Example. Find the general solution of the hyperbolic equation

$$
u_{x x}+4 u_{x y}-5 u_{y y}=0
$$
## Solution.

1. We reduce equation (5) to its canonical form. We obtain

$$
u_{\xi \eta}=0
$$

where $\xi=y-5 x, \eta=y+x$.

2. Integrating equation (6) with respect to $\eta$, we get

$$
u_{\xi}=C_{0}(\xi)
$$

Integrating equation (7) with respect to $\xi$, we get

$$
u(\xi, \eta)=\int C_{0}(\xi) d \xi+C_{2}(\eta) \equiv C_{1}(\xi)+C_{2}(\eta)
$$

3. Substituting $\xi=y-5 x$ and $\eta=y+x$ into (8), we obtain the general solution of equation (5)

$$
u(x, y)=C_{1}(y-5 x)+C_{2}(y+x)
$$

We perform a check by substituting $u(x, y)$ into equation (5).

Answer. The general solution of equation (5) is

$$
u(x, y)=C_{1}(y-5 x)+C_{2}(y+x)
$$

Conditions of the Problem. Find the general solution of the hyperbolic equation.

1. $u_{x x}+2 u_{x y}-5 u_{y y}=0$.
2. $u_{x x}-2 u_{x y}-3 u_{y y}=0$.
3. $u_{x x}+4 u_{x y}+3 u_{y y}=0$.
4. $u_{x x}-4 u_{x y}-12 u_{y y}=0$.
5. $u_{x x}+6 u_{x y}+5 u_{y y}=0$.
6. $u_{x x}-6 u_{x y}+8 u_{y y}=0$.
7. $u_{x x}+4 u_{x y}-21 u_{y y}=0$.
8. $u_{x x}-4 u_{x y}-32 u_{y y}=0$.
9. $u_{x x}-2 u_{x y}-24 u_{y y}=0$.
10. $u_{x x}+2 u_{x y}-8 u_{y y}=0$.

Answers.

1. $u=C_{1}(y-5 x)+C_{2}(y+3 x)$.
2. $u=C_{1}(y-x)+C_{2}(y+3 x)$.
3. $u=C_{1}(y-3 x)+C_{2}(y-x)$.
4. $u=C_{1}(y-2 x)+C_{2}(y+6 x)$.
5. $u=C_{1}(y-5 x)+C_{2}(y-x)$.
6. $u=C_{1}(y+2 x)+C_{2}(y+4 x)$.
7. $u=C_{1}(y-7 x)+C_{2}(y+3 x)$.
8. $u=C_{1}(y-4 x)+C_{2}(y+8 x)$.
9. $u=C_{1}(y-4 x)+C_{2}(y+6 x)$.
10. $u=C_{1}(y-4 x)+C_{2}(y+2 x)$.

## 5.3. General Solution of a Parabolic Equation

Problem Statement. Find the general solution of the parabolic equation

$$
u_{x x}+2 b u_{x y}+b^{2} u_{y y}+a u_{x}-a b u_{y}=0
$$

## PLAN OF SOLUTION.

1. We reduce equation (1) to its canonical form. We obtain

$$
b^{2} u_{\eta \eta}+a b u_{\eta}=0
$$

where $\xi=y-b x, \eta=y$ (Problem 5.1).

2. To find the general solution of equation (2), we form the corresponding characteristic equation $b^{2} k^{2}+a b k=0$ and find its roots $k_{1}=0$ and $k_{2}=-a / b$.

Therefore, the general solution of equation (2) can be written as

$$
u(\xi, \eta)=C_{1}(\xi)+C_{2}(\xi) e^{-a \eta / b}
$$

where $C_{1}(\xi)$ and $C_{2}(\xi)$ are arbitrary twice differentiable functions.

3. Substituting $\xi=y-b x$ and $\eta=y$ into (3), we obtain the general solution of equation (1):

$$
u(x, y)=C_{1}(y-b x)+C_{2}(y-b x) e^{-a y / b}
$$

We perform a check and write the answer.