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

a-
`e^x,e^(-x)`

b-
`x^(-2)`

c-
`x^3`

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\\eq 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

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)}`

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)}`