Math Problem Statement

Solve z=p^2x+q^2x by charpits method

Solution

To solve the partial differential equation z=p2x+q2xz = p^2 x + q^2 x using Charpit's method, let's follow these steps:

  1. Rewrite the PDE in terms of the variables pp and qq:

    • The given PDE is: z=p2x+q2xz = p^2 x + q^2 x
    • In Charpit's method, we have p=zxp = \frac{\partial z}{\partial x} and q=zyq = \frac{\partial z}{\partial y}.
  2. Compute the Charpit Equations:

    • Charpit's equations for a first-order PDE F(x,y,z,p,q)=0F(x, y, z, p, q) = 0 are: dxFp=dyFq=dzpFp+qFq=dpFxpFz=dqFyqFz\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{dz}{p F_p + q F_q} = \frac{dp}{-F_x - p F_z} = \frac{dq}{-F_y - q F_z}
    • For our PDE, define F(x,y,z,p,q)=zp2xq2x=0F(x, y, z, p, q) = z - p^2 x - q^2 x = 0.
  3. Calculate Partial Derivatives of FF:

    • Fp=2pxF_p = -2px
    • Fq=2qxF_q = -2qx
    • Fx=p2q2F_x = -p^2 - q^2
    • Fy=0F_y = 0
    • Fz=1F_z = 1
  4. Substitute into Charpit’s Equations:

    • The system becomes: dx2px=dy2qx=dzp(2px)+q(2qx)=dpp2+q2=dq0\frac{dx}{-2px} = \frac{dy}{-2qx} = \frac{dz}{p(-2px) + q(-2qx)} = \frac{dp}{p^2 + q^2} = \frac{dq}{0}
  5. Solve the System of Equations:

    • From dq0\frac{dq}{0}, we find that qq is constant.
    • From dx2px=dy2qx\frac{dx}{-2px} = \frac{dy}{-2qx},

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Math Problem Analysis

Mathematical Concepts

Partial Differential Equations
Charpit's Method
First-Order PDEs

Formulas

Charpit's equations: dx/F_p = dy/F_q = dz/(p F_p + q F_q) = dp/(-F_x - p F_z) = dq/(-F_y - q F_z)

Theorems

Charpit's Method for Solving First-Order Partial Differential Equations

Suitable Grade Level

Undergraduate (Math or Engineering)