Question

If $y^{1/4} + y^{-1/4} = 2x$, and $(x^2 - 1)\frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$, then $|\alpha - \beta|$ is equal to ____.

Answer 1

17

Answer 2

Let $t = y^{1/4}$. Then $y = t^4$. The given equation $y^{1/4} + y^{-1/4} = 2x$ becomes $t + \frac{1}{t} = 2x$. Differentiating $t + \frac{1}{t} = 2x$ with respect to $x$: $\frac{dt}{dx} - \frac{1}{t^2}\frac{dt}{dx} = 2$ $\frac{dt}{dx}\left(1 - \frac{1}{t^2}\right) = 2$ $\frac{dt}{dx}\left(\frac{t^2-1}{t^2}\right) = 2$ $\frac{dt}{dx} = \frac{2t^2}{t^2-1}$.

Now we find the derivatives of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = 4t^3 \cdot \frac{2t^2}{t^2-1} = \frac{8t^5}{t^2-1}$.

Differentiating $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{8t^5}{t^2-1}\right) = \frac{d}{dt}\left(\frac{8t^5}{t^2-1}\right)\frac{dt}{dx}$ $\frac{d}{dt}\left(\frac{8t^5}{t^2-1}\right) = 8 \frac{5t^4(t^2-1) - t^5(2t)}{(t^2-1)^2} = 8 \frac{5t^6 - 5t^4 - 2t^6}{(t^2-1)^2} = 8 \frac{3t^6 - 5t^4}{(t^2-1)^2} = \frac{8t^4(3t^2-5)}{(t^2-1)^2}$. So, $\frac{d^2y}{dx^2} = \frac{8t^4(3t^2-5)}{(t^2-1)^2} \cdot \frac{2t^2}{t^2-1} = \frac{16t^6(3t^2-5)}{(t^2-1)^3}$.

From $t + \frac{1}{t} = 2x$, we have $x = \frac{t^2+1}{2t}$. Then $x^2-1 = \left(\frac{t^2+1}{2t}\right)^2 - 1 = \frac{(t^2+1)^2 - 4t^2}{4t^2} = \frac{t^4+2t^2+1-4t^2}{4t^2} = \frac{t^4-2t^2+1}{4t^2} = \frac{(t^2-1)^2}{4t^2}$.

Substitute these into the given differential equation $(x^2 - 1)\frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$: $\left(\frac{(t^2-1)^2}{4t^2}\right) \left(\frac{16t^6(3t^2-5)}{(t^2-1)^3}\right) + \alpha \left(\frac{t^2+1}{2t}\right) \left(\frac{8t^5}{t^2-1}\right) + \beta t^4 = 0$. $\frac{4t^4(3t^2-5)}{t^2-1} + \frac{4\alpha t^4(t^2+1)}{t^2-1} + \beta t^4 = 0$. Divide by $t^4$: $\frac{4(3t^2-5)}{t^2-1} + \frac{4\alpha(t^2+1)}{t^2-1} + \beta = 0$. Multiply by $t^2-1$: $4(3t^2-5) + 4\alpha(t^2+1) + \beta(t^2-1) = 0$. $12t^2 - 20 + 4\alpha t^2 + 4\alpha + \beta t^2 - \beta = 0$. $(12 + 4\alpha + \beta)t^2 + (-20 + 4\alpha - \beta) = 0$. For this to hold for all $t$, the coefficients must be zero:

1. $12 + 4\alpha + \beta = 0$
2. $-20 + 4\alpha - \beta = 0$ Adding (1) and (2): $8\alpha - 8 = 0 \implies \alpha = 1$. Substituting $\alpha=1$ into (1): $12 + 4(1) + \beta = 0 \implies 16 + \beta = 0 \implies \beta = -16$. $|\alpha - \beta| = |1 - (-16)| = |1 + 16| = 17$.