00:01
In the first question of this problem, we are asked to verify that y1 of t is equal to t square and y2 of t is equal to t raised to negative 1.
00:14
Both satisfies the differential equation t square y double prime minus 2y is equal to 0.
00:23
First consider the function y1 of t is equal to t square in order to verify that this function.
00:30
Satisfies the differential equation we need to verify that the function and its derivative satisfies this equation so find the first derivative of y1 of t which is y 1 of t prime which is 2t and its second derivative y1 double prime is 2 now find y2 y2 y1 double prime minus 2 y 1 is t square times 2 minus 2 times t square which is 2 t square minus 2 t square and that is equal to 0 thus the differential equation is satisfied when y is equal to y 1 therefore y 1 of t is a solution of the differential equation this is the first required verification now consider y 2 of t which is t -raised to negative 1, find y2 prime of t, which is the first derivative of this function, which is negative 1 by t squared.
01:45
Find the second derivative of y2 of t, which is y2 double prime, and it is negative of negative 2 divided by t -cube, which is 2 divided by t -cube.
01:56
Now find t -square, y -2 double prime, minus 2 -y -2.
02:07
It is t square times 2 divided by t cube minus 2 times t inverse or t raised to negative 1.
02:17
This simplifies to 2 divided by t minus 2 divided by t since t raised to negative 1 is 1 by properties of exponents.
02:28
So that we get this is equal to 0.
02:30
Therefore when y is equal to y 2 of t, the differential equation is satisfied...