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: 2u_{x} + u_{y} = 0.

**Solution**

2u_{x} + u_{y} = 0 ——-1

Let u = XY be a solution.

Then u must satisfy the given PDE.

u_{x} = X’Y ———2

u_{y} = 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, .

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

And

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

Hence;

NB:

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. u_{x} -5u_{y} = 0

b. u_{x} + u_{y} – 2u = 0

You can access the solution using this link https://dollarupload.com/dl/164e8