Mathematics, 12.08.2019 16:10 kaitlynmoore42

Match solutions and differential equations. note: each equation may have more than one solution. select all that apply. (a) 7y'' − 7y = 0 y = ex y = x3 y = e−x y = x−2 (b) 7x2y'' + 14xy' − 14y = 0 y = ex y = x3 y = e−x y = x−2 (c) 7x2y'' − 42y = 0 y = ex y = x3 y = e−x

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Answer from: lizdeleon248

a- $e^x,e^{-x}$

b- $x^{-2}$

c- x^3

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

D^2-1=0

(D-1)(D+1)=0

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. 7x^2y''+14xy'-14 y=0

Y= y=x^3

y=3x^2

y''=6x

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, x^3 is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. 7x^2y''-42y=0

y=x^3

y'=3x^2

y''=6x

Substitute the values in the differential equation

42x^3-42x^3=0

Hence, x^3 is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- x^3

Answer from: NaVaThEBeAsT

a) $\boxed{7y''-7y=0\to y=e^{-x}}$ .

b) $\boxed{7x^2y''+14xy'-14y=0\to y=x^{-2}}$

c) $\boxed{7x^2y''-42y=0\to y=x^{-2}}$

$\boxed{7x^2y''-42y=0\to y=x^{3}}$

Step-by-step explanation:

a) The given differential equation is: 7y''-7y=0 .

The characteristic equation is: 7m^2-7=0

This implies that: $m=-1\:or\:\:m=1$

The auxiliary solution to this second order homogeneous differential equation is: $y=Ae^{m_1x}+Be^{m_2x}$

Therefore any equation of the $y=Ae^{x}+Be^{-x}$ where A and B are constants is a solution 7y''-7y=0 .

$\boxed{7y''-7y=0\to y=e^x}$ .

$\boxed{7y''-7y=0\to y=e^{-x}}$ .

b) The given differential equation is: 7x^2y''+14xy'-14y=0

The characteristic equation is given by: am(m-1)+bm+c=0 , where a=7, b=14 and c=-14

This implies that:

7m(m-1)+14m-14=0

$\implies m=-2\:or\:1$

The auxiliary equation is of the form: $y=Ax^{m_1}+Bx^{m_2}$ where A and B are constants.

Hence any equation of the form: $y=Ax^{-2}+Bx$ is a solution to

7x^2y''+14xy'-14y=0

$\boxed{7x^2y''+14xy'-14y=0\to y=x^{-2}}$

c) The given differential equation is: 7x^2y''-42y=0

The characteristic equation is given by: am(m-1)+bm+c=0 , where a=7, b=0 and c=-42

This implies that:

7m(m-1)-42=0

$\implies m=-2\:or\:3$

The auxiliary equation is of the form: $y=Ax^{m_1}+Bx^{m_2}$ where A and B are constants.

Hence any equation of the form: $y=Ax^{-2}+Bx^3$ is a solution to

7x^2y''-42y=0

$\boxed{7x^2y''-42y=0\to y=x^{-2}}$

$\boxed{7x^2y''-42y=0\to y=x^{3}}$

Answer from: Quest

Iwould say false, not positive but if prices went up for both stores, technically there will be less buyers?

Answer from: Quest

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

$ab^2+ac^2=bc^2$

$ab^2+ac^2=x^2$ )

triangle abd and cba are similar,

$\frac{ab}{bc}=\frac{bd}{ab}$

$ab^2=bc\cdot bd$

$25=bc\cdot bd$ (2)

triangle cad and cba are similar,

$\frac{ac}{bc}=\frac{cd}{ac}$

$ac^2=bc\cdot dc$

$7\times 7=bc\cdot dc$

$49=bc\cdot dc$ )

using (1), (2) and (3), we get

$25+49=x^2$

$74=x^2$

hence proved.

therefore option 2 is correct.

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