If y = (A + Bx) e^{mx} + (m - 1)^{-2} e^x,the | vedclass

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(A) Given $y = (A + Bx) e^{mx} + (m - 1)^{-2} e^x$. Let $y = y_1 + y_2$,where $y_1 = (A + Bx) e^{mx}$ and $y_2 = (m - 1)^{-2} e^x$. For $y_1 = (A + Bx

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$e^x$

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(A) Given $y = (A + Bx) e^{mx} + (m - 1)^{-2} e^x$. Let $y = y_1 + y_2$,where $y_1 = (A + Bx) e^{mx}$ and $y_2 = (m - 1)^{-2} e^x$. For $y_1 = (A + Bx) e^{mx}$,the differential equation satisfied is $\frac{d^2y_1}{dx^2} - 2m \frac{dy_1}{dx} + m^2y_1 = 0$. Now,consider $y_2 = (m - 1)^{-2} e^x$. $\frac{dy_2}{dx} = (m - 1)^{-2} e^x$ and $\frac{d^2y_2}{dx^2} = (m - 1)^{-2} e^x$. Substituting $y_2$ into the expression $\frac{d^2y_2}{dx^2} - 2m \frac{dy_2}{dx} + m^2y_2$: $= (m - 1)^{-2} e^x - 2m(m - 1)^{-2} e^x + m^2(m - 1)^{-2} e^x$ $= (m - 1)^{-2} e^x [1 - 2m + m^2]$ $= (m - 1)^{-2} e^x (m - 1)^2$ $= e^x$. Thus,$\frac{d^2y}{dx^2} - 2m \frac{dy}{dx} + m^2y = 0 + e^x = e^x$.

If $y = (A + Bx) e^{mx} + (m - 1)^{-2} e^x$,then $\frac{d^2y}{dx^2} - 2m \frac{dy}{dx} + m^2y$ is equal to:

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