51. Given that \( \frac{d y}{d x}=4+\frac{k}{x^{3}} \), where \( k \) is a constant. When \( x=-1, y=1 \) and \( \frac{d y}{d x}=14 \). a) Find the value of \( k \) b) Express \( y \) in terms of \( x \).

To find the value of \( k \), we can substitute \( x = -1 \) and \( \frac{d y}{d x} = 14 \) into the given equation \( \frac{d y}{d x} = 4 + \frac{k}{x^{3}} \). Starting with: \[ 14 = 4 + \frac{k}{(-1)^{3}} \] Simplifying the equation gives: \[ 14 = 4 - k \] Rearranging it, we find: \[ k = 4 - 14 = -10 \] Now we move on to part (b). We substitute \( k = -10 \) into the differential equation: \[ \frac{d y}{d x} = 4 - \frac{10}{x^{3}} \] Next, we integrate both sides with respect to \( x \): \[ y = \int \left( 4 - \frac{10}{x^{3}} \right) dx \] Calculating the integral, we get: \[ y = 4x + \frac{10}{2x^{2}} + C = 4x + \frac{5}{x^{2}} + C \] Now, to find the constant \( C \), we use the condition \( y(-1) = 1 \): \[ 1 = 4(-1) + \frac{5}{(-1)^{2}} + C \] This simplifies to: \[ 1 = -4 + 5 + C \] Thus, \[ C = 1 - 1 = 0 \] Finally, the expression for \( y \) in terms of \( x \) is: \[ y = 4x + \frac{5}{x^{2}} \]