InterviewSolution
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Show that `y=a*e^(2x)+b*e^(-x)` is a solution of the differential equation `(d^(2)y)/(dx^(2))-(dy)/(dx)-2y=0`. |
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Answer» `y=ae^(2x)+be^(-x)`……….`(1)` Differentiate with respect to `x` `(dy)/(dx)=a*e^(2x)*(2)+b*e^(-x)(-1)` `implies (dy)/(dx)=2a*e^(2x)-b*e^(-x)`………`(2)` Again differentiate with respect to `x` `(d^(2)y)/(dx^(2))=2a*e^(2x)*(2)-b*e^(-x)(-1)` `implies (d^(2)y)/(dx^(2))=4a*e^(2x)+b*e^(-x)`..............`(3)` Now `L.H.S=(d^(2)y)/(dx^(2))-(dy)/(dx)-2y` `=(4a*e^(2x)+be^(-x))` `-(2ae^(2x)-b*e^(-x))-2(ae^(2x)+be^(-x))` [Putting the values from `(1)`, `(2)` and `(3)`] `=4ae^(2x)+be^(-x)-2ae^(2x)+be^(-x)-2ae^(2x)-2be^(-x)` `=0=R.H.S`. `:. y=ae^(2x)+be^(-x)` is a solution of the given differential equation. |
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