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.
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:
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
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