Key Concepts and Formulas

• Bernoulli's Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$, where $n$ is a real number not equal to 0 or 1.
• Substitution for Bernoulli's Equation: To solve a Bernoulli equation, we use the substitution $v = y^{1-n}$.
• Integrating Factor: For a linear first-order differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$, the integrating factor is given by $e^{\int P(x) dx}$.

Step-by-Step Solution

Step 1: Rewrite the Differential Equation

The given differential equation is: $(1+x^2)dy = y(x-y)dx$ Divide both sides by $(1+x^2)dx$ to get: $\frac{dy}{dx} = \frac{y(x-y)}{1+x^2}$ $\frac{dy}{dx} = \frac{xy - y^2}{1+x^2}$ $\frac{dy}{dx} - \frac{x}{1+x^2}y = \frac{-y^2}{1+x^2}$ This is a Bernoulli differential equation with $n=2$.

Step 2: Apply the Bernoulli Substitution

Let $v = y^{1-n} = y^{1-2} = y^{-1}$. Then, $y = v^{-1}$. Differentiating with respect to $x$, we get: $\frac{dy}{dx} = \frac{d}{dx}(v^{-1}) = -v^{-2}\frac{dv}{dx}$ Substituting $y = v^{-1}$ and $\frac{dy}{dx} = -v^{-2}\frac{dv}{dx}$ into the equation, we have: $-v^{-2}\frac{dv}{dx} - \frac{x}{1+x^2}v^{-1} = \frac{-v^{-2}}{1+x^2}$ Multiplying by $-v^2$, we get: $\frac{dv}{dx} + \frac{x}{1+x^2}v = \frac{1}{1+x^2}$ This is now a linear first-order differential equation in terms of $v$.

Step 3: Find the Integrating Factor

The integrating factor is given by: $I.F. = e^{\int \frac{x}{1+x^2} dx}$ Let $u = 1+x^2$, then $du = 2x dx$, so $x dx = \frac{1}{2}du$. $I.F. = e^{\int \frac{1}{2u} du} = e^{\frac{1}{2} \ln|u|} = e^{\ln\sqrt{u}} = \sqrt{u} = \sqrt{1+x^2}$

Step 4: Solve the Linear Differential Equation

Multiply the linear differential equation by the integrating factor: $\sqrt{1+x^2}\frac{dv}{dx} + \frac{x\sqrt{1+x^2}}{1+x^2}v = \frac{\sqrt{1+x^2}}{1+x^2}$ $\sqrt{1+x^2}\frac{dv}{dx} + \frac{x}{\sqrt{1+x^2}}v = \frac{1}{\sqrt{1+x^2}}$ The left side is the derivative of $v\sqrt{1+x^2}$: $\frac{d}{dx}(v\sqrt{1+x^2}) = \frac{1}{\sqrt{1+x^2}}$ Integrate both sides with respect to $x$: $\int \frac{d}{dx}(v\sqrt{1+x^2}) dx = \int \frac{1}{\sqrt{1+x^2}} dx$ $v\sqrt{1+x^2} = \sinh^{-1}(x) + C$ Recall that $v = y^{-1}$, so $\frac{\sqrt{1+x^2}}{y} = \sinh^{-1}(x) + C$

Step 5: Apply the Initial Condition

Given $y(0) = 1$, so when $x=0$, $y=1$: $\frac{\sqrt{1+0^2}}{1} = \sinh^{-1}(0) + C$ $1 = 0 + C$ $C = 1$ Thus, the solution is: $\frac{\sqrt{1+x^2}}{y} = \sinh^{-1}(x) + 1$ $y = \frac{\sqrt{1+x^2}}{\sinh^{-1}(x) + 1}$

