(D^2 - D - 6)y = 0; when x=0, y=0 and y'(1)=e^3

asked by guest
on Nov 16, 2024 at 9:48 pm



You asked:

Solve the system of equations:\[\begin{aligned}\left({D}^{2} - D - 6\right) \cdot y &= 0\\x &= 0\\y &= 0\\y'\left( 1 \right) &= {e}^{3}\end{aligned}\]

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