y'''-3y''+3y'-y= e^x -x-1

asked by guest
on Jan 15, 2025 at 3:26 am



You asked:

Investigate the equation: \(y^{\prime \prime \prime} - 3 \cdot y^{\prime \prime} + 3 \cdot y^{\prime} - y = {e}^{x} - x - 1\).

MathBot Answer:

\[y{\left(x \right)} = 4 + \left(\left(\left(\frac{x}{6} + C_{3}\right) x + C_{2}\right) x + C_{1}\right) e^{x} + x\]

\(C_{1}\), \(C_{2}\), \(C_{3}\) are arbitrary constants.


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