`=ln(e^(dy/dx))=dy/dx`
and
RS`=ln(x)`

Answer 2

`y = xlnx-x + C`

Explanation:

We have:

`e^(dy/dx) = x`

Taking Natural logarithms we have:

`dy/dx = ln x`

Which is a separable Differential Equation, so we can separate the variables to get:

`y = int \ ln \ dx + C`

We can apply integration by Parts

Let `{ (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1, => v,=x ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (lnx)(1) \ dx = (lnx)(x) - int \ (x)(1/x) \ dx`
`:. int \ lnx \ dx = xlnx - int 1 \ dx`
`:. int \ lnx \ dx = xlnx - x`

Thus we have the GS:

`y = xlnx-x + C`