6.29) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text. Implicit functions and jacobians6.29.If U = x y, find dU/dt ifx' + y = t and(1)x² + y³ = ?Equations (1) and (2) define x and y as (implicit) functions of t. Then differentiating with respect to t, wehave5x* (dx/dt) + dy/t = 1(3)2x (dx/dt) + 3y²(dy/dt) = 2t(4)Solving Equations (3) and (4) simultaneously for dx/dt and dy/dt,115x*12t 3y5x2х ЗуЗу? - 2115х*у? — 2хdxdy2x2t10х*t - 2хdt1dt5x*115х*у? - 2х2х Зу10x*t - 2хdUThendtJU dxax dtJU dy= (3x²y)|ду dtЗу? - 2t15х*у? - 2х15x*y – 2x

Given U = x3yx5+y = t (1)x2+y3 = t2…

Answered: If U = x y, find dU/dt if x* + y = t…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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6.29) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.

Implicit functions and jacobians
6.29.
If U = x y, find dU/dt if
x' + y = t and
(1)
x² + y³ = ?
Equations (1) and (2) define x and y as (implicit) functions of t. Then differentiating with respect to t, we
have
5x* (dx/dt) + dy/t = 1
(3)
2x (dx/dt) + 3y²(dy/dt) = 2t
(4)
Solving Equations (3) and (4) simultaneously for dx/dt and dy/dt,
1
1
5x*
1
2t 3y
5x
2х Зу
Зу? - 21
15х*у? — 2х
dx
dy
2x
2t
10х*t - 2х
dt
1
dt
5x*
1
15х*у? - 2х
2х Зу
10x*t - 2х
dU
Then
dt
JU dx
ax dt
JU dy
= (3x²y)|
ду dt
Зу? - 2t
15х*у? - 2х
15x*y – 2x

Expert Solution

Step 1

Given

$\begin{array}{rcl} U & = & x^{3} y \\ x^{5} + y & = & t (1) \\ x^{2} + y^{3} & = & t^{2} (2) \end{array}$

Then

$\frac{\partial U}{\partial x} = 3 x^{2} y \frac{\partial U}{\partial y} = x^{3}$

Differentiate equation (1) and equation (2) with respect to $t$ , we get

$\begin{array}{rcl} 5 x^{4} (\frac{d x}{d t}) + \frac{d y}{d t} & = & 1 (3) \\ 2 x (\frac{d x}{d t}) + 3 y^{2} (\frac{d y}{d t}) & = & 2 t (4) \end{array}$

From equation (3) and equation (4) , we will find value of $\frac{d x}{d t}$ and $\frac{d y}{d t}$

We will use Cramer's rule to determine this

Equation (3) and equation (4) in the matrix form is given by $A x = B$ where

$[\begin{matrix} 5 x^{4} & 1 \\ 2 x & 3 y^{2} \end{matrix}] [\begin{matrix} \frac{d x}{d t} \\ \frac{d y}{d t} \end{matrix}] = [\begin{matrix} 1 \\ 2 t \end{matrix}]$

Then

$\begin{array}{rcl} |A| & = & |\begin{matrix} 5 x^{4} & 1 \\ 2 x & 3 y^{2} \end{matrix}| \\ = & (5 x^{4}) (3 y^{2}) - (1) (2 x) \\ = & 15 x^{4} y^{2} - 2 x \end{array}$

Then $A_{X}$ is formed by replacing the first column of matrix $A$ with the matrix $B$ and $A_{Y}$ is formed by replacing the second column of matrix $A$ with the matrix $B$

Thus

$\begin{array}{rcl} |A_{X}| & = & |\begin{matrix} 1 & 1 \\ 2 t & 3 y^{2} \end{matrix}| \\ = & (1) (3 y^{2}) - (1) (2 t) \\ = & 3 y^{2} - 2 t \\ |A_{Y}| & = & |\begin{matrix} 5 x^{4} & 1 \\ 2 x & 2 t \end{matrix}| \\ = & (5 x^{4}) (2 t) - (1) (2 x) \\ = & 10 x^{4} t - 2 x \end{array}$

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