Solving First Order PDEs by Separation of variables

Recall from ODE, that a differential equation of the form

M(y)dx + N(x)dy = 0,

can be solved using separation of variables .

i.e dx/N(x) = -dy/M(y)

Similarly most first order PDEs can also be solved by separation of variable.

Example  1.

Obtain the general solution of:  2ux + uy = 0.


2ux + uy = 0   ——-1

Let     u = XY   be a solution.

Then u must satisfy the given PDE.

ux = X’Y    ———2

uy = XY’   ——–3

Substituting  eqn 2  and eqn 3  into  eqn 1, we obtain:

2X’Y + XY’ = 0

Dividing through by XY, and separating the variables we obtain:



Since the two quantities can only be equal if the two rates of change are equal to a given constant say, image.

Hence, we have transformed the given PDE to two sets of ODEs.





Applying the laws of indices on the RHS, we obtain;






When solving First order, linear homogeneous PDEs by separation of variables with n-partial derivatives you need to obtain n-different ODEs. Moreover when the terms are more than two, then you must reduce the number of terms to two by equating to different constants.

Try the following:

Find the general solution of the following PDEs.

a. ux -5uy = 0

b. ux + uy – 2u = 0  

You can access the solution using this link


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