Question

What is the particular solution of the differential equation `x^2 dy/dx = 4e^y` with `x=2` when `y=0`?

Answer 1

The differential equation:

`x^2dy/dx =4e^y`

is separable:

`e^-ydy = 4 dx/x^2`

`int e^-ydy = 4 int dx/x^2`

`e^-y = 4/x+c`

`y=-ln ((4+cx)/x)`

`y=ln (x/(4+cx))`

The specific solution passing through `(2,0)` is determined by posing:

`0 = ln(2/(4+2c))`

and solving for `c`:

`1=2/(4+2c)`

`4+2c = 2`

`2c = -2`

`c = -1`

so that:

`y(x) = ln (x/(4-x))`

Answer 2

`y = -ln( (4-x)/x )`

Explanation:

We have:

`x^2 dy/dx = 4e^y`

This is a First Order separable equation, so we can write as:

`1/e^(y) dy/dx = 4/x^2`

and then we can "seperate the variables" to give:

`int \ e^(-y) \ dy = int \ 4x^(-2) \ dx`

Both integrals are standard ones, so we can now integrate to get:

`- e^(-y) = (4x^(-1))/(-1) + C`

`:. - e^(-y) = -4/x + C`

We now use the initial conditions `x=2` when `y=0`:

`- e^0 = -4/2 + C => -1 = -2 + C => 1`

So we get the Particular Solution:

`- e^(-y) = -4/x - 1`

`:. e^(-y) = 4/x -1`

`:. e^(-y) = (4-x)/x`

Taking logarithms, we have:

`-y = ln( (4-x)/x )`

Hence:

`y = -ln( (4-x)/x )`