In each of problems 5 through 10, verify that each given function is a solution of the differential equation.
y''-y=0, y1(t)=e^t , y2(t)=cosht

For First Solution:

is the solution of equation y''-y=0.

For 2nd Solution:

 is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution:

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

First order derivative:

2nd order Derivative:

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence is the solution of equation y''-y=0.

For 2nd Solution:

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

First order derivative:

2nd order Derivative:

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence  is the solution of equation y''-y=0.

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl