Final answer:
If y=o(t) is a solution to y' + p(t)y = 0, then y=co(t) will also be a solution for any constant c because multiplying the solution o(t) by c does not change the 'zero' result of the original equation when substituted in.
Step-by-step explanation:
If y=o(t) is a solution to the differential equation y' + p(t)y = 0, we want to show that y = co(t) is also a solution for any constant c. A solution y = o(t) implies that its derivative with respect to t, denoted as y', satisfies the differential equation such that when we substitute y and y' into the equation, we get o'(t) + p(t)o(t) = 0.
Now consider a new function y = co(t). We need to find the derivative of this new function, which is y' = c(o'(t)), since c is a constant. Substituting these into the differential equation gives us c(o'(t)) + p(t)c(o(t)) = c(o'(t) + p(t)o(t)). Since we know that o'(t) + p(t)o(t) = 0, multiplying the entire expression by c will not change this result, and we get c(0) = 0.
Therefore, the function y = co(t) will also satisfy the given differential equation, confirming that it is indeed a solution for any value of the constant c.