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Question
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
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Solution
We have
\[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0 . . . . . \left( 1 \right)\]
\[Now, \]
\[y = e^{- 3x} \]
\[ \Rightarrow \frac{dy}{dx} = - 3 e^{- 3x} \]
\[\Rightarrow\frac{d^2y}{dx^2}=9e^{-3x}\]
\[\text{Putting the values of }\frac{d^2 y}{d x^2}, \frac{dy}{dx}\text{ and y in (1), we get}\]
\[LHS = 9 e^{- 3x} - 3 e^{- 3x} - 6 e^{- 3x} \]
\[ = 0\]
\[ = RHS\]
Thus, y = e−3x is the solution of the given differential equation.
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