Answer to Question #143649 in Differential Equations for Nikhil Singh

Question

Using the separation of variable technique, solve: ∂u/∂x+ ∂u/∂y= 2(x+y)u

Answer 1

By the separation of variables technique we write $u(x,y) = \chi(x)\phi(y)$ therefore we can rewrite or equation as follows :

$\chi'(x)\phi(y) + \chi(x)\phi'(y) = 2(x+y)\chi(x)\phi(y)$

Now we will divide both parts by $u(x,y)=\chi(x)\phi(y)$ and transfer all functions of a variable x to the left, and these of y to the right:

$\frac{\chi'(x)}{\chi(x)} + \frac{\phi'(y)}{\phi(y)} = 2(x+y)$

$\frac{\chi'(x)}{\chi(x)}-2x = 2y-\frac{\phi'(y)}{\phi(y)}$

On the left we have a function of x only and on the right of y only. Therefore both parts of this equality must be a constant (as x,y are independent) that we will call $\lambda$ . Now let's solve the equations for $\phi(y), \chi(x)$ :

$\frac{\chi'(x)}{\chi(x)}-2x=\lambda$

$\chi'(x) = (2x+\lambda)\chi(x)$

$\chi(x) = A_\lambda e^{x^2+\lambda x}$

$2y-\frac{\phi'(y)}{\phi(y)}=\lambda$

$\phi'(y) = (2y-\lambda)\phi(y)$

$\phi(y)=B_\lambda e^{y^2-\lambda y}$

$u_\lambda(x,y)=\phi_\lambda(y)\times\chi_\lambda(x)=C_\lambda e^{x^2+y^2+\lambda(x-y)}$

The range of possible $\lambda$ , the constants $C_\lambda$ are determined by boundary and initial conditions, the general sum can be expressed as a linear (possibly infinite) combination of $u_\lambda(x,y)$ for permitted values of $\lambda$ .