If y = A(x) e^{int P dx} is a solution of fra | vedclass

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(D) Given the linear differential equation: $\frac{dy}{dx} + P(x) y = Q(x)$.The integrating factor $(IF)$ is $e^{\int P(x) dx}$.The general solution i

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$Q(x) e^{\int P dx}$

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(D) Given the linear differential equation: $\frac{dy}{dx} + P(x) y = Q(x)$.The integrating factor $(IF)$ is $e^{\int P(x) dx}$.The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.Given $y = A(x) e^{\int P(x) dx}$,we substitute this into the differential equation:$\frac{d}{dx} [A(x) e^{\int P(x) dx}] + P(x) [A(x) e^{\int P(x) dx}] = Q(x)$.Using the product rule:$A'(x) e^{\int P(x) dx} + A(x) P(x) e^{\int P(x) dx} + P(x) A(x) e^{\int P(x) dx} = Q(x)$.$A'(x) e^{\int P(x) dx} + 2 A(x) P(x) e^{\int P(x) dx} = Q(x)$.Wait,let us use the standard form $y = u v$ where $v = e^{\int P dx}$.Then $y' = u' v + u v' = u' e^{\int P dx} + u P e^{\int P dx}$.Substituting into $\frac{dy}{dx} + P y = Q$:$u' e^{\int P dx} + u P e^{\int P dx} + P u e^{\int P dx} = Q$ is incorrect.Correct approach: $y = A(x) e^{\int P dx}$.$\frac{dy}{dx} = A'(x) e^{\int P dx} + A(x) P(x) e^{\int P dx}$.Substitute into $\frac{dy}{dx} + P y = Q$:$A'(x) e^{\int P dx} + A(x) P(x) e^{\int P dx} + P(x) A(x) e^{\int P dx} = Q(x)$ is not the standard method.Actually,the solution is $y e^{\int P dx} = \int Q e^{\int P dx} dx + C$.Comparing $y = A(x) e^{\int P dx}$,we have $A(x) = \int Q e^{\int P dx} dx$.Therefore,$A'(x) = \frac{d}{dx} [\int Q e^{\int P dx} dx] = Q(x) e^{\int P dx}$.

If $y = A(x) e^{\int P dx}$ is a solution of $\frac{dy}{dx} + P(x) y = Q(x)$,then $A'(x) =$

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