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2 years ago

1/2x-1/4y=10

Mathematics

1 answer:

vovikov84 [41]2 years ago

6 0

Answer: (-112,-264)

Step-by-step explanation:

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John wants to magnify the image of a fire ant and crab spider so they appear to be the same length.How many times their actual s

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galben [10]

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2 years ago

(D^2 +2D +1)y=e^-x log(x) solve using method variation of parameter?

Ad libitum [116K]

First you'll need the complementary solutions. The characteristic equation for this ODE is

$D^2+2D+1=(D+1)^2=0\implies D=-1$

with multiplicity 2. This means two linearly independent solutions will be $y_1=e^{-x}$ and $y_2=xe^{-x}$ .

Via variation of parameters, the particular solution will take the form

$y_p=y_1u_1+y_2u_2$

where

$u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$

The Wronskian is

$W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}$

So, you have

$u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_1=-\displaystyle\int x\log x\,\mathrm dx$
$u_1=\dfrac14x^2-\dfrac12x^2\log x$

and

$u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_2=\displaystyle\int\log x\,\mathrm dx$
$u_2=x(\log x-1)$

So the solution to the ODE is

$y=C_1y_1+C_2y_2+y_1u_1+y_2u_2$
$y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}$
$y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)$

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2 years ago

A square has an area of 100 in2.<br><br> What is the perimeter of the square?

Travka [436]

40 in2.

To find the area you multiply the width by length so you would have to find what multiplied by what would equal 100. This gives you the answer of 10. To find the perimeter you add up all the sides and because a square has 4 sides you would add 10+10+10+10 which equals 40. Hope this helped!

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2 years ago