Question

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

Answer 1

`y= (sqrt(2Ei(x) - 2Ei(-2) + 3))/sqrt(3)`

where `Ei(x)` is the Exponential integral.

Explanation:

We have:

`(3y)(x) dy/dx = e^x` and `y(2)=1`

We can collect terms for similar variables:

`3y dy/dx = e^x/x`

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

`int \ 3y \ dy = int \ e^x/x \ 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 \ 3t \ dt = int_2^x \ e^t/t \ dt`

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

`int_1^y \ 3t \ dt = int_(-2)^(-x) \ e^(-u)/(-u) \ du`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = - int_(-2)^(-x) \ e^(-u)/(u) \ du` ..... [A]
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = - {int_(-2)^(oo) \ e^(-u)/(u) \ du - int_(-x)^(oo) \ e^(-u)/(u) \ du}`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = int_(-x)^(oo) \ e^(-u)/(u) \ du - int_(-2)^(oo) \ e^(-u)/(u) \ du`

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

`Ei(x) := int_(-x)^(oo) \ e^(-t)/(t) \ dt`

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

`[ (3t^2)/2]_1^y = Ei(x) - Ei(-2)`

`:. 3/2( y^2-1) = Ei(x) - Ei(-2)`

`:. 3y^2-3 = 2Ei(x) - 2Ei(-2)`

`:. 3y^2= 2Ei(x) - 2Ei(-2) + 3`

`:. y^2= (2Ei(x) - 2Ei(-2) + 3)/3`

Leading to the Particular Solution:

`:. y= (sqrt(2Ei(x) - 2Ei(-2) + 3))/sqrt(3)`