- #1
kJS
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And also:
y`+2y=2(1-e^-2t) Y(0)=0
y¨-2y`+y = t+e^t y(0)=1 and y`(0)=0
Please help me out here folks ;)
y`+2y=2(1-e^-2t) Y(0)=0
y¨-2y`+y = t+e^t y(0)=1 and y`(0)=0
Please help me out here folks ;)