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

Nov 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln x$

#### Explanation:

If ${e}^{\frac{\mathrm{dy}}{\mathrm{dx}}} = x$
then taking the natural log of both sides:
LS$= \ln \left({e}^{\frac{\mathrm{dy}}{\mathrm{dx}}}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$
and
RS$= \ln \left(x\right)$

Nov 4, 2017

$y = x \ln x - x + C$

#### Explanation:

We have:

${e}^{\frac{\mathrm{dy}}{\mathrm{dx}}} = x$

Taking Natural logarithms we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln x$

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

$y = \int \setminus \ln \setminus \mathrm{dx} + C$

We can apply integration by Parts

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

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us

$\int \setminus \left(\ln x\right) \left(1\right) \setminus \mathrm{dx} = \left(\ln x\right) \left(x\right) - \int \setminus \left(x\right) \left(\frac{1}{x}\right) \setminus \mathrm{dx}$
$\therefore \int \setminus \ln x \setminus \mathrm{dx} = x \ln x - \int 1 \setminus \mathrm{dx}$
$\therefore \int \setminus \ln x \setminus \mathrm{dx} = x \ln x - x$

Thus we have the GS:

$y = x \ln x - x + C$