If Ax^3+Bxy=4 (where A and B are arbitrary co | vedclass

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(A) Given the equation $Ax^3+Bxy=4$. Rearranging for $y$,we get $Bxy = 4-Ax^3$,so $y = \frac{4}{Bx} - \frac{Ax^2}{B}$. Let $C_1 = \frac{4}{B}$ and $C_

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(A) Given the equation $Ax^3+Bxy=4$. Rearranging for $y$,we get $Bxy = 4-Ax^3$,so $y = \frac{4}{Bx} - \frac{Ax^2}{B}$. Let $C_1 = \frac{4}{B}$ and $C_2 = -\frac{A}{B}$. Then $y = C_1 x^{-1} + C_2 x^2$. Now,differentiate with respect to $x$: $\frac{dy}{dx} = -C_1 x^{-2} + 2C_2 x$. Differentiate again: $\frac{d^2y}{dx^2} = 2C_1 x^{-3} + 2C_2$. Substitute these into the differential equation $F(x) \frac{d^2y}{dx^2} + G(x) \frac{dy}{dx} - 2y = 0$: $F(x)(2C_1 x^{-3} + 2C_2) + G(x)(-C_1 x^{-2} + 2C_2 x) - 2(C_1 x^{-1} + C_2 x^2) = 0$. Grouping terms by $C_1$ and $C_2$: $C_1(2F(x)x^{-3} - G(x)x^{-2} - 2x^{-1}) + C_2(2F(x) + 2xG(x) - 2x^2) = 0$. For this to hold for arbitrary constants $C_1, C_2$,the coefficients must be zero: $2F(x)x^{-3} - G(x)x^{-2} - 2x^{-1} = 0 \implies 2F(x) - xG(x) - 2x^2 = 0$. $2F(x) + 2xG(x) - 2x^2 = 0 \implies F(x) + xG(x) - x^2 = 0$. At $x=1$: $2F(1) - G(1) - 2 = 0$ (Eq $1$) $F(1) + G(1) - 1 = 0$ (Eq $2$) Adding Eq $1$ and Eq $2$: $3F(1) - 3 = 0 \implies F(1) = 1$. Substituting $F(1)=1$ into Eq $2$: $1 + G(1) - 1 = 0 \implies G(1) = 0$. Therefore,$F(1) + G(1) = 1 + 0 = 1$.

If $Ax^3+Bxy=4$ (where $A$ and $B$ are arbitrary constants) is the general solution of the differential equation $F(x) \frac{d^2 y}{d x^2}+G(x) \frac{d y}{d x}-2 y=0$,then $F(1)+G(1)=$

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