Homogeneous Linear Differential Equations (General Form of LDE)
The general s...
Question
The general solution of the differential equation d2ydx2+2dydx−5y=0 in terms of arbitrary constant K1 and K2 is
A
K1e(−1+√6)x+K2e(−1−√6)x
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B
K1e(−1−√8)x+K2e(−1−√8)x
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C
K1e(−2−√6)x+K2e(−2−√6)x
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D
K1e(−2+√8)x+K2e(−2−√8)x
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Solution
The correct option is AK1e(−1+√6)x+K2e(−1−√6)x d2ydx2+2dydx−5y=0⇒(D2+2D−5)y=0
Auxiliary equation is m2+2m−5=0 m=−1±√6
So, in terms of arbitary constants k1 and k2 we get, y=k1em1x+k2em2x y=k1e(−1+√6x)+k2e(−1−√6x)