For u=A1+B1 x we compute ∂^2 u / ∂x^2=0=∂u / ∂y . Then ∂^2 u / ∂x^2=4 ∂u / ∂y For u=A2 e^α^2 v c

For u=A1+B1 x we compute ∂^2 u / ∂x^2=0=∂u / ∂y . Then ∂^2 u...

For $u=A_{1}+B_{1} x$ we compute $\partial^{2} u / \partial x^{2}=0=\partial u / \partial y .$ Then $\partial^{2} u / \partial x^{2}=4 \partial u / \partial y$ For $u=A_{2} e^{\alpha^{2} v} \cos 2 \alpha x+B_{2} e^{\alpha^{2} y} \sin 2 \alpha x$ we compute $$\begin{aligned} \frac{\partial u}{\partial x} &=2 \alpha A_{2} e^{\alpha^{2} y} \sinh 2 \alpha x+2 \alpha B_{2} e^{\alpha^{2} y} \cosh 2 \alpha x \\ \frac{\partial^{2} u}{\partial x^{2}} &=4 \alpha^{2} A_{2} e^{\alpha^{2} y} \cosh 2 \alpha x+4 \alpha^{2} B_{2} e^{\alpha^{2} y} \sinh 2 \alpha x \end{aligned}$$ and $$\frac{\partial u}{\partial y}=\alpha^{2} A_{2} e^{\alpha^{2} y} \cosh 2 \alpha x+\alpha^{2} B_{2} e^{\alpha^{2} y} \sinh 2 \alpha x.$$ Then $\partial^{2} u / \partial x^{2}=4 \partial u / \partial y$. For $u=A_{3} e^{-\alpha^{2} y} \cosh 2 \alpha x+B_{3} e^{-\alpha^{2} y} \sinh 2 \alpha x$ we compute $\begin{aligned} \frac{\partial u}{\partial x} &=-2 \alpha A_{3} e^{-\alpha^{2} y} \sin 2 \alpha x+2 \alpha B_{3} e^{-\alpha^{2} y} \cos 2 \alpha x \\ \frac{\partial^{2} u}{\partial x^{2}} &=-4 \alpha^{2} A_{3} e^{-\alpha^{2} y} \cos 2 \alpha x-4 \alpha^{2} B_{3} e^{-\alpha^{2} y} \sin 2 \alpha x \end{aligned}$ and $$\frac{\partial u}{\partial y}=-\alpha^{2} A_{3} e^{-\alpha^{2} y} \cos 2 \alpha x-\alpha^{2} B_{3} e^{-\alpha^{2} y} \sin 2 \alpha x.$$ Then $\partial^{2} u / \partial x^{2}=4 \partial u / \partial y.$

For $u=A_{1}+B_{1} x$ we compute $\partial^{2} u / \partial x^{2}=0=\partial u / \partial y .$ Then $\partial^{2} u / \partial x^{2}=4 \partial u / \partial y$
For $u=A_{2} e^{\alpha^{2} v} \cos 2 \alpha x+B_{2} e^{\alpha^{2} y} \sin 2 \alpha x$ we compute
$$\begin{aligned}
\frac{\partial u}{\partial x} &=2 \alpha A_{2} e^{\alpha^{2} y} \sinh 2 \alpha x+2 \alpha B_{2} e^{\alpha^{2} y} \cosh 2 \alpha x \\
\frac{\partial^{2} u}{\partial x^{2}} &=4 \alpha^{2} A_{2} e^{\alpha^{2} y} \cosh 2 \alpha x+4 \alpha^{2} B_{2} e^{\alpha^{2} y} \sinh 2 \alpha x
\end{aligned}$$
and
$$\frac{\partial u}{\partial y}=\alpha^{2} A_{2} e^{\alpha^{2} y} \cosh 2 \alpha x+\alpha^{2} B_{2} e^{\alpha^{2} y} \sinh 2 \alpha x.$$
Then $\partial^{2} u / \partial x^{2}=4 \partial u / \partial y$.
For $u=A_{3} e^{-\alpha^{2} y} \cosh 2 \alpha x+B_{3} e^{-\alpha^{2} y} \sinh 2 \alpha x$ we compute
$\begin{aligned} \frac{\partial u}{\partial x} &=-2 \alpha A_{3} e^{-\alpha^{2} y} \sin 2 \alpha x+2 \alpha B_{3} e^{-\alpha^{2} y} \cos 2 \alpha x \\ \frac{\partial^{2} u}{\partial x^{2}} &=-4 \alpha^{2} A_{3} e^{-\alpha^{2} y} \cos 2 \alpha x-4 \alpha^{2} B_{3} e^{-\alpha^{2} y} \sin 2 \alpha x \end{aligned}$
and
$$\frac{\partial u}{\partial y}=-\alpha^{2} A_{3} e^{-\alpha^{2} y} \cos 2 \alpha x-\alpha^{2} B_{3} e^{-\alpha^{2} y} \sin 2 \alpha x.$$
Then $\partial^{2} u / \partial x^{2}=4 \partial u / \partial y.$

