<h2>Answer:</h2><h3>I do not see any question content, Try re-wording your problem and re-post your question. Once that is done I would love to help. </h3>
Answer:
Step-by-step explanation:
y = sin(t^2)
y' = 2tcos(t^2)
y'' = 2cos(t^2) - 4t^2sin(t^2)
so the equation become
2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0
when t=0, above eqution is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.
Number 1s answer would be 2. and number 2s answer would be 2
3 (3 + x) = 5x - 25
9 + 3x = 5x - 25
Add 25 to each side
34 + 3x = 5x
Subtract 3x from each side
34 = 2x
Divide each side by 2
17 = x
