Key Concepts and Formulas

• Linear First-Order 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 linear first-order differential equation, the integrating factor is given by $I.F. = e^{\int P(x) dx}$.
• Solution of Linear First-Order Differential Equation: The solution 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: Rewrite the given differential equation.

We are given the differential equation: $(4 + x^2)dy - 2x(x^2 + 3y + 4)dx = 0$ We want to rewrite it in the form $\frac{dy}{dx} + P(x)y = Q(x)$. First, divide both sides by $(4+x^2)dx$: $\frac{dy}{dx} - \frac{2x(x^2 + 3y + 4)}{4 + x^2} = 0$ $\frac{dy}{dx} = \frac{2x(x^2 + 3y + 4)}{4 + x^2}$ $\frac{dy}{dx} = \frac{2x^3 + 6xy + 8x}{4 + x^2}$ $\frac{dy}{dx} = \frac{2x^3 + 8x}{4 + x^2} + \frac{6xy}{4 + x^2}$ $\frac{dy}{dx} - \frac{6x}{4 + x^2}y = \frac{2x^3 + 8x}{4 + x^2}$

Step 2: Identify P(x) and Q(x).

Comparing the equation with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we have: $P(x) = -\frac{6x}{4 + x^2}$ $Q(x) = \frac{2x^3 + 8x}{4 + x^2} = \frac{2x(x^2 + 4)}{4 + x^2} = 2x$

Step 3: Calculate the Integrating Factor (I.F.).

The integrating factor is given by: $I.F. = e^{\int P(x) dx} = e^{\int -\frac{6x}{4 + x^2} dx}$ Let $u = 4 + x^2$, then $du = 2x dx$. Thus, $\int -\frac{6x}{4 + x^2} dx = -3 \int \frac{2x}{4 + x^2} dx = -3 \int \frac{du}{u} = -3 \ln|u| = -3 \ln(4 + x^2) = \ln(4 + x^2)^{-3}$ Therefore, $I.F. = e^{\ln(4 + x^2)^{-3}} = (4 + x^2)^{-3} = \frac{1}{(4 + x^2)^3}$

Step 4: Find the general solution.

The general solution is given by: $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$ $y \cdot \frac{1}{(4 + x^2)^3} = \int 2x \cdot \frac{1}{(4 + x^2)^3} dx + C$ Let $u = 4 + x^2$, then $du = 2x dx$. Thus, $\int \frac{2x}{(4 + x^2)^3} dx = \int \frac{1}{u^3} du = \int u^{-3} du = \frac{u^{-2}}{-2} + C_1 = -\frac{1}{2u^2} + C_1 = -\frac{1}{2(4 + x^2)^2} + C_1$ So, $\frac{y}{(4 + x^2)^3} = -\frac{1}{2(4 + x^2)^2} + C$ $y = -\frac{(4 + x^2)^3}{2(4 + x^2)^2} + C(4 + x^2)^3$ $y = -\frac{1}{2}(4 + x^2) + C(4 + x^2)^3$

Step 5: Apply the initial condition y(0) = 0.

The solution curve passes through the origin, so $y(0) = 0$. Substituting $x = 0$ and $y = 0$ into the general solution: $0 = -\frac{1}{2}(4 + 0^2) + C(4 + 0^2)^3$ $0 = -\frac{1}{2}(4) + C(4)^3$ $0 = -2 + 64C$ $64C = 2$ $C = \frac{2}{64} = \frac{1}{32}$

Step 6: Find the particular solution.

Substituting $C = \frac{1}{32}$ into the general solution: $y = -\frac{1}{2}(4 + x^2) + \frac{1}{32}(4 + x^2)^3$

Step 7: Calculate y(2).

We want to find $y(2)$. Substitute $x = 2$ into the particular solution: $y(2) = -\frac{1}{2}(4 + 2^2) + \frac{1}{32}(4 + 2^2)^3$ $y(2) = -\frac{1}{2}(4 + 4) + \frac{1}{32}(4 + 4)^3$ $y(2) = -\frac{1}{2}(8) + \frac{1}{32}(8)^3$ $y(2) = -4 + \frac{1}{32}(512)$ $y(2) = -4 + 16$ $y(2) = 12$

There seems to be an error in the provided answer. Let's recheck the calculations from Step 3 onwards.

Step 3 (Revised): Calculate the Integrating Factor (I.F.).

The integrating factor is given by: $I.F. = e^{\int P(x) dx} = e^{\int -\frac{6x}{4 + x^2} dx}$ Let $u = 4 + x^2$, then $du = 2x dx$. Thus, $\int -\frac{6x}{4 + x^2} dx = -3 \int \frac{2x}{4 + x^2} dx = -3 \int \frac{du}{u} = -3 \ln|u| = -3 \ln(4 + x^2) = \ln(4 + x^2)^{-3}$ Therefore, $I.F. = e^{\ln(4 + x^2)^{-3}} = (4 + x^2)^{-3} = \frac{1}{(4 + x^2)^3}$

This is correct.

