MATH SOLVE

3 months ago

Q:
# 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? Please explain you reasoning.Please 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. Thank you.

Accepted Solution

A:

Answer:a) See proof belowb)The result does not violate uniqueness of solutions.Step-by-step explanation:a)
Suppose [tex]y_1(t)[/tex] is a solution of [tex]y'+p(t)y=0[/tex]
then
[tex]y'_1+p(t)y_1=0[/tex]
If [tex]y_2(t)[/tex] is a solution of [tex]y'+p(t)y=g(t)[/tex]
then
[tex]y'_2+p(t)y_2=g(t)[/tex]
To show that [tex]y_1+y_2[/tex] is a solution of equation (2) we have to show that[tex](y_1+y_2)'+p(t)(y_1+y_2)=g(t)[/tex]
But
[tex](y_1+y_2)'+p(t)(y_1+y_2)=(y'_1+y'_2)+(p(t)y_1+p(t)y_2)=(y'_1+p(t)y_1)+(y'_2+p(t)y_2)=0+g(t)=g(t)[/tex]
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.