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.
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