Find the solution of the partial differential equation-dxazay?- 8= 0by variable separable methodMaximum file size: 100MB, maximum number of files: 1FilesYou can drag and drop files here to add them.Accepted file typesDocument files .doc.docx.epub.gdoc.odt .oth .ott pdf.rtfImage files .ai .bmp .gdraw .gif .ico .jpe .jpeg jpg .pct .pic .pict .png .svg .svgz .tif .tiffCauabu Differential Eguation

Consider the given: ∂2z∂x2−8∂2z∂y2=0. Assume z(x,y)=X(x)Y(y). Therefore, ∂2z∂x2=X''Y and…

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Math

Calculus

Find the solution of the partial differential equation- -8 0: ax2 ay? by variable separable method

Trigonometry (MindTap Course List)

8th Edition

ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter4: Graphing And Inverse Functions

Section: Chapter Questions

Problem 6GP: If your graphing calculator is capable of computing a least-squares sinusoidal regression model, use...

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Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

Question

Find the solution of the partial differential equation-
dx
az
ay?
- 8
= 0
by variable separable method
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Cauabu Differential Eguation

Expert Solution

Step 1

Consider the given:

$\frac{\partial^{2} z}{\partial x^{2}} - 8 \frac{\partial^{2} z}{\partial y^{2}} = 0$ .

Assume $z (x, y) = X (x) Y (y)$ .

Therefore, $\frac{\partial^{2} z}{\partial x^{2}} = X^{''} Y$ and $\frac{\partial^{2} z}{\partial y^{2}} = X Y^{''}$ .

So, our partial differential equation (PDE) becomes

$\begin{array}{l} X^{''} Y - 8 Y^{''} X = 0 \\ X^{''} Y = 8 Y^{''} X \\ \frac{X^{''}}{8 X} = \frac{Y^{''}}{Y} = k (Separate Variables) \end{array}$

Hence, $X^{''} - 8 k X = 0$ and $Y^{''} - k Y = 0$ .

Step 2

When $k = 0$ ,

Then equation becomes $X " = 0, Y^{''} = 0$ .

Implies

$X (x) = c_{1} x + c_{2}, Y (y) = c_{3} y + c_{4}$

Therefore,

$\begin{array}{l} z (x, y) = X (x) Y (y) \\ = (c_{1} x + c_{2}) (c_{3} y + c_{4}) \end{array}$

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