Question

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

Answer 1

Answer:

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 2

Answer:

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

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

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

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