This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 1: Rewrite the differential equation in standard linear form. The given equation is: (x) (dy)/(dx) = y + (x) Divide by (x) (which is equivalent to multiplying by (x)): (dy)/(dx) = (y)/((x)) + ((x))/((x)) (dy)/(dx) = y (x) + (x) (x) Rearrange into the standard form (dy)/(dx) + P(x)y = Q(x): (dy)/(dx) - (x) y = (x) (x) Here, P(x) = -(x) and Q(x) = (x) (x). Step 2: Calculate the integrating factor I(x). The integrating factor is given by I(x) = e^ P(x) dx. I(x) = e^ -(x) dx I(x) = e^-(x) Step 3: Multiply the standard form equation by the integrating factor. e^-(x) ( (dy)/(dx) - (x) y ) = e^-(x) ((x) (x)) The left side of the equation is the derivative of the product y · I(x): (d)/(dx) (y e^-(x)) = (x) (x) e^-(x) Step 4: Integrate both sides with respect to x. (d)/(dx) (y e^-(x)) dx = (x) (x) e^-(x) dx y e^-(x) = (x) (x) e^-(x) dx To evaluate the integral on the right side, use a substitution. Let u = (x), so du = (x) dx. The integral becomes: u e^-u du This integral requires integration by parts, v dw = vw - w dv. Let v = u and dw = e^-u du. Then dv = du and w = -e^-u. u e^-u du = u(-e^-u) - (-e^-u) du = -u e^-u + e^-u du = -u e^-u - e^-u + C Factor out -e^-u: = -e^-u(u+1) + C Substitute back u = (x): = -e^-(x)((x)+1) + C Step 5: Substitute the result of the integral back into the equation and solve for y. y e^-(x) = -e^-(x)((x)+1) + C Divide both sides by e^-(x): y = -((x)+1) + C e^(x) y = C e^(x) - (x) - 1 The final answer is y = C e^(x) - (x) - 1.