Key Concepts and Formulas

• Bernoulli Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ can be transformed into a linear differential equation using the substitution $v = y^{1-n}$.
• Integrating Factor: For a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$, the integrating factor is given by $I.F. = e^{\int P(x) dx}$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.

Step-by-Step Solution

Step 1: Rewrite the given differential equation in Bernoulli form.

The given differential equation is $\frac{dy}{dx} = \frac{y}{x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3$ Expanding the right side, we get $\frac{dy}{dx} = \frac{y}{x} + \frac{y}{x} \cdot x{y^2}(1 + {{\log }_e}x) = \frac{y}{x} + {y^3}(1 + {{\log }_e}x)$ Rearranging the terms to match the Bernoulli form, we have $\frac{dy}{dx} - \frac{1}{x}y = (1 + {{\log }_e}x)y^3$ Here, $P(x) = -\frac{1}{x}$, $Q(x) = 1 + \log_e x$, and $n = 3$.

Step 2: Apply the substitution to transform the Bernoulli equation into a linear equation.

We use the substitution $v = y^{1-n} = y^{1-3} = y^{-2} = \frac{1}{y^2}$. Differentiating $v$ with respect to $x$, we get $\frac{dv}{dx} = \frac{d}{dx}(y^{-2}) = -2y^{-3}\frac{dy}{dx}$ So, $\frac{dy}{dx} = -\frac{1}{2}y^3 \frac{dv}{dx}$. Substituting this into the Bernoulli equation: $-\frac{1}{2}y^3 \frac{dv}{dx} - \frac{1}{x}y = (1 + {{\log }_e}x)y^3$ Divide the equation by $y^3$: $-\frac{1}{2}\frac{dv}{dx} - \frac{1}{x}y^{-2} = (1 + {{\log }_e}x)$ Substitute $v = y^{-2}$: $-\frac{1}{2}\frac{dv}{dx} - \frac{1}{x}v = (1 + {{\log }_e}x)$ Multiply by $-2$ to get the linear form: $\frac{dv}{dx} + \frac{2}{x}v = -2(1 + {{\log }_e}x)$

Step 3: Solve the linear differential equation using the integrating factor.

The integrating factor is given by $I.F. = e^{\int P(x) dx}$, where $P(x) = \frac{2}{x}$. $\int \frac{2}{x} dx = 2 \int \frac{1}{x} dx = 2 \log_e x = \log_e (x^2)$ So, $I.F. = e^{\log_e (x^2)} = x^2$. The general solution of the linear differential equation is $v \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$ $v x^2 = \int -2(1 + {{\log }_e}x)x^2 dx + C = -2 \int (x^2 + x^2 {{\log }_e}x) dx + C$ We evaluate the integral $\int x^2 \log_e x \, dx$ using integration by parts. Let $u = \log_e x$ and $dv = x^2 dx$. Then $du = \frac{1}{x} dx$ and $v = \frac{x^3}{3}$. $\int x^2 \log_e x \, dx = \frac{x^3}{3} \log_e x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \log_e x - \frac{1}{3} \int x^2 dx = \frac{x^3}{3} \log_e x - \frac{x^3}{9}$ Therefore, $\int (x^2 + x^2 \log_e x) dx = \frac{x^3}{3} + \frac{x^3}{3} \log_e x - \frac{x^3}{9} = \frac{2x^3}{9} + \frac{x^3}{3} \log_e x = \frac{x^3}{9} (2 + 3 \log_e x)$ Substituting this back into the equation for $v x^2$: $v x^2 = -2 \left( \frac{2x^3}{9} + \frac{x^3}{3} \log_e x \right) + C = -\frac{4x^3}{9} - \frac{2x^3}{3} \log_e x + C$ $v = -\frac{4x}{9} - \frac{2x}{3} \log_e x + \frac{C}{x^2}$ Since $v = \frac{1}{y^2}$, we have $\frac{1}{y^2} = -\frac{2x^3}{9}(2 + 3\log_e x) \frac{1}{x^2} + \frac{C}{x^2} = \frac{C}{x^2} -\frac{2x}{9}(2 + 3\log_e x)$ $\frac{1}{y^2} = \frac{C}{x^2} - \frac{2x(2+3\log_e x)}{9} = \frac{9C - 2x^3(2+3\log_e x)}{9x^2}$ $y^2 = \frac{9x^2}{9C - 2x^3(2+3\log_e x)}$

Step 4: Apply the initial condition to find the constant C.

Given $y(1) = 3$, we have $y^2(1) = 9$. Substituting $x = 1$ and $y^2 = 9$ into the equation: $9 = \frac{9(1)^2}{9C - 2(1)^3(2+3\log_e 1)} = \frac{9}{9C - 2(2+0)} = \frac{9}{9C - 4}$ So, $9(9C - 4) = 9$, which means $9C - 4 = 1$, and $9C = 5$, so $C = \frac{5}{9}$. Substituting $C = \frac{5}{9}$ into the equation for $y^2$: $y^2 = \frac{9x^2}{9(\frac{5}{9}) - 2x^3(2+3\log_e x)} = \frac{9x^2}{5 - 2x^3(2+3\log_e x)}$ Then $\frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2+3\log_e x)}$ Since $3\log_e x = \log_e x^3$, we have $\frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2+\log_e x^3)}$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when applying the substitution and integrating factor.
• Integration by Parts: Remember the correct formula and choose $u$ and $dv$ strategically.
• Logarithm Properties: Use logarithm properties correctly to simplify expressions.

4. Summary

We solved the given Bernoulli differential equation by first transforming it into a linear differential equation using a suitable substitution. Then, we found the integrating factor and solved the linear equation. Finally, we applied the initial condition to determine the constant of integration and obtain the solution for $y^2(x)$. We then divided the result by 9 to arrive at the final expression.

5. Final Answer

The final answer is $\boxed{\frac{x^2}{5 - 2{x^3}(2 + {{\log }_e}{x^3})}}$, which corresponds to option (A).