Key Concepts and Formulas

• Implicit Differentiation: Differentiating an equation where $y$ is not explicitly defined as a function of $x$, using the chain rule $\frac{d}{dx}[f(y)] = f'(y) \frac{dy}{dx}$.
• Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.
• Second-Order Differential Equations: Equations involving a function and its first and second derivatives.

Step-by-Step Solution

Step 1: Rewrite the given equation

We are given $2x = y^{\frac{1}{5}} + y^{-\frac{1}{5}}$. This equation relates $x$ and $y$ implicitly. We will use this to find relationships between $x$, $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$.

Step 2: Differentiate with respect to x (first derivative)

Differentiating both sides of $2x = y^{\frac{1}{5}} + y^{-\frac{1}{5}}$ with respect to $x$, we get: $2 = \frac{1}{5}y^{-\frac{4}{5}}\frac{dy}{dx} - \frac{1}{5}y^{-\frac{6}{5}}\frac{dy}{dx}$ $2 = \frac{1}{5}\frac{dy}{dx} \left(y^{-\frac{4}{5}} - y^{-\frac{6}{5}}\right)$ $10 = \frac{dy}{dx} \left(\frac{1}{y^{\frac{4}{5}}} - \frac{1}{y^{\frac{6}{5}}}\right)$ $10 = \frac{dy}{dx} \left(\frac{y^{\frac{2}{5}} - 1}{y^{\frac{6}{5}}}\right)$ $\frac{dy}{dx} = \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}} - 1}$

Step 3: Differentiate with respect to x (second derivative)

Differentiating $\frac{dy}{dx} = \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}} - 1}$ with respect to $x$, we use the quotient rule: $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. $\frac{d^2y}{dx^2} = \frac{(y^{\frac{2}{5}} - 1) \frac{d}{dx}(10y^{\frac{6}{5}}) - 10y^{\frac{6}{5}} \frac{d}{dx}(y^{\frac{2}{5}} - 1)}{(y^{\frac{2}{5}} - 1)^2}$ $\frac{d^2y}{dx^2} = \frac{(y^{\frac{2}{5}} - 1) (10 \cdot \frac{6}{5} y^{\frac{1}{5}} \frac{dy}{dx}) - 10y^{\frac{6}{5}} (\frac{2}{5}y^{-\frac{3}{5}} \frac{dy}{dx})}{(y^{\frac{2}{5}} - 1)^2}$ $\frac{d^2y}{dx^2} = \frac{12 (y^{\frac{2}{5}} - 1) y^{\frac{1}{5}} \frac{dy}{dx} - 4y^{\frac{3}{5}} \frac{dy}{dx}}{(y^{\frac{2}{5}} - 1)^2}$ $\frac{d^2y}{dx^2} = \frac{dy}{dx} \cdot \frac{12y^{\frac{3}{5}} - 12y^{\frac{1}{5}} - 4y^{\frac{3}{5}}}{(y^{\frac{2}{5}} - 1)^2}$ $\frac{d^2y}{dx^2} = \frac{dy}{dx} \cdot \frac{8y^{\frac{3}{5}} - 12y^{\frac{1}{5}}}{(y^{\frac{2}{5}} - 1)^2}$ $\frac{d^2y}{dx^2} = \frac{dy}{dx} \cdot \frac{4y^{\frac{1}{5}}(2y^{\frac{2}{5}} - 3)}{(y^{\frac{2}{5}} - 1)^2}$

Step 4: Substitute $\frac{dy}{dx}$ into the equation for $\frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} = \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}} - 1} \cdot \frac{4y^{\frac{1}{5}}(2y^{\frac{2}{5}} - 3)}{(y^{\frac{2}{5}} - 1)^2}$ $\frac{d^2y}{dx^2} = \frac{40y^{\frac{7}{5}}(2y^{\frac{2}{5}} - 3)}{(y^{\frac{2}{5}} - 1)^3}$

