Math Problem Statement

 solve pde d^2-7dd^1+12d^12 = e^x-y it is a question from 2nd and higher order pde

Solution

Let's first analyze the given equation, which seems to be a second-order partial differential equation (PDE):

2x27x+122y2=exy\frac{\partial^2}{\partial x^2} - 7 \frac{\partial}{\partial x} + 12 \frac{\partial^2}{\partial y^2} = e^{x-y}

Here’s the breakdown:

  1. Left-hand side terms: The terms involve partial derivatives:

    • 2x2\frac{\partial^2}{\partial x^2}: Second derivative with respect to xx.
    • x\frac{\partial}{\partial x}: First derivative with respect to xx.
    • 2y2\frac{\partial^2}{\partial y^2}: Second derivative with respect to yy.
  2. Right-hand side: It is simply an exponential function of xx and yy, namely exye^{x-y}.

Solution Strategy

To solve this PDE, we might consider separation of variables or use specific methods like the method of undetermined coefficients or trial solutions for specific forms (e.g., exponential functions). Given that the right-hand side is exye^{x-y}, we might look for a solution of the form u(x,y)=Aexyu(x, y) = A e^{x-y}.

Let’s proceed by testing this form of solution.

Step 1: Substituting the trial solution

Let u(x,y)=Aexyu(x, y) = A e^{x-y}, where AA is a constant to be determined.

Now, we calculate the required derivatives:

  1. First derivatives: ux=Aexy\frac{\partial u}{\partial x} = A e^{x-y} uy=Aexy\frac{\partial u}{\partial y} = -A e^{x-y}

  2. Second derivatives: 2ux2=Aexy\frac{\partial^2 u}{\partial x^2} = A e^{x-y} 2uy2=Aexy\frac{\partial^2 u}{\partial y^2} = A e^{x-y}

Step 2: Substituting into the PDE

Now, substitute these derivatives into the original PDE:

2ux27ux+122uy2=exy\frac{\partial^2 u}{\partial x^2} - 7 \frac{\partial u}{\partial x} + 12 \frac{\partial^2 u}{\partial y^2} = e^{x-y}

This becomes:

Aexy7Aexy+12Aexy=exyA e^{x-y} - 7 A e^{x-y} + 12 A e^{x-y} = e^{x-y}

Simplify the left-hand side:

(17+12)Aexy=exy(1 - 7 + 12) A e^{x-y} = e^{x-y}

6Aexy=exy6 A e^{x-y} = e^{x-y}

Step 3: Solving for AA

To satisfy this equation, we must have 6A=16A = 1, so:

A=16A = \frac{1}{6}

Thus, the solution to the PDE is:

u(x,y)=16exyu(x, y) = \frac{1}{6} e^{x-y}

Conclusion

The solution to the given PDE is:

u(x,y)=16exyu(x, y) = \frac{1}{6} e^{x-y}

Let me know if you need any more details or clarification!

Related Questions:

  1. What are the general methods for solving second-order linear partial differential equations?
  2. How do separation of variables work in the context of PDEs?
  3. What is the role of trial solutions in solving PDEs?
  4. How would you solve a PDE with a non-exponential right-hand side function?
  5. How can boundary conditions influence the solution of a PDE?

Tip: When solving PDEs, always check for the form of the solution based on the right-hand side. For exponential functions, trial solutions with similar forms are often very effective.

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Math Problem Analysis

Mathematical Concepts

Partial Differential Equations (PDE)
Second-order PDE
Exponential Functions

Formulas

Second-order partial derivatives
Trial solution method
Method of undetermined coefficients

Theorems

Separation of Variables
Undetermined Coefficients Theorem

Suitable Grade Level

Undergraduate (Advanced Mathematics)