The differential equation $5 \frac{d y}{d x}-2 x^{7} y^{4}=\frac{5 y}{x}$ can be solved using the substitution. Select the correct answer. $u=y$ $u=y^{-6}$ $u=y^{2}$ $u=y^{-3}$

Step 1 ：First, we rewrite the given differential equation as \(\frac{dy}{dx} = \frac{2x^7y^4 + \frac{5y}{x}}{5}\)

Step 2 ：Now, we make the substitution \(u = y^{-3}\) which implies \(y = u^{-\frac{1}{3}}\)

Step 3 ：Differentiating both sides with respect to x, we get \(\frac{dy}{dx} = -\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dx}\)

Step 4 ：Substitute the expressions for y and \(\frac{dy}{dx}\) in the given differential equation: \(-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dx} = \frac{2x^7(u^{-\frac{1}{3}})^4 + \frac{5(u^{-\frac{1}{3}})}{x}}{5}\)

Step 5 ：Simplify the equation: \(-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dx} = \frac{2x^7u^{\frac{4}{3}} + 5u^{\frac{1}{3}}}{5x}\)

Step 6 ：Multiply both sides by \(-3u^{\frac{4}{3}}x\): \(xu\frac{du}{dx} = -6x^8u + 15x^2\)

Step 7 ：Now, we have a first-order linear differential equation in u: \(x\frac{du}{dx} + 6x^8u = 15x^2\)

Step 8 ：The integrating factor is \(e^{\int 6x^7dx} = e^{\frac{3}{4}x^8}\)

Step 9 ：Multiply the differential equation by the integrating factor: \(xe^{\frac{3}{4}x^8}\frac{du}{dx} + 6x^8e^{\frac{3}{4}x^8}u = 15x^2e^{\frac{3}{4}x^8}\)

Step 10 ：The left side of the equation is now the derivative of the product of u and the integrating factor: \(\frac{d}{dx}(ue^{\frac{3}{4}x^8}) = 15x^2e^{\frac{3}{4}x^8}\)

Step 11 ：Integrate both sides with respect to x: \(\int\frac{d}{dx}(ue^{\frac{3}{4}x^8})dx = \int 15x^2e^{\frac{3}{4}x^8}dx\)

Step 12 ：Integrating the left side, we get: \(ue^{\frac{3}{4}x^8} = \int 15x^2e^{\frac{3}{4}x^8}dx + C\)

Step 13 ：Now, we have found the general solution for u, and we can substitute back y: \(y^{-3}e^{\frac{3}{4}x^8} = \int 15x^2e^{\frac{3}{4}x^8}dx + C\)

Step 14 ：The correct substitution is \(\boxed{u=y^{-3}}\)