If y'' - 3y' + 2y = 0 where y(0) = 1 and y'(0 | vedclass

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(D) The given differential equation is $y'' - 3y' + 2y = 0$. The characteristic equation is $m^2 - 3m + 2 = 0$. Factoring the quadratic,we get $(m - 1

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(D) The given differential equation is $y'' - 3y' + 2y = 0$. The characteristic equation is $m^2 - 3m + 2 = 0$. Factoring the quadratic,we get $(m - 1)(m - 2) = 0$,so the roots are $m = 1$ and $m = 2$. The general solution is $y(x) = Ae^x + Be^{2x}$. Differentiating with respect to $x$,we get $y'(x) = Ae^x + 2Be^{2x}$. Using the initial conditions: At $x = 0$,$y(0) = A + B = 1$. At $x = 0$,$y'(0) = A + 2B = 0$. Subtracting the first equation from the second,we get $B = -1$. Substituting $B = -1$ into $A + B = 1$,we get $A = 2$. Thus,the specific solution is $y(x) = 2e^x - e^{2x}$. Now,evaluate $y$ at $x = \log_{e} 2$: $y(\log_{e} 2) = 2e^{\log_{e} 2} - e^{2\log_{e} 2} = 2(2) - (e^{\log_{e} 2})^2 = 4 - (2)^2 = 4 - 4 = 0$.

If $y'' - 3y' + 2y = 0$ where $y(0) = 1$ and $y'(0) = 0$,then the value of $y$ at $x = \log_{e} 2$ is

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