Problem 5
Question
Verify that the given function \(y\) satisfies the given differential equation. In each expression for \(y(x)\) the letter \(C\) denotes a constant. $$ \frac{d y}{d x}=x+y, y(x)=C e^{x}-x-1 $$
Step-by-Step Solution
Verified Answer
Yes, the function satisfies the differential equation.
1Step 1: Find the Derivative of y(x)
To verify that the function satisfies the differential equation, we first need to find the derivative of the function \( y(x) = Ce^x - x - 1 \). Using the rules of differentiation, the derivative is \( \frac{d}{dx}(y) = Ce^x - 1 \) since the derivative of \(-x\) is \(-1\) and the derivative of \(-1\) is \(0\).
2Step 2: Substitute y(x) into the Differential Equation
Next, we substitute \( y(x) = Ce^x - x - 1 \) into the given differential equation \( \frac{dy}{dx} = x + y \). So, we have: \( x + y = x + Ce^x - x - 1 \). This simplifies to \( Ce^x - 1 \).
3Step 3: Compare with Computed Derivative
Now, we compare the expression from step 2 with the derivative found in step 1. Both expressions, \( Ce^x - 1 \), are identical. This confirms that the derivative of \( y(x) \) matches the right-hand side of the differential equation with the substitution of \( y(x) \).
Key Concepts
Verification SolutionDifferentiationInitial Condition
Verification Solution
Solving a differential equation often requires verifying that a proposed solution is indeed correct. This involves ensuring that a given function satisfies the original differential equation. Verification provides confidence in the solution's accuracy.
When tasked to verify, you start by expressing the function clearly. For instance, with a function like \( y(x) = Ce^x - x - 1 \), identifying this correctly as a candidate solution is crucial.
You then need to perform necessary operations to demonstrate that substituting this function back into the original differential equation yields a true statement. This often includes differentiation and substitution. In our example, after differentiating \( y(x) \), the form \( Ce^x - 1 \) arises. Substituting \( y(x) \) into the right side of the differential equation shows that the left-hand side matches the right-hand side.
This is a stark indication that our original function is indeed a solution to the differential equation given. When both sides of the equation equate, it verifies the solution's validity.
When tasked to verify, you start by expressing the function clearly. For instance, with a function like \( y(x) = Ce^x - x - 1 \), identifying this correctly as a candidate solution is crucial.
You then need to perform necessary operations to demonstrate that substituting this function back into the original differential equation yields a true statement. This often includes differentiation and substitution. In our example, after differentiating \( y(x) \), the form \( Ce^x - 1 \) arises. Substituting \( y(x) \) into the right side of the differential equation shows that the left-hand side matches the right-hand side.
This is a stark indication that our original function is indeed a solution to the differential equation given. When both sides of the equation equate, it verifies the solution's validity.
Differentiation
Differentiation is a fundamental mathematical tool used to find the rate at which a function is changing at any given point. It's especially vital in solving or verifying differential equations.
In the context of differential equations, we often start with a function, like \( y(x) = Ce^x - x - 1 \).
The process of differentiation involves finding the derivative of this function with respect to \( x \), noted as \( \frac{dy}{dx} \). Using differentiation rules, the derivative of \( Ce^x \) is \( Ce^x \), the derivative of \(-x\) is \(-1\), and the derivative of a constant \(-1\) is 0.
Therefore, the complete derivative in this example is \( Ce^x - 1 \). Differentiation allows us to express how the function \( y(x) \) changes as \( x \) changes, which is crucial for analyzing and solving differential equations.
In the context of differential equations, we often start with a function, like \( y(x) = Ce^x - x - 1 \).
The process of differentiation involves finding the derivative of this function with respect to \( x \), noted as \( \frac{dy}{dx} \). Using differentiation rules, the derivative of \( Ce^x \) is \( Ce^x \), the derivative of \(-x\) is \(-1\), and the derivative of a constant \(-1\) is 0.
Therefore, the complete derivative in this example is \( Ce^x - 1 \). Differentiation allows us to express how the function \( y(x) \) changes as \( x \) changes, which is crucial for analyzing and solving differential equations.
Initial Condition
Initial conditions in the context of differential equations offer essential information that enables the solution to be uniquely determined. It refers to the values of the solution and its derivatives at a specific point.
Typically, an initial condition can specify \( y(x_0) = y_0 \), which is where \( y \) assumes a particular value when \( x = x_0 \).
While our current exercise doesn’t specify an initial condition, such conditions are often essential for pinpointing one particular solution out of many possible general solutions. For example, if we were given that \( y(0) = 2 \), we could substitute \( x = 0 \) and \( y = 2 \) into the solution \( y(x) = Ce^x - x - 1 \) to solve for \( C \).
Typically, an initial condition can specify \( y(x_0) = y_0 \), which is where \( y \) assumes a particular value when \( x = x_0 \).
While our current exercise doesn’t specify an initial condition, such conditions are often essential for pinpointing one particular solution out of many possible general solutions. For example, if we were given that \( y(0) = 2 \), we could substitute \( x = 0 \) and \( y = 2 \) into the solution \( y(x) = Ce^x - x - 1 \) to solve for \( C \).
- This ensures the solution is specifically tailored to the given condition.
Other exercises in this chapter
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