y'' -2y' +y = (x-1)*e^x

asked by guest
on Nov 29, 2024 at 6:38 am



You asked:

Investigate the equation: \(y^{\prime \prime} - 2 \cdot y^{\prime} + y = \left(x - 1\right) \cdot {e}^{x}\).

MathBot Answer:

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

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


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