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Show that y = ae2x + be–x is a solution of the differential equation \(\frac{d^2y}{dx^2}-\frac{dy}{dx}-2y=0.\) |
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Answer» The differential equation is \(\frac{d^2y}{dx^2}-\frac{dy}{dx}-2y=0\) and the function that is to be proven as solution is y = ae2x + be–x, now we need to find the value of \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}.\) \(\frac{dy}{dx}\) = 2ae2x – be–x \(\frac{d^2y}{dx^2}\) = = 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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