Problem:
2y'' - 3y' + 6y = 0
Solution:
The solutions to the characteristic polynomial are $ \frac{3 \pm \sqrt{39}i}{4} $
Skipping all the derivation we have already done in the last post, the answer is
$ y = Ae^{\frac{3}{4}}\cos\frac{\sqrt{39}}{4}x + Be^{\frac{3}{4}}\sin\frac{\sqrt{39}}{4}x $
2y'' - 3y' + 6y = 0
Solution:
The solutions to the characteristic polynomial are $ \frac{3 \pm \sqrt{39}i}{4} $
Skipping all the derivation we have already done in the last post, the answer is
$ y = Ae^{\frac{3}{4}}\cos\frac{\sqrt{39}}{4}x + Be^{\frac{3}{4}}\sin\frac{\sqrt{39}}{4}x $
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