Step 5: Find $x^2 - 1$

We have $2x = y^{\frac{1}{5}} + y^{-\frac{1}{5}}$. Squaring both sides: $4x^2 = y^{\frac{2}{5}} + 2 + y^{-\frac{2}{5}}$ $4x^2 - 4 = y^{\frac{2}{5}} - 2 + y^{-\frac{2}{5}}$ $4(x^2 - 1) = y^{\frac{2}{5}} - 2 + y^{-\frac{2}{5}}$ $4(x^2 - 1) = (y^{\frac{1}{5}} - y^{-\frac{1}{5}})^2$ Taking the square root: $2\sqrt{x^2 - 1} = y^{\frac{1}{5}} - y^{-\frac{1}{5}}$ Squaring again: $4(x^2-1)=(y^{\frac{1}{5}}-y^{-\frac{1}{5}})^2 = y^{\frac{2}{5}}-2+y^{-\frac{2}{5}}$ Also $4x^2 = (y^{\frac{1}{5}} + y^{-\frac{1}{5}})^2 = y^{\frac{2}{5}} + 2 + y^{-\frac{2}{5}}$. Thus, $4(x^2-1) = y^{\frac{2}{5}} - 2 + y^{-\frac{2}{5}}$, and $y^{\frac{2}{5}} + y^{-\frac{2}{5}} = 4x^2 - 2$. Then $y^{\frac{2}{5}} + \frac{1}{y^{\frac{2}{5}}} = 4x^2-2$. Multiplying by $y^{\frac{2}{5}}$, we get $(y^{\frac{2}{5}})^2 -(4x^2-2)y^{\frac{2}{5}}+1=0$. $y^{\frac{2}{5}}=\frac{4x^2-2 \pm \sqrt{(4x^2-2)^2-4}}{2}=\frac{4x^2-2 \pm \sqrt{16x^4-16x^2}}{2}=2x^2-1 \pm 2x\sqrt{x^2-1}$.

However, a simpler expression for $x^2-1$ is needed. From $2x = y^{\frac{1}{5}} + y^{-\frac{1}{5}}$, we have $4x^2 = y^{\frac{2}{5}} + 2 + y^{-\frac{2}{5}}$. Thus, $4x^2 - 4 = y^{\frac{2}{5}} - 2 + y^{-\frac{2}{5}}$, or $4(x^2-1) = (y^{\frac{1}{5}} - y^{-\frac{1}{5}})^2$.

Step 6: Substitute into the given differential equation

We have $(x^2 - 1) \frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$. Substituting the expressions we found for $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$: $(x^2 - 1) \left( \frac{40y^{\frac{7}{5}}(2y^{\frac{2}{5}} - 3)}{(y^{\frac{2}{5}} - 1)^3} \right) + \lambda x \left( \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}} - 1} \right) + ky = 0$ We also have $4(x^2-1) = (y^{\frac{1}{5}}-y^{-\frac{1}{5}})^2 = y^{\frac{2}{5}}-2+y^{-\frac{2}{5}} = y^{\frac{2}{5}}-2+\frac{1}{y^{\frac{2}{5}}} = \frac{y^{\frac{4}{5}}-2y^{\frac{2}{5}}+1}{y^{\frac{2}{5}}} = \frac{(y^{\frac{2}{5}}-1)^2}{y^{\frac{2}{5}}}$ $x^2-1 = \frac{(y^{\frac{2}{5}}-1)^2}{4y^{\frac{2}{5}}}$ Substituting this into the differential equation: $\frac{(y^{\frac{2}{5}}-1)^2}{4y^{\frac{2}{5}}} \left( \frac{40y^{\frac{7}{5}}(2y^{\frac{2}{5}} - 3)}{(y^{\frac{2}{5}} - 1)^3} \right) + \lambda x \left( \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}} - 1} \right) + ky = 0$ $\frac{10y(2y^{\frac{2}{5}} - 3)}{(y^{\frac{2}{5}}-1)} + \lambda x \left( \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}} - 1} \right) + ky = 0$ Since $2x = y^{\frac{1}{5}} + y^{-\frac{1}{5}}$, $x=\frac{y^{\frac{1}{5}}+y^{-\frac{1}{5}}}{2}$. Multiplying by $y^{\frac{2}{5}}-1$, we have $10y(2y^{\frac{2}{5}} - 3) + \lambda \frac{y^{\frac{1}{5}}+y^{-\frac{1}{5}}}{2} 10y^{\frac{6}{5}} + ky(y^{\frac{2}{5}}-1)=0$. $20y^{\frac{7}{5}}-30y+\lambda (y^{\frac{7}{5}}+y) + ky^{\frac{7}{5}}-ky=0$ $(20+\lambda+k)y^{\frac{7}{5}} + (\lambda-30-k)y = 0$ For this equation to hold true for all $y$, the coefficients of $y^{\frac{7}{5}}$ and $y$ must be zero. Thus, $20 + \lambda + k = 0$ and $\lambda - 30 - k = 0$. Adding the two equations gives $2\lambda - 10 = 0$, so $\lambda = 5$. Substituting $\lambda = 5$ into $20 + \lambda + k = 0$, we get $20 + 5 + k = 0$, so $k = -25$. Therefore, $\lambda + k = 5 - 25 = -20$.