Step 6: Find β

We are given $y(2\sqrt{2}) = \beta$. Substituting $x = 2\sqrt{2}$ into the solution: $\beta = \frac{\sqrt{1+(2\sqrt{2})^2}}{\sinh^{-1}(2\sqrt{2}) + 1} = \frac{\sqrt{1+8}}{\sinh^{-1}(2\sqrt{2}) + 1} = \frac{3}{\sinh^{-1}(2\sqrt{2}) + 1}$ $\beta^{-1} = \frac{\sinh^{-1}(2\sqrt{2}) + 1}{3}$ $3\beta^{-1} = \sinh^{-1}(2\sqrt{2}) + 1$ $\sinh^{-1}(2\sqrt{2}) = 3\beta^{-1} - 1$ We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$. So, $\sinh(3\beta^{-1} - 1) = 2\sqrt{2}$ $\frac{e^{3\beta^{-1} - 1} - e^{-(3\beta^{-1} - 1)}}{2} = 2\sqrt{2}$ $e^{3\beta^{-1} - 1} - e^{-3\beta^{-1} + 1} = 4\sqrt{2}$ Multiply by $e^{3\beta^{-1}-1}$: $e^{2(3\beta^{-1} - 1)} - 4\sqrt{2}e^{3\beta^{-1} - 1} - e^0 = 0$ Let $z = e^{3\beta^{-1} - 1}$. Then $z^2 - 4\sqrt{2}z - 1 = 0$. Using the quadratic formula: $z = \frac{4\sqrt{2} \pm \sqrt{(4\sqrt{2})^2 - 4(1)(-1)}}{2} = \frac{4\sqrt{2} \pm \sqrt{32+4}}{2} = \frac{4\sqrt{2} \pm \sqrt{36}}{2} = \frac{4\sqrt{2} \pm 6}{2} = 2\sqrt{2} \pm 3$ Since $z = e^{3\beta^{-1} - 1} > 0$, we take the positive root: $z = 3 + 2\sqrt{2}$ $e^{3\beta^{-1} - 1} = 3 + 2\sqrt{2}$ $e^{3\beta^{-1}}e^{-1} = 3 + 2\sqrt{2}$ $e^{3\beta^{-1}} = e(3 + 2\sqrt{2})$ Taking the natural logarithm of both sides: $3\beta^{-1} = 1 + \ln(3+2\sqrt{2})$

Let's re-examine the options. We have $\beta = \frac{3}{\sinh^{-1}(2\sqrt{2})+1}$. From here, we can write: $\frac{1}{\beta} = \frac{\sinh^{-1}(2\sqrt{2})+1}{3}$, $\frac{3}{\beta} = \sinh^{-1}(2\sqrt{2})+1$. We know $\sinh^{-1}(x) = \ln(x + \sqrt{x^2+1})$. Thus, $\sinh^{-1}(2\sqrt{2}) = \ln(2\sqrt{2} + \sqrt{8+1}) = \ln(2\sqrt{2} + 3)$. Therefore, $\frac{3}{\beta} = \ln(3+2\sqrt{2}) + 1$. So $e^{3/\beta} = e^{\ln(3+2\sqrt{2}) + 1} = e \cdot e^{\ln(3+2\sqrt{2})} = e(3+2\sqrt{2})$.

Therefore, $e^{3 \beta^{-1}}=e(3+2 \sqrt{2})$

Step 7: Match with the options given

The final solution matches with option (C).

Common Mistakes & Tips

• Sign Errors: Be careful with negative signs when substituting and simplifying the differential equation.
• Integration: Remember the constant of integration and use the initial condition to find its value.
• Hyperbolic Functions: Knowing the relationship between inverse hyperbolic functions and logarithms can be helpful.

Summary

We solved the Bernoulli differential equation by using a suitable substitution to transform it into a linear first-order differential equation. We then found the integrating factor and solved the resulting linear equation. Finally, we applied the given initial condition to find the constant of integration and determined the value of $\beta$ using the given boundary condition. The final result $e^{3 \beta^{-1}}=e(3+2 \sqrt{2})$ matched with option (C).

Final Answer

The final answer is \boxed{e^{3 \beta^{-1}}=e(3+2 \sqrt{2})}, which corresponds to option (C).