8 An Introduction to Partial Differential Equations with MATLAB©R
and our general solution is
u=ysinx+f 1 (x)+g(y).Finally, sincef,f 1 andgare arbitrary, we drop the subscript:
u=ysinx+f(x)+g(y),wherefandgare arbitrary functions.
Example 5Find the general solution ofux+2u=y. Remember that we
would solve theODE dydx+2y= 5 by using the integrating factore^2 x.We
may do the same here:
ux+2u=y⇒∂
∂x(e^2 xu)=ye^2 x⇒ e^2 xu=1
2
ye^2 x+f(y)or
u=1
2
y+e−^2 xf(y)for arbitraryf.
Example 6Find the general solution ofuxxz=x+y−z.Wehave
uxx=xz+yz−z^2
2
+f(x, y).Then,
ux=x^2 z
2+xyz−xz^2
2+f 1 (x, y)+g(y,z)and
u=x^3 z
6+
x^2 yz
2−
x^2 z^2
4+f 2 (x, y)+xg(y,z)+h(y,z).Example 7There aremanyother types of PDEs that we may solve at this
point. For example, the PDEuyy+u= 0 looks like the ODEy′′+y=0.
Since the latter has solutiony=c 1 cosx+c 2 sinx, the former will have general
solution
u=f(x)cosy+g(x)siny
for arbitraryf andg. Similarly, the PDEu^2 ux=xwill behave like the
separable ODEy^2 dydx=x, which has solutiony=^3
√
3
2 x(^2) +c. Therefore, the
PDE’s solution is
u=
3
√
3
2
x^2 +f(y)for arbitraryf. (However, we must remember that, while we may multiply
and divide by the differentialsdxanddyin the ODE, this generally isnot
true of the “numerator” and “denominator” in thepartialderivative∂u∂x.)