Find the general solution of the given first-order linear differential equation. $\frac { d y } { d x } + \frac { y } { 20 x } = \sqrt [ 20 ] { x ^ { 19 } } e ^ { x }$
A) $y = \frac { e ^ { x } } { x ^ { 1 / 20 } } + C$
B) $y = \frac { e ^ { x } ( x - 1 ) + C } { x ^ { 20 } }$
C) $y = x ^ { 1 / 20 } e ^ { x } ( x - 1 ) + C$
D) $y = \frac { e ^ { x } ( x - 1 ) + C } { x ^ { 1 / 20 } }$

Question

Quizplus · Accepted Answer

The given differential equation is a first-order linear differential equation. To solve it, we first find the integrating factor (IF), which is $e^{\int P(x) dx}$, where $P(x) = \frac{1}{20x}$. The integrating factor is $e^{\int \frac{1}{20x} dx} = e^{\ln(x^{1/20})} = x^{1/20}$. Multiplying the entire equation by the integrating factor, we integrate both sides with respect to $x$ to find the general solution. The correct solution involves the integrating factor, the given functions, and an arbitrary constant $C$, leading to the form $y = \frac { e ^ { x } ( x - 1 ) + C } { x ^ { 1 / 20 } }$.

Find the General Solution of the Given First-Order Linear Differential dydx+y20x=x1920ex\frac { d y } { d x } + \frac { y } { 20 x } = \sqrt [ 20 ] { x ^ { 19 } } e ^ { x }dxdy​+20xy​=20x19​ex

Find the General Solution of the Given First-Order Linear Differential $\frac { d y } { d x } + \frac { y } { 20 x } = \sqrt [ 20 ] { x ^ { 19 } } e ^ { x }$