Question

Solve: z(p-q)=z^2 + (x+y^2)

Answer 1

Answer on Question #51248 - Math - Differential Equations

Solve: $z(p - q) = z^2 + (x + y^2)$

Solution:

$p$ denotes $\frac{\partial z}{\partial x'}$ and $q$ denotes $\frac{\partial z}{\partial y}$. Therefore,

$$z \frac{\partial z}{\partial x} - z \frac{\partial z}{\partial y} = z^2 + (x + y^2)$$

The general solution of this differential equation is given by

$$F[u_1(x, y, z), u_2(x, y, z)] = 0,$$

where $F(u_1, u_2)$ is continuously differentiable function, $u_1(x, y, z)$ and $u_2(x, y, z)$ are two independent first integrals of autonomous system

$$\frac{dx}{z} = -\frac{dy}{z} = \frac{dz}{z^2 + (x + y^2)}$$

Let's first consider the following equation

$$\frac{dx}{z} = -\frac{dy}{z}$$ $$d(x + y) = 0$$

This gives one of the first integrals

$$u_1(x, y, z) = x + y$$

Let's now consider the another equation

$$-\frac{dy}{z} = \frac{dz}{z^2 + (x + y^2)}$$ $$z^2 + (x + y^2) = -z z'$$

Using $u_1 = x + y$, we obtain

$$z^2 + u_1 - y + y^2 = -\frac{1}{2}(z^2)'$$

Let's first find the solution of homogenous equation:

$$z^2 = -\frac{1}{2}(z^2)'$$

Its solution is given by

$$z_p^2 = u_2 e^{-2y},$$

where $u_{2}$ is some constant.

The particular solution is given by

$$z _ {p} ^ {2} = - 1 - u _ {1} + 2 y - y ^ {2}$$

Indeed

$$z _ {p} ^ {2} + \frac {1}{2} \left(z _ {p} ^ {2}\right) ^ {\prime} = - 1 - u _ {1} + 2 y - y ^ {2} + \frac {1}{2} (2 - 2 y) = - u _ {1} + y - y ^ {2}$$

Therefore, the general solution is given by

$$z ^ {2} = u _ {2} e ^ {- 2 y} - 1 - u _ {1} + 2 y - y ^ {2}$$

which is equivalent to

$$\begin{array}{l} u _ {2} = e ^ {2 y} \left(z ^ {2} + 1 + u _ {1} - 2 y + y ^ {2}\right) = e ^ {2 y} \left(z ^ {2} + 1 + (x + y) - 2 y + y ^ {2}\right) = \\ = e ^ {2 y} \left(z ^ {2} + 1 + x - y + y ^ {2}\right) \\ \end{array}$$

This gives another first integral. Therefore, the solution of the initial equation is given by

$$F [ x + y, e ^ {2 y} (z ^ {2} + 1 + x - y + y ^ {2}) ] = 0$$

**Answer:** $F[x + y, e^{2y}(z^2 + 1 + x - y + y^2)] = 0$, where $F(u_1, u_2)$ is continuously differentiable function.

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