7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.
u(t) = C₁t⁵ + C₂t⁻⁴ - t³; t²u''(t) - 20u(t) = 14t³

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

Logistic growth in India The population of India was 435 million in 1960 (t=0) and 487 million in 1965 (t=5). The projected population for 2050 is 1.57 billion.

e. Discuss some possible shortcomings of this model. Why might the carrying capacity be either greater than or less than the value predicted by the model?

Direction fields Consider the direction field for the equation y′=y(2−y) shown in the figure and initial conditions of the form y(0)=A.

d. For what values of A are the corresponding solutions decreasing, for t≥0

11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.

y′(t) = -3y + 9, y(0) = 4

What are the assumptions underlying the predator-prey model discussed in this section?

39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form

a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t). 

Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems. 

t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6