Solving a Differential Equation Using the Method of Integrating Factors | A Step-by-Step Guide

y’=(2x+3)(sinx)+(cosx)(x²+3x+5)

To solve the given differential equation y’ = (2x+3)(sinx) + (cosx)(x²+3x+5), we will use the method of integrating factors

To solve the given differential equation y’ = (2x+3)(sinx) + (cosx)(x²+3x+5), we will use the method of integrating factors.

First, let’s write the equation in the standard form: y’ – (2x+3)sinx – (x²+3x+5)cosx = 0.

Now, let the integrating factor be denoted as μ(x). We can find μ(x) by multiplying the coefficient of y’, which is 1, with the integrating factor: μ(x) = e^(∫(2x+3)sinx dx).

Next, we need to calculate ∫(2x+3)sinx dx. Applying integration by parts, let u = (2x+3) and dv = sinx dx.

Differentiating u with respect to x, we get du = 2 dx.
Integrating dv, we get v = -cosx.

Using the integration by parts formula, we have ∫(2x+3)sinx dx = (2x+3)(-cosx) – ∫(-cosx)(2) dx.
Simplifying this expression, we get ∫(2x+3)sinx dx = -(2x+3)cosx + 2∫cosx dx.

Evaluating ∫cosx dx, we find ∫cosx dx = sinx + C, where C is the constant of integration.

Plugging this result back into ∫(2x+3)sinx dx, we have ∫(2x+3)sinx dx = -(2x+3)cosx + 2sinx + 2C.

Now, substituting this expression back into the integrating factor μ(x), we have μ(x) = e^(∫(2x+3)sinx dx) = e^(-(2x+3)cosx + 2sinx + 2C) = e^{-2xcosx – 3cosx + 2sinx + 2C}.

Multiplying μ(x) to both sides of the differential equation, we get μ(x)y’ – μ(x)(2x+3)sinx – μ(x)(x²+3x+5)cosx = 0.

Rewriting the equation using the product rule for differentiation, we have (μ(x)y)’ = 0.

Integrating both sides of the equation with respect to x, we find ∫(μ(x)y)’ dx = ∫0 dx, which simplifies to μ(x)y = C, where C is the constant of integration.

Finally, solving for y, we have y = C/μ(x).

So, the general solution to the given differential equation y’ = (2x+3)(sinx) + (cosx)(x²+3x+5) is y = C/e^{-2xcosx – 3cosx + 2sinx + 2C}, where C is an arbitrary constant.

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