If c and d are arbitrary constants,then y=e^{ | vedclass

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(B) The given solution is $y=e^{2 x}(c \cosh \sqrt{2} x+d \sinh \sqrt{2} x)$.This is of the form $y=e^{\alpha x}(c \cosh \beta x+d \sinh \beta x)$,whi

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$y^{\prime \prime}-4 y^{\prime}+2 y=0$

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(B) The given solution is $y=e^{2 x}(c \cosh \sqrt{2} x+d \sinh \sqrt{2} x)$.This is of the form $y=e^{\alpha x}(c \cosh \beta x+d \sinh \beta x)$,which corresponds to the auxiliary equation roots $m = \alpha \pm \beta$.Here,$\alpha = 2$ and $\beta = \sqrt{2}$.The roots are $m = 2 \pm \sqrt{2}$.The characteristic equation is $(m - (2 + \sqrt{2}))(m - (2 - \sqrt{2})) = 0$.$(m - 2 - \sqrt{2})(m - 2 + \sqrt{2}) = 0$.$(m - 2)^2 - (\sqrt{2})^2 = 0$.$m^2 - 4m + 4 - 2 = 0$.$m^2 - 4m + 2 = 0$.Replacing $m^2$ with $y^{\prime \prime}$ and $m$ with $y^{\prime}$,we get the differential equation $y^{\prime \prime} - 4y^{\prime} + 2y = 0$.

If $c$ and $d$ are arbitrary constants,then $y=e^{2 x}(\cosh \sqrt{2} x+d \sinh \sqrt{2} x)$ is the general solution of the differential equation

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