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

##### 1 Answer

# 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=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

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

# :. 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)) #