21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.
y'(t) = t lnt + 1

25–28. Two steps of Euler’s method For the following initial value problems, compute the first two approximations u1 and u2 given by Euler’s method using the given time step.

y′(t) = 2−y, y(0) = 1; Δt = 0.1

12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.

y(x) = sin y, y(−2) = 1/2

5–10. First-order linear equations Find the general solution of the following equations.

y'(x) = −y + 2

21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.

P′(t) = 0.05P(1−P/800); P(0) = 100, P(0) = 400, P(0) = 700

45–48. General first-order linear equations Consider the general first-order linear equation y'(t)+a(t)y(t)=f(t). This equation can be solved, in principle, by defining the integrating factor p(t)=exp(∫a(t)dt). Here is how the integrating factor works. Multiply both sides of the equation by p (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes

p(t)(y′(t) + a(t)y(t)) = d/dt(p(t)y(t)) = p(t)f(t).

Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor.

y′(t) + (2t)/(t² + 1)y(t) = 1 + 3t², y(1) = 4

33–42. Solving initial value problems Solve the following initial value problems.

y'(x) = 4 sec² 2x, y(0) = 8