# How to find the general solution of differential equation (3y)(x)dy/dx = e^x when x=2 and y= 1 ?

Apr 4, 2018

$y = \frac{\sqrt{2 E i \left(x\right) - 2 E i \left(- 2\right) + 3}}{\sqrt{3}}$

where $E i \left(x\right)$ is the Exponential integral.

#### Explanation:

We have:

$\left(3 y\right) \left(x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x}$ and $y \left(2\right) = 1$

We can collect terms for similar variables:

$3 y \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / x$

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

$\int \setminus 3 y \setminus \mathrm{dy} = \int \setminus {e}^{x} / x \setminus \mathrm{dx}$

The LHS integral is a standard function, and the RHS is non-integrable using standard elementary function. We can apply the initial conditions and change the variable of integration, to get:

${\int}_{1}^{y} \setminus 3 t \setminus \mathrm{dt} = {\int}_{2}^{x} \setminus {e}^{t} / t \setminus \mathrm{dt}$

By applying a substitution to the RHS integral, $u = - t$ and evaluating the LHS integral then we have:

${\int}_{1}^{y} \setminus 3 t \setminus \mathrm{dt} = {\int}_{- 2}^{- x} \setminus {e}^{- u} / \left(- u\right) \setminus \mathrm{du}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - {\int}_{- 2}^{- x} \setminus {e}^{- u} / \left(u\right) \setminus \mathrm{du}$ ..... [A]
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \left\{{\int}_{- 2}^{\infty} \setminus {e}^{- u} / \left(u\right) \setminus \mathrm{du} - {\int}_{- x}^{\infty} \setminus {e}^{- u} / \left(u\right) \setminus \mathrm{du}\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\int}_{- x}^{\infty} \setminus {e}^{- u} / \left(u\right) \setminus \mathrm{du} - {\int}_{- 2}^{\infty} \setminus {e}^{- u} / \left(u\right) \setminus \mathrm{du}$

And using the definition of the Exponential integral , we have:

$E i \left(x\right) : = {\int}_{- x}^{\infty} \setminus {e}^{- t} / \left(t\right) \setminus \mathrm{dt}$

So utilising this definition, and integrating [A] we have:

${\left[\frac{3 {t}^{2}}{2}\right]}_{1}^{y} = E i \left(x\right) - E i \left(- 2\right)$

$\therefore \frac{3}{2} \left({y}^{2} - 1\right) = E i \left(x\right) - E i \left(- 2\right)$

$\therefore 3 {y}^{2} - 3 = 2 E i \left(x\right) - 2 E i \left(- 2\right)$

$\therefore 3 {y}^{2} = 2 E i \left(x\right) - 2 E i \left(- 2\right) + 3$

$\therefore {y}^{2} = \frac{2 E i \left(x\right) - 2 E i \left(- 2\right) + 3}{3}$

$\therefore y = \frac{\sqrt{2 E i \left(x\right) - 2 E i \left(- 2\right) + 3}}{\sqrt{3}}$