Question
Use the integrating factor explained on page 47 to solve the following: Cy-Acos(x) y = B/be-4ax where a, b, A, B, C are constants. Solve the differential equation. With the aid of the appropriate integrating factor, there is a standard technique for solving the linear first-order equation dy/dx + P(x)y = Q(x) on an interval on which the coefficient functions P(x) and Q(x) arc continuous. We multiply each side in Eq. (3) by the integrating factor rho(x) = e P(x)dx. The result is e P(x)dx dy/dx + P(x)e P(x)dx y = Q(x)e P(x)dx. Because Dx [ P(x)dx] = P(x), the left-hand side is the derivative of the product , so Eq. (5) is equivalent to Dx [y(x) e P(x)dx] = Q(x)e P(x)dx. Integration of both sides of this equation gives y(x)e P(x)dx = (Q(x)e P(x)dx) dx + C. Finally, solving for y, we obtain the general solution of the linear first-order equation in (3): y(x) =e- P(x)dx[ (Q(x)e P(x)dx)dx + C]. This formula should not be memorized. In a specific problem it generally is simpler to use the method by which we developed the formula. That is, in order to solve an equation that can be written in the form in Eq. (3) with the coefficient functions P(x) and Q(x) displayed explicitly, you should attempt to carry out the following steps. Begin by calculating the integrating factor rho (x) = e P(x)dx . Then multiply both sides of the differential equation by rho(x). Next, recognize the left-hand side of the resulting equation as the derivative of a product: Dx [rho (x)y(x)] = rho (x) Q(x). Finally, integrate this equation, rho (x) y(x) = rho (x)Q(x)dx + C, then solve for y to obtain the general solution of the original differential equation.
Explanation / Answer
P(x) = -Acos(x)
Q(x) = B/b e^(-4ax)
integrating factor roh = e^(integration of P(x) dx)
= e^[-Asin(x)]
so final solution:
roh * y(x) = integration [roh*Q(x) dx] + C
or e^[-Asin(x)] * y(x) = integration [ e^[-Asin(x)]* B/b e^(-4ax)dx] + C
or e^[-Asin(x)] * y(x) = B/b* integration [ e^[-Asin(x) - 4ax] dx] + C
