Question

Obtain the differential equation by eliminating the arbitrary constants from the given equation, $y=Ae^{3x}+B\cdotp e^{-3x}$

Answer 1

To obtain the differential equation by eliminating the arbitrary constants from the given equation, $y=Ae^{3x}+B\cdotp e^{-3x}$

we had to differentiate it twice because there are two arbitrary constants

$y = A {e}^{3x} + B {e}^{ - 3x}...eq1 \\ \\ \frac{dy}{dx} = 3A {e}^{3x} - 3B {e}^{ - 3x} ...eq2 \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = 9A {e}^{3x} + 9B {e}^{ - 3x} ...eq3 \\ \\ taking \: 9 \: common \: from \: R.H.S. \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = 9(A {e}^{3x} + B {e}^{ - 3x}) \\ \\ place \: value \: from \: eq1 \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = 9 \: y \\ \\ or \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } - 9 \: y = 0$
is the differential equation.