1. The problem gives a differential equation: $$\frac{d w}{d x} = -kW^{3} T^{-\frac{1}{2}}.$$\n\n2. We want to solve this differential equation for $w$ as a function of $x$. However, it looks like $T$ and $k$ are constants or parameters, and $W$ depends on $x$.\n\n3. The equation is separable, meaning we can write it as $$\frac{d w}{d x} = -k W^{3} T^{-\frac{1}{2}}.$$\n\n4. Assuming $k$ and $T$ are constants, and $W$ is actually $w$, rewrite as $$\frac{d w}{d x} = -k w^{3} T^{-\frac{1}{2}}.$$\n\n5. Separate variables: $$\frac{d w}{w^{3}} = -k T^{-\frac{1}{2}} d x.$$\n\n6. Integrate both sides: $$\int w^{-3} d w = \int -k T^{-\frac{1}{2}} d x.$$\n\n7. Compute left integral: $$\int w^{-3} d w = \int w^{-3} d w = \frac{w^{-2}}{-2} + C_1 = -\frac{1}{2 w^{2}} + C_1.$$\n\n8. Compute right integral: $$\int -k T^{-\frac{1}{2}} d x = -k T^{-\frac{1}{2}} x + C_2.$$\n\n9. Set constants: $$-\frac{1}{2 w^{2}} = -k T^{-\frac{1}{2}} x + C,$$ where $C = C_2 - C_1$.\n\n10. Multiply both sides by $-2$:\n$$\frac{1}{w^{2}} = 2 k T^{-\frac{1}{2}} x - 2 C.$$\n\n11. Let $C' = 2 C$, rewrite: $$\frac{1}{w^{2}} = 2 k T^{-\frac{1}{2}} x - C'.$$\n\n12. Finally, solve for $w$: $$w = \pm \frac{1}{\sqrt{2 k T^{-\frac{1}{2}} x - C'}}.$$\n\nThis is the general solution for $w$ in terms of $x$, with constants collected in $C'$ determined by initial conditions.