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′′′−3⋅y′′+3⋅y′−y=ex−x−1y^{\prime \prime \prime} - 3 \cdot y^{\prime \prime} + 3 \cdot y^{\prime} - y = {e}^{x} - x - 1.

MathBot Answer:

y(x)=4+(((x6+C3)x+C2)x+C1)ex+xy{\left(x \right)} = 4 + \left(\left(\left(\frac{x}{6} + C_{3}\right) x + C_{2}\right) x + C_{1}\right) e^{x} + x

C1C_{1}, C2C_{2}, C3C_{3} are arbitrary constants.

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

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