Question

What is the particular solution of the differential equation? : `dy/dx + 1/x = e^y/x` with `y(1)=ln2`

Answer 1

`y = ln(2/(2-x))`

Explanation:

We have:

`dy/dx + 1/x = e^y/x` with `y(1)=ln2` ..... [A]

This is a First Order non-linear Differential Equation. The `e^y` term is problematic, so Let us try a substitution of the form `e^y=t` to eliminate the problematic term. Then we have,

`e^y=t => y=lnt => dy/dt = 1/t`

As we are changing variable,m we change the initial conditions accordingly:

When `y=ln2 => t=2`

Substituting into the original differential equation [A], we get:

`:. dy/dt dt/dx + 1/x = e^y/x`
`:. 1/t dt/dx + 1/x = t/x`
`:. 1/t dt/dx = (t-1)/x`
`:. 1/(t(t-1)) dt/dx = 1/x`

This has reduced the DE [A] to a First Order Separable equation, so we can "seperate the variables" to get:

`int \ 1/(t(t-1)) \ dt = int \ 1/x \ dx` ..... [B]

The LHS integral is trivial, and for the RHS integral we must decompose the integrand into partial fractions:

`1/(t(t-1)) -= A/t + B/(t-1)`
`" " = (A(t-1) + Bt)/ ( t(t-1) )`

Leading to the identity:

`1 -= A(t-1) + Bt`

We can find the coefficients `A` and `B` via direct substitution (in practice we use the Cover Up method).

Put `t=0 => 1=A(-1)+0 => A=-1`
Put `t=1 => 1=0+B \ \ \ \ \ \ \ \ \ \=> B=1`

Using this result, we can now integrate [B] to get the general solution:

`int \ 1/(t-1) -1/t \ dt = int \ 1/x \ dx`

`:. ln|t-1| - ln|t| = ln |x| + C`

Using the initial condition `t=2` when `x=1` we have:

`:. ln1 - ln2 = ln 1+ C => C=-ln2`

And so the particular solution:

`ln|t-1| - ln|t| = ln |x| -ln2`

`:. ln|(t-1)/t| = ln |x/2|`

`:. (t-1)/t = x/2`

`:. 2t-2 = tx`

`:. t(2-x)=2`
`:. t = 2/(2-x)`

Restoring the substitution, we have:

`y = lnt`
`\ \ = ln(2/(2-x))`