If y = At^2 + frac{B}{t} (A, B are parameters | vedclass

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(D) Given the general solution $y = At^2 + Bt^{-1}$.First derivative: $y' = 2At - Bt^{-2}$.Second derivative: $y'' = 2A + 2Bt^{-3}$.Substitute $y, y',

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(D) Given the general solution $y = At^2 + Bt^{-1}$.First derivative: $y' = 2At - Bt^{-2}$.Second derivative: $y'' = 2A + 2Bt^{-3}$.Substitute $y, y', y''$ into the differential equation $f(t) y'' + g(t) y' + h(t) y = 0$:$f(t)(2A + 2Bt^{-3}) + g(t)(2At - Bt^{-2}) + h(t)(At^2 + Bt^{-1}) = 0$.Group terms with $A$ and $B$:$A[2f(t) + 2t g(t) + t^2 h(t)] + B[2t^{-3} f(t) - t^{-2} g(t) + t^{-1} h(t)] = 0$.Since this holds for any $A$ and $B$,the coefficients must be zero:$2f(t) + 2t g(t) + t^2 h(t) = 0$ $(1)$$2t^{-3} f(t) - t^{-2} g(t) + t^{-1} h(t) = 0 \implies 2f(t) - t g(t) + t^2 h(t) = 0$ $(2)$Subtracting $(2)$ from $(1)$: $3t g(t) = 0 \implies g(t) = 0$.Substituting $g(t) = 0$ into $(1)$: $2f(t) + t^2 h(t) = 0$.

If $y = At^2 + \frac{B}{t}$ ($A, B$ are parameters) is the general solution of the differential equation $f(t) y''(t) + g(t) y'(t) + h(t) y = 0$,then $2 f(t) + t^2 h(t) =$

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