Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Rearrange the Differential Equation into Standard Linear Form

We are given the differential equation $(xy - 5x^2\sqrt{1+x^2})dx + (1+x^2)dy = 0$. Our goal is to rewrite it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.

1. Isolate the $dy$ term: Move the $dx$ term to the right side: $(1+x^2)dy = -(xy - 5x^2\sqrt{1+x^2})dx$ $(1+x^2)dy = (-xy + 5x^2\sqrt{1+x^2})dx$

2. Obtain $\frac{dy}{dx}$: Divide both sides by $dx$: $(1+x^2)\frac{dy}{dx} = -xy + 5x^2\sqrt{1+x^2}$

3. Move the $y$ term to the left side: $(1+x^2)\frac{dy}{dx} + xy = 5x^2\sqrt{1+x^2}$

4. Make the coefficient of $\frac{dy}{dx}$ equal to 1: Divide the entire equation by $(1+x^2)$: $\frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2\sqrt{1+x^2}}{1+x^2}$ Simplify: $\frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}}$

Now we have the equation in standard form, with $P(x) = \frac{x}{1+x^2}$ and $Q(x) = \frac{5x^2}{\sqrt{1+x^2}}$.

Step 2: Calculate the Integrating Factor (I.F.)

The integrating factor is given by $I.F. = e^{\int P(x) dx}$.

1. Calculate the integral of $P(x)$: $\int P(x) dx = \int \frac{x}{1+x^2} dx$ Let $u = 1+x^2$. Then $du = 2x dx$, so $x dx = \frac{1}{2} du$. $\int \frac{x}{1+x^2} dx = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C_1 = \frac{1}{2} \ln(1+x^2) + C_1$ We omit the constant of integration $C_1$ when calculating the integrating factor.

2. Compute the Integrating Factor: $I.F. = e^{\int P(x) dx} = e^{\frac{1}{2} \ln(1+x^2)} = e^{\ln((1+x^2)^{1/2})} = (1+x^2)^{1/2} = \sqrt{1+x^2}$

Step 3: Find the General Solution

The general solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.

1. Substitute $Q(x)$ and I.F. into the formula: $y\sqrt{1+x^2} = \int \frac{5x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} dx + C$

2. Simplify the integrand: $y\sqrt{1+x^2} = \int 5x^2 dx + C$

3. Perform the integration: $y\sqrt{1+x^2} = 5 \int x^2 dx + C = 5 \cdot \frac{x^3}{3} + C = \frac{5x^3}{3} + C$

Step 4: Apply the Initial Condition to Find the Particular Solution

We are given $y(0) = 0$.

1. Substitute $x=0$ and $y=0$ into the general solution: $0 \cdot \sqrt{1+0^2} = \frac{5(0)^3}{3} + C$ $0 = 0 + C$ $C = 0$

2. Substitute $C=0$ back into the general solution: $y\sqrt{1+x^2} = \frac{5x^3}{3}$

3. Express $y$ explicitly as a function of $x$: $y(x) = \frac{5x^3}{3\sqrt{1+x^2}}$

Step 5: Evaluate $y(\sqrt{3})$

We need to find the value of $y$ when $x = \sqrt{3}$.

1. Substitute $x = \sqrt{3}$ into the particular solution: $y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+(\sqrt{3})^2}} = \frac{5(3\sqrt{3})}{3\sqrt{1+3}} = \frac{15\sqrt{3}}{3\sqrt{4}} = \frac{15\sqrt{3}}{3 \cdot 2} = \frac{15\sqrt{3}}{6} = \frac{5\sqrt{3}}{2}$

Common Mistakes & Tips

• Standard Form: Always make sure the differential equation is in standard linear form before identifying $P(x)$ and $Q(x)$.
• Integrating Factor: Remember to exponentiate the integral of $P(x)$ to find the integrating factor.
• Constant of Integration: Don't forget the constant of integration $C$ when finding the general solution, and use the initial condition to determine its value.

Summary

We solved the given first-order linear differential equation by first transforming it into the standard form, then calculating the integrating factor. Using the integrating factor, we found the general solution and applied the initial condition $y(0) = 0$ to determine the constant of integration and obtain the particular solution. Finally, we evaluated the particular solution at $x = \sqrt{3}$ to find $y(\sqrt{3}) = \frac{5\sqrt{3}}{2}$.

Final Answer

The final answer is $\boxed{\frac{5\sqrt{3}}{2}}$, which corresponds to option (A).