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Show that each function \(y=f(x)\) is a solution of the accompanying differential equation. $$y=\frac{1}{x} \int_{1}^{x^{\prime}} \frac{e^{t}}{t} d t, \quad x^{2} y^{\prime}+x y=e^{x}$$

Short Answer

Expert verified
The function satisfies the differential equation.

Step by step solution

01

Differentiate the Function

Given the function \( y = \frac{1}{x} \int_{1}^{x} \frac{e^{t}}{t} \, dt \), we need to find its derivative \( y' \). Start by using the product rule on terms \( \frac{1}{x} \) and \( \int_{1}^{x} \frac{e^{t}}{t} \, dt \). Let \( u = \frac{1}{x} \) and \( v = \int_{1}^{x} \frac{e^{t}}{t} \, dt \). The product rule gives \( (uv)' = u'v + uv' \). Calculate \( u' = -\frac{1}{x^2} \), and use the Fundamental Theorem of Calculus to obtain \( v' = \frac{e^{x}}{x} \). Substitute these expressions into the derivative formula: \[ y' = \left(-\frac{1}{x^2}\right)\left(\int_{1}^{x}\frac{e^{t}}{t} \, dt\right) + \left(\frac{1}{x}\right)\left(\frac{e^{x}}{x}\right) \] Simplify to get: \[ y' = -\frac{1}{x^2} \int_{1}^{x} \frac{e^{t}}{t} \, dt + \frac{e^{x}}{x^2} \]
02

Substitute into the Differential Equation

Now that we have the derivative \( y' \), substitute \( y \) and \( y' \) back into the differential equation \( x^2 y' + x y = e^{x} \). Substituting \( y' = -\frac{1}{x^2} \int_{1}^{x} \frac{e^{t}}{t} \, dt + \frac{e^{x}}{x^2} \) into \( x^2 y' \) gives: \[ x^2 y' = x^2 \left(-\frac{1}{x^2} \int_{1}^{x} \frac{e^{t}}{t} \, dt + \frac{e^{x}}{x^2}\right) = -\int_{1}^{x} \frac{e^{t}}{t} \, dt + e^{x} \] Substituting \( y = \frac{1}{x} \int_{1}^{x} \frac{e^{t}}{t} \, dt \) into \( x y \) gives: \[ x y = \int_{1}^{x} \frac{e^{t}}{t} \, dt \] Adding these results: \[ x^2 y' + x y = -\int_{1}^{x} \frac{e^{t}}{t} \, dt + e^{x} + \int_{1}^{x} \frac{e^{t}}{t} \, dt = e^{x} \]
03

Conclusion

The terms \(-\int_{1}^{x} \frac{e^{t}}{t} \, dt\) and \(+\int_{1}^{x} \frac{e^{t}}{t} \, dt\) cancel each other out, leaving us with \(e^{x}\). This confirms that \(x^2 y' + x y = e^{x}\), so the function \(y = \frac{1}{x} \int_{1}^{x} \frac{e^{t}}{t} \, dt\) satisfies the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule in Calculus
The product rule is a fundamental concept in calculus used to differentiate products of two functions. Suppose you have two functions, let's call them \(u\) and \(v\), which are both differentiable. When taking the derivative of the product \(u \cdot v\), you can't simply take the derivative of each function and multiply those. Instead, you use the product rule, which states:
  • \((uv)' = u'v + uv'\)
The derivative of the first function, \(u\), is multiplied by the second function \(v\), and added to the product of \(u\) with the derivative of \(v\).
This rule is crucial in solving the differential equation in this example, as it involves a product of \(\frac{1}{x}\) and an integral. By carefully applying the product rule, we are able to find the derivative of the function effectively, which is necessary for subsequent steps in confirming that it solves the differential equation given.
Understanding the Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of integration with differentiation. There are two main parts to this theorem:
  • The first part says that if you have an antiderivative \(F\) of a function \(f\), then the definite integral of \(f\) over any interval \([a, b]\) can be calculated as \(F(b) - F(a)\).
  • The second part focuses on differentiating an integral function. It states that if \(f\) is continuous over \([a, b]\), and you define \(F(x) = \int_a^x f(t) \, dt\), then \(F'(x) = f(x)\).
In this exercise, the theorem helps us differentiate the integral \(\int_{1}^{x} \frac{e^{t}}{t} \, dt\). By applying the theorem, you find that the derivative with respect to \(x\) is \(\frac{e^{x}}{x}\).
This result plays a pivotal role in solving the equation, as it provides the derivative of the integral portion of the function \(y\).
The Concept of Integration
Integration is a key process in calculus, essentially the inverse of differentiation. While differentiation breaks down a function into its rate of change or slope, integration combines or accumulates all of these rates to reconstruct the original function or find the area under a curve.
  • There are two primary types of integrals: indefinite and definite.
  • An indefinite integral results in a family of functions that differ by a constant, often represented as \(\int f(x) \, dx = F(x) + C\).
  • A definite integral, such as \(\int_a^b f(x) \, dx\), results in a numeric value representing the area under function \(f\) from \(a\) to \(b\).
In our example, integration is used within the function \(y\) to describe a cumulative process of the function \(\frac{e^t}{t}\) from 1 to \(x\).
The ability to understand and compute this integral is essential to both evaluating \(y\) in the differential equation and facilitating further differentiation.

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