Show that the substitution z = sinh−1 x transforms the equation (1 + x2) d2ydx2 + x dydx = 4yinto d2ydz2 = 4y.

Sure, let's go through this step by step.

1. First, we need to find the derivatives of x in terms of z. Given z = sinh^(-1)x, we can differentiate both sides with respect to x to get dz/dx = 1/√(1 + x^2).

2. Next, we differentiate again to find the second derivative, d^2z/dx^2. Using the quotient rule, we get d^2z/dx^2 = -x/(1 + x^2)^(3/2).

3. Now, we substitute these derivatives into the original equation. The term d^2y/dx^2 becomes d^2y/dz^2 * (dz/dx)^2 + dy/dz * d^2z/dx^2. Substituting the derivatives we found, this simplifies to d^2y/dz^2 * 1/(1 + x^2) - x/(1 + x^2)^(3/2) * dy/dz.

4. The term x * dy/dx becomes x * dy/dz * dz/dx, which simplifies to x * dy/dz / √(1 + x^2).

5. Substituting these terms back into the original equation, we get (1 + x^2) * [d^2y/dz^2 * 1/(1 + x^2) - x/(1 + x^2)^(3/2) * dy/dz] + x * dy/dz / √(1 + x^2) = 4y.

6. Simplifying, we see that the x^2 terms cancel out, leaving us with d^2y/dz^2 - dy/dz + dy/dz = 4y.

7. This simplifies to d^2y/dz^2 = 4y, which is the desired result.