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Activity 7.4.4

Avg difficulty 2.5

Solve each of the following differential equations or initial value problems.

  1. dydt−(2−t)y=2−t\frac{dy}{dt} - (2-t) y = 2-t

  2. 1tdydt=et2−2y\frac{1}{t}\frac{dy}{dt} = e^{t^2-2y}

  3. y′=2y+2y' = 2y+2,  y(0)=2y(0)=2

  4. y′=2y2y' = 2y^2,  y(−1)=2y(-1) = 2

  5. dydt=−2tyt2+1\frac{dy}{dt} = \frac{-2ty}{t^2 + 1},  y(0)=4y(0) = 4

Boelkins, Active Calculus, §7.4 Separable differential equations, Activity 7.4.4

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.