Step 7: Re-examine the derivatives

Let's go back to Step 2. We have $2x = y^{1/5} + y^{-1/5}$. Let $y^{1/5} = t$. Then $2x = t + 1/t$. Differentiating with respect to $x$: $2 = (1/5)y^{-4/5} dy/dx - (1/5)y^{-6/5} dy/dx = (dy/dx) (1/5) (y^{-4/5} - y^{-6/5})$. Then, $dy/dx = 10/(y^{-4/5} - y^{-6/5}) = 10 y^{6/5} / (y^{2/5} - 1)$. $d^2y/dx^2 = 10 [ (6/5)y^{1/5} (dy/dx) (y^{2/5}-1) - y^{6/5} (2/5) y^{-3/5} (dy/dx) ] / (y^{2/5}-1)^2 = (2dy/dx) [ 6y^{1/5}(y^{2/5}-1) - 2 y^{3/5} ] / (y^{2/5}-1)^2 = (2dy/dx) [ 6y^{3/5} - 6y^{1/5} - 2y^{3/5}]/(y^{2/5}-1)^2 = (4dy/dx) [2y^{3/5}-3y^{1/5}]/(y^{2/5}-1)^2$. $d^2y/dx^2 = (40 y^{6/5}/(y^{2/5}-1) ) [2y^{3/5}-3y^{1/5}]/(y^{2/5}-1)^2 = 40 y^{1/5} (2y^{2/5}-3) y^{6/5} / (y^{2/5}-1)^3 = 40 y^{7/5} (2y^{2/5}-3) / (y^{2/5}-1)^3$.

$x^2 - 1 = (t+1/t)^2/4 - 1 = (t^2 + 2 + 1/t^2)/4 - 1 = (t^2-2+1/t^2)/4 = (t-1/t)^2/4 = (y^{1/5} - y^{-1/5})^2/4$. Then, $x^2-1 = (y^{2/5} + y^{-2/5} - 2)/4 = (y^{4/5} + 1 - 2y^{2/5})/4y^{2/5} = (y^{2/5}-1)^2/4y^{2/5}$.

Then, $(x^2-1) d^2y/dx^2 = ( (y^{2/5}-1)^2/4y^{2/5} ) (40 y^{7/5} (2y^{2/5}-3) / (y^{2/5}-1)^3 ) = 10y (2y^{2/5}-3) / (y^{2/5}-1)$. $\lambda x dy/dx = \lambda (y^{1/5}+y^{-1/5})/2 ( 10y^{6/5}/(y^{2/5}-1) ) = 5\lambda (y^{7/5}+y^{1/5})/(y^{2/5}-1)$.

Then, $10y (2y^{2/5}-3) / (y^{2/5}-1) + 5\lambda (y^{7/5}+y^{1/5})/(y^{2/5}-1) + ky = 0$. $10y (2y^{2/5}-3) + 5\lambda (y^{7/5}+y^{1/5}) + ky (y^{2/5}-1) = 0$. $20y^{7/5} - 30y + 5\lambda y^{7/5} + 5\lambda y^{1/5} + ky^{7/5} - ky = 0$. $(20+5\lambda+k) y^{7/5} - (30+k)y + 5\lambda y^{1/5} = 0$. Then, $20+5\lambda+k = 0$, $30+k=0$, and $\lambda = 0$. $k = -30$. $20 + 0 - 30 = 0$ which is false. Let's re-examine our calculation of $x^2-1$: $4x^2 = (y^{1/5}+y^{-1/5})^2 = y^{2/5}+2+y^{-2/5}$. Then $4(x^2-1) = y^{2/5}-2+y^{-2/5} = (y^{1/5}-y^{-1/5})^2$. Then $2 \sqrt{x^2-1} = y^{1/5}-y^{-1/5}$.

From the given equation $(x^2-1) \frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$. Comparing this with the standard form of Chebyshev's differential equation $(1-x^2)y'' - xy' + n^2y = 0$, if we put $x = \cos \theta$, then $y = A \cos (n \theta) + B \sin(n \theta)$. In our case, $2x = y^{1/5} + y^{-1/5}$. If we let $y = z^5$, then $2x = z + 1/z$. So $z = x + \sqrt{x^2 - 1}$.

Let $y^{1/5} = e^t$, then $y^{-1/5} = e^{-t}$. Then $2x = e^t + e^{-t} = 2 \cosh t$. So $x = \cosh t$. Then $y = e^{5t}$. $dy/dx = dy/dt dt/dx = 5e^{5t} dt/dx$. $x = \cosh t$, $dx/dt = \sinh t$. $dy/dx = 5e^{5t}/ \sinh t$. $d^2y/dx^2 = d/dx (5e^{5t}/ \sinh t) = d/dt (5e^{5t}/ \sinh t) dt/dx = [25 e^{5t} \sinh t - 5e^{5t} \cosh t]/\sinh^2 t (1/\sinh t) = 5e^{5t} [5 \sinh t - \cosh t] / \sinh^3 t$.

We have $(x^2-1) = \sinh^2 t$, and $x = \cosh t$. Then $\sinh^2 t ( 5e^{5t} [5 \sinh t - \cosh t] / \sinh^3 t ) + \lambda \cosh t ( 5e^{5t}/ \sinh t ) + k e^{5t} = 0$. $5e^{5t} [5 \sinh t - \cosh t] / \sinh t + 5\lambda \cosh t e^{5t}/ \sinh t + k e^{5t} = 0$. $5 [5 \sinh t - \cosh t] + 5\lambda \cosh t + k \sinh t = 0$. $(25+k) \sinh t + (5\lambda-5) \cosh t = 0$. Then $25+k=0$ and $5\lambda - 5 = 0$. So $k=-25$ and $\lambda = 1$. $\lambda + k = 1 - 25 = -24$.

Step 8: Correcting the Error

There was an error in Step 6. Let's go back to $(x^2 - 1) \frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$ and use the expressions $\frac{dy}{dx} = \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}} - 1}$ $\frac{d^2y}{dx^2} = \frac{40y^{\frac{7}{5}}(2y^{\frac{2}{5}} - 3)}{(y^{\frac{2}{5}} - 1)^3}$ $x^2-1 = \frac{(y^{\frac{2}{5}}-1)^2}{4y^{\frac{2}{5}}}$ Substitute into the original equation: $\frac{(y^{\frac{2}{5}}-1)^2}{4y^{\frac{2}{5}}} \frac{40y^{\frac{7}{5}}(2y^{\frac{2}{5}}-3)}{(y^{\frac{2}{5}}-1)^3} + \lambda \frac{(y^{\frac{1}{5}}+y^{\frac{-1}{5}})}{2} \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}}-1} + ky = 0$ $\frac{10y(2y^{\frac{2}{5}}-3)}{y^{\frac{2}{5}}-1} + \lambda \frac{(y^{\frac{1}{5}}+y^{\frac{-1}{5}})}{2} \frac{10y^{\frac{6}{5}}}{y^{\frac{2}{5}}-1} + ky = 0$ Multiply by $y^{\frac{2}{5}}-1$: $10y(2y^{\frac{2}{5}}-3) + 5\lambda (y^{\frac{1}{5}}+y^{\frac{-1}{5}})(y^{\frac{6}{5}}) + ky(y^{\frac{2}{5}}-1) = 0$ $20y^{\frac{7}{5}}-30y + 5\lambda (y^{\frac{7}{5}}+y) + ky^{\frac{7}{5}}-ky = 0$ $(20+5\lambda+k)y^{\frac{7}{5}} + (5\lambda-30-k)y = 0$ So, we have $20+5\lambda+k=0$ and $5\lambda-30-k=0$. Adding gives $5\lambda+20+5\lambda-30 = 0$, so $10\lambda = 10$, and $\lambda=1$. Then $20+5+k=0$, so $k=-25$. Then $\lambda+k=1-25=-24$.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with algebraic manipulations, especially when dealing with fractional exponents and applying the chain rule multiple times.
• Simplification: Simplify expressions as much as possible after each differentiation step to make the subsequent steps easier.
• Recognizing Patterns: Look for patterns that might simplify the calculations, such as recognizing the expression for $x^2 - 1$ in terms of $y$.

Summary

The problem involves implicit differentiation and solving a second-order differential equation. The key is to carefully differentiate the given equation twice with respect to $x$, expressing the derivatives in terms of $y$ and its derivatives. Then, substitute these expressions into the given differential equation and solve for $\lambda$ and $k$. The final answer is $\lambda + k = -24$.

Final Answer

The final answer is \boxed{-24}, which corresponds to option (B).