Key Concepts

Partial Differential Equations

Partial differential equations involve functions of multiple variables and their partial derivatives. They are used to describe a wide range of phenomena in physics, engineering, and mathematics. In this context, verifying that a function satisfies a given PDE means checking that its computed partial derivatives obey the relation defined by the equation.

Partial Derivatives

Computing partial derivatives involves differentiating a function with respect to one variable while holding the other variables constant. This process is fundamental when working with functions of several variables and is essential for analyzing and verifying solutions to partial differential equations.

Chain Rule in Differentiation

The chain rule is a technique used to differentiate composite functions. When functions involve exponentials, trigonometric, or hyperbolic functions that depend on a combination of variables, the chain rule allows one to effectively compute the necessary derivatives, ensuring that each part of the composition is correctly differentiated with respect to its independent variables.

Separation of Variables

Separation of variables is a method used to solve partial differential equations by assuming that the solution can be written as a product of functions, each depending on a single variable. This approach reduces the PDE into simpler ordinary differential equations, which can then be solved individually. The examples provided illustrate how different forms of separated solutions can be verified against the given differential equation.

Verification of Candidate Solutions

Verification of candidate solutions involves substituting the assumed solution into the given partial differential equation and checking that all derivative computations satisfy the equation. This process ensures that the proposed functions are indeed valid solutions of the PDE, highlighting the importance of careful differentiation and algebraic manipulation.

Transcript

00:01 For this problem, we are told to let z equal x to the power of 4 times y cubed minus x to the power of 8 plus y to the power of 4.

00:08 And we are asked to compute the three partial derivatives shown here.

00:12 So for the first, we can see that we want to find dy of dx of dx of z.

00:19 Dx of z will be equal to 4x cubed, y cubed, minus 8x to the power of 7.

00:27 Differentiating that with respect to x we'll give us 12 x squared y cubed minus 8 times 7 that is going to be 56 so minus 56 x to the power of 6 then differentiating that with respect to y we'll get 36 x squared y squared.

00:52 Then for the second derivative, or for the second derivative at we are asked to calculate, it will be dx, d .y of partial with respect to x.

01:04 So it will be dx, d .y of 4x cubed, y cubed, minus 8x to the power of 7, which then would be equal to dx of 12x squared, y squared, which then differentiating with respect to x one moment here oh whoops that should be 12x cubed y squared then differentiating this with respect to x gives us 36x squared y squared and then lastly we want d x of dx of the partial derivative with respect to y which would be three would be three x to the power of four y squared plus 4 y cubed, then differentiating with respect to x, we'll have dx is equal to 12x cubed, y squared.

02:05 Then differentiating with respect to x again, we'll get 36x squared, y squared.

02:12 Then for part b, oops, let me get more room on the screen here, for part b, we are first asked to compute d cubed z by dx, d -y, d -y, d -y, which would be equal to partial with respect to x of the partial with respect to y of d -y.

02:33 We've already found d -y, so that is 3x to the power of 4, y squared, plus 4 y -cubed.

02:41 Partial of that with respect to y will be equal to 6x to the power of 4 plus 12 y squared...

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