The solution of the initial value problem that's given will be 2yt + t² - 3t + 4y² - y = 14.
It should be noted that the objective is to solve (2y + 2t − 3) dt + (8y + 2t − 1) dy = 0, y(−1) = 2.
(2y + 2t − 3)dt + (8y + 2t − 1)dy = 0
2ydt + 2tdt - 3dt + 8ydy + 2tdy - dy = 0
2(ydt + tdt) + (2t - 3)dt + (8y - 1)dy = 0
2d(yt) + (2t - 3)dt + (8y - 1)dy = 0
Integrating both sides will be
2yt + t² - 3t + 4y² - y = 0
Then, apply the initial condition, y(-1) = 2.
2(2)(-1) + (-1)² - 3(-1) + 4(2)² - (2) = C
-4 + 1 + 3 + 16 - 2 = C
C = 14
Therefore, the solution will be 2yt + t² - 3t + 4y² - y = 14.
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