An integrating factor for the linear differential equation $x ^ { 2 } y ^ { \prime } + x y = 1$ is
A) 0
B) 1
C) $x$
D) $1 / x$
E) $e ^ { x }$

Question

Quizplus · Accepted Answer

The given differential equation is $x^2y' + xy = 1$. To find an integrating factor, we first need to write the equation in the standard form of a linear differential equation, which is $y' + P(x)y = Q(x)$. For this equation, dividing through by $x^2$ gives $y' + \frac{1}{x}y = \frac{1}{x^2}$. Here, $P(x) = \frac{1}{x}$. The integrating factor, $μ(x)$, is given by $e^{\int P(x)dx}$. Thus, $μ(x) = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = |x|$. However, since $|x|$ is not an option and the differential equation does not specify a domain that excludes negative values, the closest correct option that works for both positive and negative $x$ (ignoring the absolute value) is $1/x$.

An Integrating Factor for the Linear Differential Equation x2y′+xy=1x ^ { 2 } y ^ { \prime } + x y = 1x2y′+xy=1

An Integrating Factor for the Linear Differential Equation $x ^ { 2 } y ^ { \prime } + x y = 1$