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(A) Given the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$. Substitute $y = z^{\alpha}$,then $dy = \alpha z^{\alpha - 1} dz$. Substituting the

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$\alpha = -1$

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(A) Given the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$. Substitute $y = z^{\alpha}$,then $dy = \alpha z^{\alpha - 1} dz$. Substituting these into the equation: $(x^2(z^{\alpha})^2 - 1)(\alpha z^{\alpha - 1} dz) + 2x(z^{\alpha})^3 dx = 0$ $(x^2 z^{2\alpha} - 1)(\alpha z^{\alpha - 1} dz) + 2x z^{3\alpha} dx = 0$ $\alpha x^2 z^{3\alpha - 1} dz - \alpha z^{\alpha - 1} dz + 2x z^{3\alpha} dx = 0$ For the equation to be homogeneous,the sum of powers of $x$ and $z$ in each term must be equal. The powers are: Term $1$: $x^2 z^{3\alpha - 1} \rightarrow 2 + (3\alpha - 1) = 3\alpha + 1$ Term $2$: $z^{\alpha - 1} \rightarrow 0 + (\alpha - 1) = \alpha - 1$ Term $3$: $x z^{3\alpha} \rightarrow 1 + 3\alpha = 3\alpha + 1$ Equating the powers: $3\alpha + 1 = \alpha - 1$ $2\alpha = -2 \Rightarrow \alpha = -1$.

The substitution $y = z^{\alpha}$ transforms the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$ into a homogeneous differential equation for

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