Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (IF): For a first-order linear differential equation, the integrating factor is given by $IF = e^{\int P(x) dx}$.
• General Solution: The general solution of a first-order linear differential equation is given by $y \cdot IF = \int Q(x) \cdot IF \, dx + C$.

Step-by-Step Solution

Step 1: Rewrite the Differential Equation in Standard Linear Form

We are given the differential equation $(1-x^2)dy = [xy + (x^3+2)\sqrt{3(1-x^2)}]dx$. Our goal is to rewrite this in the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Divide both sides by $(1-x^2)dx$: $\frac{dy}{dx} = \frac{xy}{1-x^2} + \frac{(x^3+2)\sqrt{3(1-x^2)}}{1-x^2}$

Rearrange to isolate $\frac{dy}{dx}$ and the $y$ term on the left: $\frac{dy}{dx} - \frac{x}{1-x^2}y = \frac{(x^3+2)\sqrt{3(1-x^2)}}{1-x^2}$

Simplify the right-hand side. Since $1-x^2 > 0$, we have $\sqrt{1-x^2} > 0$, and we can write $1-x^2 = (\sqrt{1-x^2})^2$: $\frac{dy}{dx} - \frac{x}{1-x^2}y = \frac{(x^3+2)\sqrt{3}\sqrt{1-x^2}}{(\sqrt{1-x^2})^2} = \frac{\sqrt{3}(x^3+2)}{\sqrt{1-x^2}}$

Comparing this to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify: $P(x) = -\frac{x}{1-x^2}$ $Q(x) = \frac{\sqrt{3}(x^3+2)}{\sqrt{1-x^2}}$

Step 2: Calculate the Integrating Factor (IF)

The integrating factor is $IF = e^{\int P(x) dx}$. We need to compute $\int P(x) dx = \int -\frac{x}{1-x^2} dx$.

Let $u = 1-x^2$. Then $du = -2x \, dx$, so $-\frac{1}{2}du = x \, dx$. $\int -\frac{x}{1-x^2} dx = \int \frac{1}{2u} du = \frac{1}{2} \ln|u| + C' = \frac{1}{2} \ln(1-x^2) + C'$ Since $-1 < x < 1$, $1-x^2 > 0$, so we can drop the absolute value. Also, we can ignore the constant $C'$ when computing the integrating factor. $\int P(x) dx = \frac{1}{2} \ln(1-x^2) = \ln(\sqrt{1-x^2})$

Therefore, the integrating factor is: $IF = e^{\ln(\sqrt{1-x^2})} = \sqrt{1-x^2}$

Step 3: Find the General Solution

The general solution is given by $y \cdot IF = \int Q(x) \cdot IF \, dx + C$. Substituting the expressions for $IF$ and $Q(x)$: $y\sqrt{1-x^2} = \int \frac{\sqrt{3}(x^3+2)}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} \, dx + C$

The $\sqrt{1-x^2}$ terms cancel: $y\sqrt{1-x^2} = \int \sqrt{3}(x^3+2) \, dx + C$

Integrate the right-hand side: $y\sqrt{1-x^2} = \sqrt{3} \left(\frac{x^4}{4} + 2x\right) + C$

Step 4: Use the Initial Condition to Find the Constant C

We are given $y(0) = 0$. Substitute $x=0$ and $y=0$ into the general solution: $0 \cdot \sqrt{1-0^2} = \sqrt{3} \left(\frac{0^4}{4} + 2(0)\right) + C$ $0 = 0 + C$ $C = 0$

Step 5: Write the Particular Solution and Evaluate $y\left(\frac{1}{2}\right)$

Since $C=0$, the particular solution is: $y\sqrt{1-x^2} = \sqrt{3} \left(\frac{x^4}{4} + 2x\right)$

Now, we evaluate $y\left(\frac{1}{2}\right)$. Substitute $x=\frac{1}{2}$ into the particular solution: $y\left(\frac{1}{2}\right) \sqrt{1-\left(\frac{1}{2}\right)^2} = \sqrt{3} \left(\frac{\left(\frac{1}{2}\right)^4}{4} + 2\left(\frac{1}{2}\right)\right)$ $y\left(\frac{1}{2}\right) \sqrt{1-\frac{1}{4}} = \sqrt{3} \left(\frac{\frac{1}{16}}{4} + 1\right)$ $y\left(\frac{1}{2}\right) \sqrt{\frac{3}{4}} = \sqrt{3} \left(\frac{1}{64} + 1\right)$ $y\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2} = \sqrt{3} \left(\frac{65}{64}\right)$ $y\left(\frac{1}{2}\right) = \frac{2}{\sqrt{3}} \cdot \sqrt{3} \cdot \frac{65}{64} = 2 \cdot \frac{65}{64} = \frac{65}{32}$

