Question: p2x + q2y= z solve by charpit method.
1. **State the problem:** Solve the first-order partial differential equation $$p2x + q2y = z$$ using Charpit's method, where $$p = \frac{\partial z}{\partial x}$$ and $$q = \frac{\partial z}{\partial y}$$.
2. **Recall Charpit's method:** For a PDE of the form $$F(x,y,z,p,q) = 0$$, Charpit's equations are:
$$\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{dz}{pF_p + qF_q} = -\frac{dp}{F_x + pF_z} = -\frac{dq}{F_y + qF_z}$$
where subscripts denote partial derivatives.
3. **Identify $$F$$ and its derivatives:**
Given $$F = p2x + q2y - z = 0$$.
Calculate partial derivatives:
- $$F_p = 2x$$
- $$F_q = 2y$$
- $$F_z = -1$$
- $$F_x = 2p$$
- $$F_y = 2q$$
4. **Write Charpit's system:**
$$\frac{dx}{2x} = \frac{dy}{2y} = \frac{dz}{p(2x) + q(2y)} = -\frac{dp}{2p - p} = -\frac{dq}{2q - q}$$
Simplify denominators in $$dp$$ and $$dq$$ terms:
- $$F_x + pF_z = 2p + p(-1) = 2p - p = p$$
- $$F_y + qF_z = 2q + q(-1) = 2q - q = q$$
So,
$$\frac{dx}{2x} = \frac{dy}{2y} = \frac{dz}{2xp + 2yq} = -\frac{dp}{p} = -\frac{dq}{q}$$
5. **Solve the first equality:**
$$\frac{dx}{2x} = \frac{dy}{2y}$$
Cross-multiplied:
$$\frac{dx}{x} = \frac{dy}{y}$$
Integrate both sides:
$$\ln|x| = \ln|y| + C_1$$
Or equivalently:
$$\frac{y}{x} = C_2$$
6. **Solve the second equality:**
$$-\frac{dp}{p} = -\frac{dq}{q}$$
Simplify:
$$\frac{dp}{p} = \frac{dq}{q}$$
Integrate:
$$\ln|p| = \ln|q| + C_3$$
Or:
$$\frac{p}{q} = C_4$$
7. **Solve the third equality:**
$$\frac{dx}{2x} = -\frac{dp}{p}$$
Rewrite:
$$\frac{dx}{2x} = -\frac{dp}{p}$$
Cross-multiplied:
$$\frac{dx}{x} = -2 \frac{dp}{p}$$
Integrate:
$$\ln|x| = -2 \ln|p| + C_5$$
Rewrite:
$$\ln|x| + 2 \ln|p| = C_5$$
$$\ln|x p^2| = C_5$$
Or:
$$x p^2 = C_6$$
8. **Similarly, solve:**
$$\frac{dy}{2y} = -\frac{dq}{q}$$
Gives:
$$y q^2 = C_7$$
9. **Summary of integrals:**
- $$\frac{y}{x} = k_1$$
- $$\frac{p}{q} = k_2$$
- $$x p^2 = k_3$$
- $$y q^2 = k_4$$
10. **Use these to find the general solution:**
Recall $$p = \frac{\partial z}{\partial x}$$ and $$q = \frac{\partial z}{\partial y}$$.
From $$\frac{p}{q} = k_2$$, write $$p = k_2 q$$.
From $$x p^2 = k_3$$ and $$y q^2 = k_4$$, substitute $$p = k_2 q$$:
$$x (k_2 q)^2 = k_3 \Rightarrow x k_2^2 q^2 = k_3$$
But $$y q^2 = k_4$$, so:
$$q^2 = \frac{k_4}{y}$$
Substitute back:
$$x k_2^2 \frac{k_4}{y} = k_3 \Rightarrow k_2^2 = \frac{k_3 y}{k_4 x}$$
11. **The solution involves implicit relations between $$x,y,z$$ and constants.**
**Final answer:** The general solution to $$p2x + q2y = z$$ by Charpit's method is implicitly given by the relations:
$$\frac{y}{x} = C_1, \quad \frac{p}{q} = C_2, \quad x p^2 = C_3, \quad y q^2 = C_4$$
where $$p = \frac{\partial z}{\partial x}$$ and $$q = \frac{\partial z}{\partial y}$$, and $$C_1, C_2, C_3, C_4$$ are constants determined by initial/boundary conditions.