Step 4 (Revised): Find the general solution.

The general solution is given by: $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$ $y \cdot \frac{1}{(4 + x^2)^3} = \int 2x \cdot \frac{1}{(4 + x^2)^3} dx + C$ Let $u = 4 + x^2$, then $du = 2x dx$. Thus, $\int \frac{2x}{(4 + x^2)^3} dx = \int \frac{1}{u^3} du = \int u^{-3} du = \frac{u^{-2}}{-2} + C_1 = -\frac{1}{2u^2} + C_1 = -\frac{1}{2(4 + x^2)^2} + C_1$ So, $\frac{y}{(4 + x^2)^3} = -\frac{1}{2(4 + x^2)^2} + C$ $y = -\frac{(4 + x^2)^3}{2(4 + x^2)^2} + C(4 + x^2)^3$ $y = -\frac{1}{2}(4 + x^2) + C(4 + x^2)^3$

This is correct.

Step 5 (Revised): Apply the initial condition y(0) = 0.

The solution curve passes through the origin, so $y(0) = 0$. Substituting $x = 0$ and $y = 0$ into the general solution: $0 = -\frac{1}{2}(4 + 0^2) + C(4 + 0^2)^3$ $0 = -\frac{1}{2}(4) + C(4)^3$ $0 = -2 + 64C$ $64C = 2$ $C = \frac{2}{64} = \frac{1}{32}$

This is correct.

Step 6 (Revised): Find the particular solution.

Substituting $C = \frac{1}{32}$ into the general solution: $y = -\frac{1}{2}(4 + x^2) + \frac{1}{32}(4 + x^2)^3$

This is correct.

Step 7 (Revised): Calculate y(2).

We want to find $y(2)$. Substitute $x = 2$ into the particular solution: $y(2) = -\frac{1}{2}(4 + 2^2) + \frac{1}{32}(4 + 2^2)^3$ $y(2) = -\frac{1}{2}(4 + 4) + \frac{1}{32}(4 + 4)^3$ $y(2) = -\frac{1}{2}(8) + \frac{1}{32}(8)^3$ $y(2) = -4 + \frac{1}{32}(512)$ $y(2) = -4 + 16$ $y(2) = 12$

The calculation remains the same. The solution $y(2) = 12$ does not match the provided answer of 4. Let's review the original differential equation for any errors.

The differential equation is $(4 + x^2)dy - 2x(x^2 + 3y + 4)dx = 0$. We rewrite it as $\frac{dy}{dx} = \frac{2x(x^2 + 3y + 4)}{4 + x^2}$. $\frac{dy}{dx} = \frac{2x^3 + 6xy + 8x}{4 + x^2}$. $\frac{dy}{dx} - \frac{6x}{4 + x^2}y = \frac{2x^3 + 8x}{4 + x^2} = \frac{2x(x^2 + 4)}{4 + x^2} = 2x$.

The equation is indeed $\frac{dy}{dx} - \frac{6x}{4 + x^2}y = 2x$.

$P(x) = -\frac{6x}{4 + x^2}$, $Q(x) = 2x$. $I.F. = e^{\int P(x)dx} = e^{\int -\frac{6x}{4 + x^2}dx} = e^{-3\ln(4 + x^2)} = (4 + x^2)^{-3}$. $y(I.F.) = \int Q(x)(I.F.) dx + C$. $y(4 + x^2)^{-3} = \int 2x(4 + x^2)^{-3} dx + C$. $y(4 + x^2)^{-3} = \int (4 + x^2)^{-3} d(4 + x^2)/1 + C$. $y(4 + x^2)^{-3} = \frac{(4 + x^2)^{-2}}{-2} + C$. $y = -\frac{1}{2}(4 + x^2) + C(4 + x^2)^3$. $y(0) = 0 = -\frac{1}{2}(4) + C(4)^3$. $0 = -2 + 64C$. $C = \frac{2}{64} = \frac{1}{32}$. $y = -\frac{1}{2}(4 + x^2) + \frac{1}{32}(4 + x^2)^3$. $y(2) = -\frac{1}{2}(8) + \frac{1}{32}(8)^3 = -4 + \frac{512}{32} = -4 + 16 = 12$.

There appears to be an error with the given correct answer. The correct calculation yields $y(2) = 12$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when integrating and applying the integrating factor. A small sign error can propagate through the entire solution.
• Integration Techniques: Make sure you are comfortable with u-substitution and other integration techniques necessary to find the integral of $Q(x) \cdot (I.F.)$.
• Checking the Solution: After finding the particular solution, it's a good practice to plug it back into the original differential equation to verify that it satisfies the equation and the initial condition.

Summary

We solved the given first-order differential equation by identifying it as a linear first-order differential equation. We calculated the integrating factor, found the general solution, and then used the initial condition $y(0) = 0$ to find the particular solution. Finally, we evaluated $y(2)$ using the particular solution. The calculated value is $y(2) = 12$.

Final Answer

The final answer is \boxed{12}.