Therefore, $y\left(\frac{1}{2}\right) = \frac{65}{32}$. We are given that $y\left(\frac{1}{2}\right) = \frac{m}{n}$, where $m$ and $n$ are coprime. Thus, $m=65$ and $n=32$. Therefore, $m+n = 65+32 = 97$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when finding $P(x)$ and $Q(x)$. A small sign error can propagate through the entire solution.
• Forgetting the Constant of Integration: Always remember to add the constant of integration $C$ after performing the integration in the general solution.
• Simplifying Expressions: Take the time to simplify expressions, especially when dealing with square roots and fractions. It makes the calculations easier and reduces the chance of errors.

Summary

We solved a first-order linear differential equation by rewriting it in standard form, finding the integrating factor, and then using the integrating factor to find the general solution. We used the initial condition to determine the constant of integration, which gave us the particular solution. Finally, we evaluated the particular solution at $x=\frac{1}{2}$ to find $y\left(\frac{1}{2}\right) = \frac{65}{32}$. Since $m=65$ and $n=32$, we have $m+n = 65+32 = 97$. The problem asks for $m+n$. There must be an error in the given correct answer. Let's recalculate everything.

$y\sqrt{1-x^2} = \sqrt{3} \left(\frac{x^4}{4} + 2x\right)$ $y\left(\frac{1}{2}\right) \sqrt{1-\left(\frac{1}{2}\right)^2} = \sqrt{3} \left(\frac{\left(\frac{1}{2}\right)^4}{4} + 2\left(\frac{1}{2}\right)\right)$ $y\left(\frac{1}{2}\right) \sqrt{1-\frac{1}{4}} = \sqrt{3} \left(\frac{\frac{1}{16}}{4} + 1\right)$ $y\left(\frac{1}{2}\right) \sqrt{\frac{3}{4}} = \sqrt{3} \left(\frac{1}{64} + 1\right)$ $y\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2} = \sqrt{3} \left(\frac{65}{64}\right)$ $y\left(\frac{1}{2}\right) = \frac{2}{\sqrt{3}} \cdot \sqrt{3} \cdot \frac{65}{64} = 2 \cdot \frac{65}{64} = \frac{65}{32}$

Therefore, $y\left(\frac{1}{2}\right) = \frac{65}{32}$. We are given that $y\left(\frac{1}{2}\right) = \frac{m}{n}$, where $m$ and $n$ are coprime. Thus, $m=65$ and $n=32$. Therefore, $m+n = 65+32 = 97$.

The final answer is 97. However, the correct answer is given as 1. Let us go back to the original equation: $\frac{dy}{dx} - \frac{x}{1-x^2}y = \frac{\sqrt{3}(x^3+2)}{\sqrt{1-x^2}}$ $y\sqrt{1-x^2} = \sqrt{3} \left(\frac{x^4}{4} + 2x\right) + C$ $y(0) = 0 \implies C = 0$ $y\sqrt{1-x^2} = \sqrt{3} \left(\frac{x^4}{4} + 2x\right)$ $x=\frac{1}{2}$ $y\left(\frac{1}{2}\right) \sqrt{1-\frac{1}{4}} = \sqrt{3} \left(\frac{\frac{1}{16}}{4} + 1\right)$ $y\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2} = \sqrt{3} \left(\frac{1}{64} + 1\right)$ $y\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2} = \sqrt{3} \left(\frac{65}{64}\right)$ $y\left(\frac{1}{2}\right) = 2 \cdot \frac{65}{64} = \frac{65}{32}$ $m=65$, $n=32$, $m+n=97$.

It seems there is an error in the given answer.

The question is likely wrong. It is possible the initial condition is incorrect. If $y(1/2) = 1/1$, then $m=1, n=1$, $m+n = 2$. Suppose $y(1/2) = 1$. $1 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \left(\frac{1}{64} + C\right)$ $\frac{1}{2} = \frac{1}{64} + C$ $C = \frac{31}{64}$ If the answer were 1, then we need $m+n=1$, which is not possible.

Let's assume the initial condition is wrong. If the answer is 1, then $y(1/2) = 1$.

The question is incorrect.

Final Answer

The final answer is \boxed{97}. There is likely an error in the given correct answer. There is no way to arrive at the correct answer of 1 based on the given information.