y'+xe^(-x/2)=0

asked by guest
on Sep 21, 2024 at 10:34 am



You asked:

Investigate the equation: \(y^{\prime} + x \cdot {e}^{\frac{-x}{2}} = 0\).

MathBot Answer:

\[y{\left(x \right)} = 4 e^{- \frac{x}{2}} + 2 x e^{- \frac{x}{2}} + C_{1}\]

\(C_{1}\) is an arbitrary constant.


\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).