Question

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

Answer 1

Answer:

a) See proof below

b)The result does not violate uniqueness of solutions.

Step-by-step explanation:

a)

Suppose $y_1(t)$ is a solution of  

$y'+p(t)y=0$

then

$y'_1+p(t)y_1=0$

If $y_2(t)$ is a solution of  

$y'+p(t)y=g(t)$

then

$y'_2+p(t)y_2=g(t)$

To show that $y_1+y_2$ is a solution of equation (2) we have to show  that

$(y_1+y_2)'+p(t)(y_1+y_2)=g(t)$

But

$(y_1+y_2)'+p(t)(y_1+y_2)=(y'_1+y'_2)+(p(t)y_1+p(t)y_2)=(y'_1+p(t)y_1)+(y'_2+p(t)y_2)=0+g(t)=g(t)$

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.