Verified step by step guidance
07:33
05:03 07:3312–16. 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
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.
(t² + 1)y′(t) + 2ty = 3t², y(2) = 8
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
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
5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.
u'(x) = e²ˣ⁻ᵘ
33–42. Solving initial value problems Solve the following initial value problems.
y'(x) = 4 sec² 2x, y(0) = 8
