Use the integrating factor method to find the general solution of the linear first order ODExdydx−y=x2ex x\frac{dy}{dx} - y =x^2 e^x .The general solution of the ODE can be written as

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First, rewrite the ODE in standard linear form. Starting from x dy/dx − y = x^2 e^x, divide both sides by x (assuming x ≠ 0) to obtain dy/dx − (1/x) y = x e^x.

Next, identify the integrating factor μ(x) = exp(∫ −1/x dx) = exp(−ln|x|) = 1/x (valid for x ≠ 0).

Multiply every term by μ(x) = 1/x to get (1/x) dy/dx − (1/x^2) y = e^x. Observe that the left-hand side is the derivative of y/x with respect to x: d/dx (y/x) = (x dy/dx − y)/x^2, which matches after simplification.

Thus, d/dx (y/x) = e^x. Integrate both sides with respect to x: y/x = ∫ e^x dx = e^x + C, where C is the constant of integration.

Solve for y: y = x (e^x + C) = x e^x + C x.

A few clarifications: the solution is valid for x ≠ 0 due to the integrating factor step and division by x. The general solution family is y(x) = x e^x + C x, where C is an arbitrary constant. If one were to consider the case x = 0 separately, the differential equation at x = 0 would need a limit-based or piecewise treatment, but the standard general solution on intervals not containing zero is y = x e^x + C x.

Use the integrating factor method to find the general solution of the linear first order ODExdydx−y=x2ex x\frac{dy}{dx} - y =x^2 e^x .The general solution of the ODE can be written as

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