AnswerQuestion: What is the solution to the differential equation $\frac{dy}{dx} = 2x$ with the initial condition $y(0) = 1$?

["Answer Question: Solving $\frac{dy}{dx} = 2x$ with $y(0) = 1$ – Step-by-Step Guide", "Differential equations are fundamental tools in mathematics, science, and engineering, helping to model change in real-world systems. One of the most common types is the first-order ordinary differential equation (ODE), where the derivative of an unknown function is known. A classic example is:", "$$\n\frac{dy}{dx} = 2x,\quad y(0) = 1\n$$", "This question asks: What is the solution to this differential equation given the initial condition?", "In this article, we’ll walk through the complete solution using fundamental techniques in calculus, explaining each step clearly for students, self-learners, and professionals seeking to strengthen their understanding.", "---", "### Understanding the Differential Equation", "The equation\n$$\n\frac{dy}{dx} = 2x\n$$\ntells us that the rate of change of $y$ with respect to $x$ is a straight line: $2x$. To find $y(x)$, we need to reconstruct the function whose slope at any point $x$ matches this expression.", "---", "### Step 1: Integrate Both Sides", "To solve such an equation, integration is the key operation. The general solution comes from integrating both sides with respect to $x$:", "$$\n\int \frac{dy}{dx} , dx = \int 2x , dx\n$$", "Since $\frac{dy}{dx}$ is treated as the derivative of $y$, integrating the left side adds $y$ as a constant:", "$$\ny = \int 2x , dx\n$$", "---", "### Step 2: Compute the Integral", "Compute the right-hand side:", "$$\n\int 2x , dx = 2 \cdot \frac{x^2}{2} + C = x^2 + C\n$$", "So, the general solution is:", "$$\ny(x) = x^2 + C\n$$", "Here, $C$ is the constant of integration, determined uniquely by the initial condition.", "---", "### Step 3: Apply the Initial Condition", "We are given $y(0) = 1$. Substitute $x = 0$ and $y = 1$ into the general solution:", "$$\n1 = (0)^2 + C \quad \Rightarrow \quad C = 1\n$$", "---", "### Step 4: Write the Final Solution", "Substitute $C = 1$ back into the general form:", "$$\ny(x) = x^2 + 1\n$$", "This is the unique solution satisfying both the differential equation and the initial condition.", "---", "### Why Is This Solution Correct?", "By differentiating $y(x) = x^2 + 1$, we get:", "$$\n\frac{dy}{dx} = 2x\n$$", "which matches the given ODE. At $x = 0$, $y(0) = 0^2 + 1 = 1$, satisfying the initial condition. No other function satisfies both criteria, ensuring uniqueness.", "---", "### Applications of the Solution", "This simple ODE models scenarios such as:", "- The position of an object under constant acceleration in physics\n- The area under a linear velocity function yielding position\n- Quadratic growth processes in economics or biology", "Understanding how to solve and interpret such equations is crucial for solving real-world problems in science and engineering.", "---", "### Summary", "The differential equation $\frac{dy}{dx} = 2x$ with initial condition $y(0) = 1$ leads to the solution:", "$$\n\boxed{y(x) = x^2 + 1}\n$$", "Mastering this type of problem builds a strong foundation for solving more complex differential equations and deepens comprehension of calculus applications. Use this step-by-step guide to confidently tackle similar problems in your learning journey.", "---", "Keywords: differential equation solution, $\frac{dy}{dx} = 2x$, initial condition $y(0) = 1$, calculus step-by-step, solving ODEs, first-order differential equation, integration to solve, tanh of tan activation not applicable, application of integrals, foundational math skills.", "---", "Further Reading: Explore first-order ODEs, boundary value problems, and applications in physics and engineering to reinforce your understanding."]