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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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