Solving linear PDEs

PDE

General solution for a linear PDE:

$Pp+Qq=R$

$$ \begin{align*} F(u,v)=0\\ u(x,y,z)=c_{1}; v(x,y,z)=c_{2}\\ \text{constants found by auxilary eq:}\\ \frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R} \end{align*} $$

Cases #

When we can find two cs #

One of the variables is either absent or cancels out from the set of auxilary equation

When we can find only one c #

We use the $c_1$ found and substitute in another relation and solve and replace back $c_1$ by its function in order to find $c_2$.

Lagrangian multipliers #

$$ \begin{align*} \frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}=\frac{P_1dx+Q_1dy+R_1dz}{P_1P+Q_1Q+R_1R} \end{align*} $$

• if the denominator is 0, we can just integrate the numerator to find a constant.
• or we can choose $P_1,Q_1,R_1$ such that the numerator is an exact differential of the denominator