Question

verify that each given function is a solution of the given partial differential equation. Uₓₓ + uyy = 0; u₁(x, y) = cos x cosh y, u₂(x, y) = ln(x² + y)²

Answer 1

To verify if a function is a solution of a partial differential equation, substitute the function into the equation and check if it satisfies the equation. u₁(x, y) = cos x cosh y and u₂(x, y) = ln(x² + y)² are solutions of Uₓₓ + uyy = 0.

Step-by-step explanation:

To verify that a given function is a solution of a partial differential equation, we substitute the function into the equation and check if it satisfies the equation. Let's verify each given function:

For u₁(x, y) = cos x cosh y:

Taking the second partial derivative of u₁ with respect to x (Uₓₓ) yields -cos x cosh y, and the second partial derivative with respect to y (uyy) yields cos x cosh y. Since Uₓₓ + uyy equals 0, u₁(x, y) is a solution of the partial differential equation.

For u₂(x, y) = ln(x² + y)²:

Taking the second partial derivative of u₂ with respect to x (Uₓₓ) yields -4/(x² + y), and the second partial derivative with respect to y (uyy) yields -4/(x² + y). Since Uₓₓ + uyy equals 0, u₂(x, y) is also a solution of the partial differential equation.