$y^{'} = 6 e^{2 x} + 6 e^{3 x}, y^{''} = 12 e^{2 x} + 18 e^{3 x} .$ Substitute in $L . H . S .$ for $y, y^{'}, y^{''} .$

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Solution

Given,

$y^{''} = 12 e^{2 x} + 18 e^{3 x}$

integrating on both sides, we get,

$y^{'} = 6 e^{2 x} + 6 e^{3 x}$

again integrating on both sides, we get,

$y = 3 e^{2 x} + 2 e^{3 x}$

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y′=6e2x+6e3x, y′′=12e2x+18e3x. Substitute in L.H.S. for y, y′, y′′.

Given,y′′=12e2x+18e3xintegrating on both sides, we get,y′=6e2x+6e3xagain integrating on both sides, we get,y=3e2x+2e3x

$y^{'} = 6 e^{2 x} + 6 e^{3 x}, y^{''} = 12 e^{2 x} + 18 e^{3 x} .$ Substitute in $L . H . S .$ for $y, y^{'}, y^{''} .$

Given,

$y^{''} = 12 e^{2 x} + 18 e^{3 x}$

integrating on both sides, we get,

$y^{'} = 6 e^{2 x} + 6 e^{3 x}$

again integrating on both sides, we get,

$y = 3 e^{2 x} + 2 e^{3 x}$