Solve the differential equation xy' = y + xe^(9y/x) by making the change of variable v = y/x.

Question

Solve the differential equation
xy' = y + xe^(9y/x)
by making the change of variable v = y/x. 

 

Please help and show your work! Thanks in advance!

Answer 1

This can be kind of tricky so let's go through this step-by-step. 

 

1. Let's start by isolating y' on the left side of the equation.

 

xy'=y+xe^(9y/x)
xdy/dx = y + xe^(9y/x)
dy/dx =y/x + e^(9y/x)

 

2. Next, let's rearrange

let v = y/x
y=vx

 

If v = y/x then
y = vx and y ' = v+ xv ' (chain rule) and now substitute in y ' = (y/x)+ e^(y/x) you get
v+ xv ' = v + e^v 

so

v+ xv ' = v + e^9v

 

3. Next lets simplify our equation and make it more manageable. 

 

v+ xv ' = v + e^9v
v - v +xv'=e^(9v)
xv' = e^(9v)
xdv = e^(9v)dx

 

4. Now let's integrate!

 

dv...........dx
∫--------- =∫ ----------
..e^(9v)..........x
...................dx
∫ e^(-9v)dx =∫ ----------
....................x

 

and/or

 

v'e^-9v = 1/x
(e^-9v) ' = 1/x

-9e^(-v) = ln(x) +C 

e^(-v) = - 1/9[ln(x) + C]

Now let's substitute back for v=y/x

e^(-y/x)= - ln(x) - C

-y/x = ln[-ln(x) - C]
y=-1/9xln(9[-ln(x) -C])