The differential equation is and the function that is to be proven as solution is
y = ae2x + be–x, now we need to find the value of and .
= 2ae2x – be–x
= 4ae2x + be–x
Putting these values in the equation, we get,
4ae2x + be–x –(2ae2x – be–x) – 2(ae2x + be–x) = 0,
0 = 0
As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.
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