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 (IF): For a linear first-order differential equation, the integrating factor is given by $\text{IF} = e^{\int P(x) \, dx}$.
• General Solution: The general solution of a linear first-order differential equation is given by $y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \, dx + C$, where $C$ is the constant of integration.

Step-by-Step Solution

Step 1: Transform the Given Differential Equation into Standard Linear Form

We are given the differential equation $(t+1) dx = (2x + (t+1)^4) dt$. Our goal is to rewrite this equation in the standard linear form $\frac{dx}{dt} + P(t)x = Q(t)$.

• Divide both sides by $dt$ to isolate $\frac{dx}{dt}$: $(t+1) \frac{dx}{dt} = 2x + (t+1)^4$
• Divide both sides by $(t+1)$ to make the coefficient of $\frac{dx}{dt}$ equal to 1: $\frac{dx}{dt} = \frac{2x}{t+1} + \frac{(t+1)^4}{t+1}$
• Simplify the equation: $\frac{dx}{dt} = \frac{2x}{t+1} + (t+1)^3$
• Rearrange the equation to match the standard form: $\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$
• Identify $P(t)$ and $Q(t)$: Comparing with the standard form, we have: $P(t) = -\frac{2}{t+1}$ $Q(t) = (t+1)^3$

Step 2: Calculate the Integrating Factor (IF)

The integrating factor is given by $\text{IF} = e^{\int P(t) \, dt}$.

• Substitute $P(t)$: $\text{IF} = e^{\int -\frac{2}{t+1} \, dt}$
• Evaluate the integral: $\int -\frac{2}{t+1} \, dt = -2 \int \frac{1}{t+1} \, dt = -2 \ln|t+1| = -2 \ln(t+1)$ Since we are looking for $x(1)$ given $x(0) = 2$, we can assume $t+1>0$ and drop the absolute value.
• Simplify the integrating factor using logarithm and exponential properties: $\text{IF} = e^{-2 \ln(t+1)} = e^{\ln(t+1)^{-2}} = (t+1)^{-2} = \frac{1}{(t+1)^2}$

Step 3: Find the General Solution

The general solution is given by $x \cdot \text{IF} = \int Q(t) \cdot \text{IF} \, dt + C$.

• Substitute $x$, IF, and $Q(t)$: $x \cdot \frac{1}{(t+1)^2} = \int (t+1)^3 \cdot \frac{1}{(t+1)^2} \, dt + C$
• Simplify the integrand: $\frac{x}{(t+1)^2} = \int (t+1) \, dt + C$
• Perform the integration: $\int (t+1) \, dt = \frac{t^2}{2} + t$
• Write the general solution: $\frac{x}{(t+1)^2} = \frac{t^2}{2} + t + C$
• Express $x(t)$ explicitly: $x(t) = \left(\frac{t^2}{2} + t + C\right)(t+1)^2$

Step 4: Apply the Initial Condition to Find C

We are given the initial condition $x(0) = 2$.

• Substitute $t=0$ and $x=2$: $2 = \left(\frac{0^2}{2} + 0 + C\right)(0+1)^2$
• Solve for $C$: $2 = (0 + 0 + C)(1)^2$ $2 = C$
• Write the particular solution: $x(t) = \left(\frac{t^2}{2} + t + 2\right)(t+1)^2$

Step 5: Calculate $x(1)$

We need to find the value of $x(1)$.

• Substitute $t=1$: $x(1) = \left(\frac{1^2}{2} + 1 + 2\right)(1+1)^2$
• Perform the arithmetic: $x(1) = \left(\frac{1}{2} + 1 + 2\right)(2)^2$ $x(1) = \left(\frac{1}{2} + 3\right)(4)$ $x(1) = \left(\frac{7}{2}\right)(4)$ $x(1) = 14$

Common Mistakes & Tips:

• Sign of P(t): Be careful with the sign of $P(t)$ when identifying it from the standard form. A wrong sign will lead to an incorrect integrating factor and ultimately a wrong answer.
• Constant of Integration: Always remember to add the constant of integration, $C$, after evaluating the integral in the general solution.
• Simplification: Simplify the expressions as much as possible before integrating and substituting values to avoid arithmetic errors.

Summary

We solved the given first-order linear differential equation by first transforming it into the standard form, then finding the integrating factor. We used the integrating factor to find the general solution and applied the initial condition to determine the constant of integration, resulting in the particular solution. Finally, we evaluated the particular solution at $t=1$ to find $x(1)$. The final answer is 14.

The final answer is \boxed{14}.