Verify that y1 (t) = 1 and y2 (t) = t ^1/2 are solutions of the differential equation:

yy'' + (y') ^ 2 = 0, t > 0. (3)

Then show that for any nonzero constants c1 and c2, c1 + c2t^1/2 is not a solution of this equation.

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^ (1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y') ² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1') ² = (1) (0) + (0) ² = 0

Hence, y1 is a solution.

y2 = t^ (1/2)

y2' = (1/2) t^ (-1/2)

y2'' = (-1/4) t^ (-3/2)

So,

y2y2'' + (y2') ² = t^ (1/2) * (-1/4) t^ (-3/2) + [ (1/2) t^ (-1/2) ]² = (-1/4) t^ (-1) + (1/4) t^ (-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^ (1/2) is a solution, then y = c1 + c2t^ (1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^ (1/2)

y' = (1/2) c2t^ (-1/2)

y'' = (-1/4) c2t (-3/2)

yy'' + (y') ² = [c1 + c2t^ (1/2) ][ (-1/4) c2t (-3/2) ] + [ (1/2) c2t^ (-1/2) ]²

= (-1/4) c1c2t (-3/2) + (-1/4) (c2) ²t (-3/2) + (1/4) (c2) ²t^ (-1)

= (-1/4) c1c2t (-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.