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Solve the following differential equation by using integrating factors. You may assume that x > -3.

(x + 3)y' = 9x + y

User Shyim
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Final answer:

To solve the differential equation (x + 3)y' = 9x + y using integrating factors, we write the equation in the form y' + P(x)y = Q(x), find the integrating factor I(x), use it to transform the equation, and then solve for the general solution y(x) = (x + 3)∫(Q(x)/|x + 3|)dx + C.

Step-by-step explanation:

To solve the differential equation (x + 3)y' = 9x + y using integrating factors, we begin by writing the equation in the form y' + P(x)y = Q(x), where P(x) = 1/(x + 3) and Q(x) = 9x + y/(x + 3). The integrating factor is then given by the formula I(x) = e^(∫P(x)dx) = e^(ln|x + 3|) = |x + 3|. Multiplying the original equation by the integrating factor, we have |x + 3|y' + |x + 3|P(x)y = |x + 3|Q(x). Now, the left side of the equation is a product rule derivative, so we can write it as d(|x + 3|y)/dx = |x + 3|Q(x). Integrating both sides with respect to x and adding the constant of integration, we obtain the general solution y(x) = (x + 3)∫(Q(x)/|x + 3|)dx + C, where C is the constant of integration.

User Jonathan Protzenko
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