Let y = y1(t) be a solution of y' + p(t)y = 0 (1)

and let y = y2(t) be a solution of y' + p(t)y = g(t) (2)

a) show that y = y1(t) + y2(t) is also a solution of equation (2).

b) suppose that p(t), g(t) in the above problem are continuous for all t. do the results of the above problem violate uniqueness of solutions? why or why not? explain you reasoning.

answer both parts a and b of this problem completely. i will rate highly for a complete answer that is done in a timely manner that is correct. you.

a) See proof below

b)The result does not violate uniqueness of solutions.

Step-by-step explanation:

a)

Suppose is a solution of  

then

If is a solution of  

then

To show that is a solution of equation (2) we have to show  that

But

which is what we wanted to show.

b)

The results of the above problem does not violate uniqueness of solutions, because the solution is unique for each specific given initial condition, but there are no initial conditions stated.

12m

step-by-step explanation:

square root means have of a number so have of 3 is 1.5 then multiply it by 8 and you get your answer!

98%

step-by-step explanation: