If y = sinh(ksin⁻¹x), show that (1-x²)y'' - xy' - k²y = 0

Step 1: Find the first derivative, y'. Given y = (k ^-1 x). We use the chain rule. Let u = k ^-1 x. Then y = (u). The derivative of (u) with respect to u is (u). The derivative of k ^-1 x with respect to x is k · (1)/(sqrt(1-x^2)). So, y' = (dy)/(dx) = (d)/(du)((u)) · (du)/(dx) = (k ^-1 x) · (k)/(sqrt(1-x^2)). y' = k (k ^-1 x)sqrt(1-x^2) Step 2: Rearrange the first derivative to simplify the next differentiation. Multiply both sides by sqrt(1-x^2): sqrt(1-x^2) y' = k (k ^-1 x) Step 3: Differentiate both sides of the rearranged equation with respect to x. On the left side, use the product rule: (d)/(dx)(uv) = u'v + uv'. Let u = sqrt(1-x^2) and v = y'. (du)/(dx) = (1)/(2)(1-x^2)^-1/2(-2x) = (-x)/(sqrt(1-x^2)). (dv)/(dx) = y''. So, the left side becomes: (-x)/(sqrt(1-x^2)) y' + sqrt(1-x^2) y'' On the right side, use the chain rule: (d)/(dx)(k (k ^-1 x)) = k · (k ^-1 x) · (d)/(dx)(k ^-1 x) = k · (k ^-1 x) · (k)/(sqrt(1-x^2)) = k^2 (k ^-1 x)sqrt(1-x^2) Step 4: Equate the differentiated left and right sides. (-x y')/(sqrt(1-x^2)) + sqrt(1-x^2) y'' = k^2 (k ^-1 x)sqrt(1-x^2) Step 5: Substitute y back into the equation. Recall that y = (k ^-1 x). Substitute this into the equation: (-x y')/(sqrt(1-x^2)) + sqrt(1-x^2) y'' = (k^2 y)/(sqrt(1-x^2)) Step 6: Multiply the entire equation by sqrt(1-x^2) to clear the denominators. -x y' + (1-x^2) y'' = k^2 y Step 7: Rearrange the terms to match the desired form. (1-x^2) y'' - x y' - k^2 y = 0 This proves the given differential equation. That's 2 down. 3 left today — send the next one.