Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts (b) and (C), write your answer in explicit form.

a. y' = t²+7/y⁴-4y³

b. y' = (cos²y) ln t

c. (t²+t) y' + y² = ty²+, y (1) = - 1

(a) (y^5) / 5 + y^4 = (t^3) / 3 + 7t + C

(b) y = arctan (t (lnt - 1) + C)

(c) y = - 1/ln|0.09 (t + 1) ²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7) / (y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3) dy = (t^2 + 7) dt

Integrate both sides

(y^5) / 5 + y^4 = (t^3) / 3 + 7t + C

(b) dy/dt = (cos²y) lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t (lnt - 1) + C

y = arctan (t (lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y (1) = - 1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y² (t - 1)

(t² + t) / (t - 1) dy/dt = y²

Separating the variables

(t - 1) dt / (t² + t) = dy/y²

tdt / (t² + t) - dt / (t² + t) = dy/y²

dt / (t + 1) - dt / (t (t + 1)) = dy/y²

dt / (t + 1) - dt/t + dt / (t + 1) = dy/y²

Integrate both sides

ln (t + 1) - lnt + ln (t + 1) + lnC = - 1/y

2ln (t + 1) - lnt + lnC = - 1/y

ln|C (t + 1) ²/t| = - 1/y

y = - 1/ln|C (t + 1) ²/t|

Apply y (1) = - 1

-1 = ln|C (1 + 1) ²/1|

-1 = ln (4C)

4C = e^ (-1)

C = (1/4) e^ (-1) ≈ 0.09

y = - 1/ln|0.09 (t + 1) ²/t|