Question:

Step 1: Rewrite the given differential equation. The given differential equation is $\dfrac{dy}{dx}=\dfrac{y}{x}(1+xy^2(1+\log_e x))$. First, distribute the $\frac{y}{x}$ term on the right side: $$ \dfrac{dy}{dx} = \dfrac{y}{x} + \dfrac{y}{x} \cdot xy^2(1+\log_e x) $$ $$ \dfrac{dy}{dx} = \dfrac{y}{x} + y^3(1+\log_e x) $$ Rearrange the terms to get it in a standard form: $$ \dfrac{dy}{dx} - \dfrac{1}{x}y = y^3(1+\log_e x) $$ Step 2: Convert the Bernoulli equation to a linear differential equation. The equation is a Bernoulli differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)y^n$, where $n=3$. To convert it into a linear differential equation, divide the entire equation by $y^3$: $$ \dfrac{1}{y^3}\dfrac{dy}{dx} - \dfrac{1}{x}\dfrac{1}{y^2} = (1+\log_e x) $$ Let $z = y^{-2} = \dfrac{1}{y^2}$. Differentiate $z$ with respect to $x$: $$ \dfrac{dz}{dx} = -2y^{-3}\dfrac{dy}{dx} $$ $$ \dfrac{1}{y^3}\dfrac{dy}{dx} = -\dfrac{1}{2}\dfrac{dz}{dx} $$ Substitute $z$ and $\dfrac{dz}{dx}$ into the equation: $$ -\dfrac{1}{2}\dfrac{dz}{dx} - \dfrac{1}{x}z = (1+\log_e x) $$ Multiply by $-2$ to get it in the standard linear form $\dfrac{dz}{dx} + P(x)z = Q(x)$: $$ \dfrac{dz}{dx} + \dfrac{2}{x}z = -2(1+\log_e x) $$ Step 3: Find the integrating factor. This is a linear first-order differential equation. The integrating factor (IF) is given by $e^{\int P(x) dx}$. Here, $P(x) = \dfrac{2}{x}$. $$ \text{IF} = e^{\int \frac{2}{x} dx} = e^{2\log_e |x|} = e^{\log_e x^2} = x^2 \quad (\text{since } x>0) $$ Step 4: Solve the linear differential equation. Multiply the linear equation by the integrating factor $x^2$: $$ x^2\dfrac{dz}{dx} + x^2\left(\dfrac{2}{x}\right)z = -2x^2(1+\log_e x) $$ $$ x^2\dfrac{dz}{dx} + 2xz = -2x^2(1+\log_e x) $$ The left side is the derivative of the product $z \cdot \text{IF}$: $$ \dfrac{d}{dx}(zx^2) = -2x^2(1+\log_e x) $$ Integrate both sides with respect to $x$: $$ zx^2 = \int -2x^2(1+\log_e x) dx + C $$ $$ zx^2 = -2 \int x^2(1+\log_e x) dx + C $$ Step 5: Evaluate the integral $\int x^2(1+\log_e x) dx$. We use integration by parts, $\int u dv = uv - \int v du$. Let $u = (1+\log_e x)$ and $dv = x^2 dx$. Then $du = \dfrac{1}{x} dx$ and $v = \dfrac{x^3}{3}$. $$ \int x^2(1+\log_e x) dx = (1+\log_e x)\left(\dfrac{x^3}{3}\right) - \int \dfrac{x^3}{3} \cdot \dfrac{1}{x} dx $$ $$ = \dfrac{x^3}{3}(1+\log_e x) - \int \dfrac{x^2}{3} dx $$ $$ = \dfrac{x^3}{3}(1+\log_e x) - \dfrac{x^3}{9} $$ Factor out $\dfrac{x^3}{9}$: $$ = \dfrac{x^3}{9} \left( 3(1+\log_e x) - 1 \right) $$ $$ = \dfrac{x^3}{9} (3 + 3\log_e x - 1) $$ $$ = \dfrac{x^3}{9} (2 + 3\log_e x) $$ Step 6: Substitute the integral back and find the general solution for $y(x)$. Substitute the result of the integral back into the equation for $zx^2$: $$ zx^2 = -2 \left[ \dfrac{x^3}{9} (2 + 3\log_e x) \right] + C $$ $$ zx^2 = -\dfrac{2x^3}{9} (2 + 3\log_e x) + C $$ Now, substitute $z = \dfrac{1}{y^2}$: $$ \dfrac{x^2}{y^2} = -\dfrac{2x^3}{9} (2 + 3\log_e x) + C $$ Step 7: Use the initial condition to find the constant $C$. The initial condition is $y(1)=3$. Substitute $x=1$ and $y=3$ into the general solution: $$ \dfrac{1^2}{3^2} = -\dfrac{2(1)^3}{9} (2 + 3\log_e 1) + C $$ Since $\log_e 1 = 0$: $$ \dfrac{1}{9} = -\dfrac{2}{9} (2 + 0) + C $$ $$ \dfrac{1}{9} = -\dfrac{4}{9} + C $$ Solving for $C$: $$ C = \dfrac{1}{9} + \dfrac{4}{9} = \dfrac{5}{9} $$ Substitute $C = \dfrac{5}{9}$ back into the solution: $$ \dfrac{x^2}{y^2} = -\dfrac{2x^3}{9} (2 + 3\log_e x) + \dfrac{5}{9} $$ $$ \dfrac{x^2}{y^2} = \dfrac{5 - 2x^3(2 + 3\log_e x)}{9} $$ Step 8: Express the solution in the required form $\dfrac{y^2(x)}{9}$. From the previous step, we have: $$ \dfrac{x^2}{y^2} = \dfrac{5 - 2x^3(2 + 3\log_e x)}{9} $$ Invert both sides: $$ \dfrac{y^2}{x^2} = \dfrac{9}{5 - 2x^3(2 + 3\log_e x)} $$ Multiply both sides by $x^2$: $$ y^2 = \dfrac{9x^2}{5 - 2x^3(2 + 3\log_e x)} $$ Finally, divide by 9: $$ \dfrac{y^2(x)}{9} = \dfrac{x^2}{5 - 2x^3(2 + 3\log_e x)} $$ Using the logarithm property $n\log_e x = \log_e x^n$, we can write $3\log_e x = \log_e x^3$: $$ \dfrac{y^2(x)}{9} = \dfrac{x^2}{5 - 2x^3(2 + \log_e x^3)} $$ The final answer is $\dfrac{y^2(x)}{9} = \dfrac{x^2}{5 - 2x^3(2 + \log_e x^3)}$. This matches Option 1. The final answer is $\boxed{\dfrac{x^2}{5-2x^3(2+\log_e x^3)}}$.