Verify that the given function y is a solution of the ODE for any value of C?

Verify that the given function y is a solution of the ODE for any value of C. a.) y+y'=1, y(t)=1+Ce^(-t) b.) yy'=t, y(t)=Sqrt(t^(2) + C) c.) y'+6y=0, y(t)=Ce^(-6t)

Verify that the given function y is a solution of the ODE for any value of C?
This is pretty easy, just plug the given equation into the ODE and make sure it works. I'll do (a) and (b), see if you can do (c) for yourself.





(a): y + y' = 1, y(t) = 1 + Ce^(-t).


Well, we're given y, we obviously need to find y' to make sure the ODE works. So differentiate:


y'(t) = 0 + Ce^(-t).(-1) = -Ce^(-t)


Then the LHS of the ODE is y + y' = 1 + Ce^(-t) + (-Ce^(-t))


= 1 as required.





(b) y y' = t, y(t) = √(t^2 + C)


Again we need to differentiate so we can substitute the expression for y' into the equation of the ODE:


y' = d/dt (t^2 + C)^(1/2)


= (1/2) (t^2 + C)^(-1/2) . (2t)


= t / √(t^2 + C)


So substituting into the LHS of the ODE gives


y y' = [√(t^2 + C)] . [t / √(t^2 + C)]


= t as required.





Note that (c) is quite similar to (a). As I said, I'll let you do that one for yourself.