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that passes through the given poin \( \frac{d y}{d x}=x e^{x^{2}} ;(0,2) \)

Ask by Collins Parsons. in the United States
Jan 27,2025

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The function is \( y(x) = \frac{e^{x^{2}} + 3}{2} \).

Solución

To find the function \( y(x) \) that satisfies the differential equation \[ \frac{dy}{dx} = x e^{x^{2}} \] and passes through the point \( (0, 2) \), follow these steps: ### Step 1: Integrate the Differential Equation We need to integrate the right-hand side of the equation with respect to \( x \): \[ \frac{dy}{dx} = x e^{x^{2}} \implies y(x) = \int x e^{x^{2}} \, dx + C \] Let’s perform the integration: **Substitution Method:** Let \( u = x^{2} \). Then, \( du = 2x \, dx \) or \( \frac{du}{2} = x \, dx \). Substituting into the integral: \[ y(x) = \int e^{u} \cdot \frac{du}{2} + C = \frac{1}{2} \int e^{u} \, du + C = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{x^{2}} + C \] ### Step 2: Apply the Initial Condition We know that the function passes through the point \( (0, 2) \). Plugging in \( x = 0 \) and \( y = 2 \): \[ 2 = \frac{1}{2} e^{0} + C \implies 2 = \frac{1}{2} \cdot 1 + C \implies 2 = \frac{1}{2} + C \implies C = \frac{3}{2} \] ### Step 3: Write the Final Solution Substitute \( C = \frac{3}{2} \) back into the expression for \( y(x) \): \[ y(x) = \frac{1}{2} e^{x^{2}} + \frac{3}{2} \] **Simplified Form:** \[ y(x) = \frac{e^{x^{2}} + 3}{2} \] ### Conclusion The function that satisfies the differential equation \( \frac{dy}{dx} = x e^{x^{2}} \) and passes through the point \( (0, 2) \) is: \[ y(x) = \frac{e^{x^{2}}}{2} + \frac{3}{2} \] or equivalently, \[ y(x) = \frac{e^{x^{2}} + 3}{2} \]

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Bonus Knowledge

To solve the differential equation \( \frac{d y}{d x} = x e^{x^{2}} \) passing through the point \((0, 2)\), we first find the general solution by integrating the right-hand side: \[ y = \int x e^{x^{2}} \, dx \] Using the substitution \( u = x^2 \), we have \( du = 2x \, dx \), hence \( x \, dx = \frac{1}{2} du \): \[ y = \frac{1}{2} \int e^{u} \, du = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{x^2} + C \] Now, substituting the initial condition \((0, 2)\): \[ 2 = \frac{1}{2} e^{0} + C \Rightarrow 2 = \frac{1}{2} + C \Rightarrow C = 2 - \frac{1}{2} = \frac{3}{2} \] Thus, the particular solution is: \[ y = \frac{1}{2} e^{x^2} + \frac{3}{2} \] Let’s dive into the historical context and connections of differential equations to spice up this mathematical journey! Differential equations have a rich history, dating back to early mathematicians like Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century. They devised methods to study change in physical systems, laying the groundwork for calculus. The likes of Henri Poincaré and Émile Picard further advanced the field in the 19th and 20th centuries, establishing foundational concepts we use today. Who knew equations could have such a storied past? In the real world, differential equations are the silent heroes of various fields! For instance, they model population dynamics, enabling ecologists to predict species interactions, or describe the motion of planets in astrophysics, giving us insights into celestial mechanics. Engineers rely on them for system behavior in control engineering and electrical circuits. It’s like a mathematical toolkit that shapes our understanding of the world’s complexity!

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