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Consider the following linear differential equation,

sin (x) yʹ-cos (x) y=6 cos (x)
(a) Calculate the Integrating factor
I(x)=

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Final answer:

The Integrating factor for the linear differential equation sin(x)y' - cos(x)y = 6cos(x) is calculated as I(x) = sin(x), obtained by evaluating the indefinite integral of the function -cot(x).

Step-by-step explanation:

To calculate the Integrating factor (IF) for the given linear differential equation sin(x)y' - cos(x)y = 6cos(x), we will use the standard form of a first-order linear differential equation.

The standard form of a first-order linear differential equation is y' + P(x)y = Q(x), for some functions P(x) and Q(x). The Integrating factor, I(x), is given by e^{∫ P(x)dx}. In our differential equation, P(x) = -cos(x)/sin(x). Therefore, we must evaluate the indefinite integral ∫ -cot(x)dx for our Integrating factor.

The indefinite integral of -cot(x) is ln(sin(x)). Thus, the Integrating factor is I(x) = e^{ln(sin(x))} = sin(x).

User Rehan Yousaf
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