Chapter 2: Problem 53
In Problems 51-54 find a continuous solution satisfying the given differential equation and the indicated initial condition. Use a graphing utility to graph the solution curve. $$ \frac{d y}{d x}+2 x y=f(x), \quad f(x)=\left\\{\begin{array}{ll} x, & 0 \leq x<1 \\ 0, & x \geq 1 \end{array}, \quad y(0)=2\right. $$
Short Answer
Step by step solution
Understanding the Differential Equation
Solving for 0 ≤ x < 1
Integrating for 0 ≤ x < 1
Applying Initial Condition for 0 ≤ x < 1
Solving for x ≥ 1
Simplifying the Solution for x ≥ 1
Continuous Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Function
- \( f(x) = x \) for \( 0 \leq x < 1 \)
- \( f(x) = 0 \) for \( x \geq 1 \)
Integrating Factor
\[\mu(x) \frac{dy}{dx} + \mu(x)P(x) y = \mu(x)Q(x)\]
Here, the left-hand side becomes the derivative of \( \mu(x)y \). This makes the equation easy to integrate with respect to \( x \). In our case, for \( 0 \leq x < 1 \), the integrating factor is \( \mu(x) = e^{x^2} \). It turns our differential equation into a solvable form, leading to a solution valid for \( 0 \leq x < 1 \). By applying the integrating factor, we ensure that the solution process is streamlined and efficient.
Separable Differential Equation
- \( \frac{dy}{y} = -2x \, dx \)
The power of the separable differential equation method is its simplicity, allowing us to directly isolate and integrate the variables, resulting in straightforward solutions under appropriate conditions.