Chapter 6: Problem 39
Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. . \(\left\\{\begin{array}{l}y^{\prime}=x y \\ y(0)=-1\end{array}\right.\)
Short Answer
Expert verified
Solution: \( y = -e^{\frac{x^2}{2}} \) satisfies both the differential equation and the initial condition.
Step by step solution
01
Recognize the Type of Differential Equation
The given differential equation is \( y' = x y \). This is a first-order, linear, separable differential equation, where \( y' \) denotes the derivative of \( y \) with respect to \( x \).
02
Separate Variables
To solve the separable differential equation, rearrange the terms to isolate \( y \) and \( x \) on opposite sides: \( \frac{dy}{y} = x \, dx \).
03
Integrate Both Sides
Integrate both sides of the equation. The left side with respect to \( y \) and the right side with respect to \( x \): \( \int \frac{1}{y} \, dy = \int x \, dx \). This gives \( \ln|y| = \frac{x^2}{2} + C \), where \( C \) is the constant of integration.
04
Solve for \( y \)
To express \( y \) explicitly, exponentiate both sides of the equation to eliminate the natural logarithm: \( |y| = e^{\frac{x^2}{2} + C} \). This can be rewritten as \( y = A e^{\frac{x^2}{2}} \), where \( A = e^C \) is a constant.
05
Apply Initial Condition
Use the initial condition \( y(0) = -1 \) to find the constant \( A \). Substitute \( x = 0 \) and \( y = -1 \) into the equation: \(-1 = A e^0 \). Since \( e^0 = 1 \), we find \( A = -1 \). Therefore, \( y = -e^{\frac{x^2}{2}} \).
06
Verify the Solution
Verify that the function \( y = -e^{\frac{x^2}{2}} \) satisfies both the differential equation and the initial condition. The derivative \( y' = \frac{d}{dx}[-e^{\frac{x^2}{2}}] = -e^{\frac{x^2}{2}} \cdot x = x \cdot (-e^{\frac{x^2}{2}}) \), which matches \( y' = x y \). The initial condition \( y(0) = -1 \) holds true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Condition
In differential equations, an "initial condition" is a crucial component. It gives us specific information about the value of the function at a certain point. For example, for the differential equation \( y' = xy \), it is paired with an initial condition \( y(0) = -1 \). This condition tells us that when the variable \( x \) is zero, the value of \( y \) should be -1.
Initial conditions are essential in solving differential equations because they help determine the particular solution from a family of solutions. When you integrate a differential equation, the result typically includes a constant of integration \( C \). The initial condition allows us to solve for \( C \) and find the unique solution that satisfies the condition. In our example, after determining \( C \), we find that \( y = -e^{\frac{x^2}{2}} \), which satisfies \( y(0) = -1 \).
The process of applying initial conditions involves:
Initial conditions are essential in solving differential equations because they help determine the particular solution from a family of solutions. When you integrate a differential equation, the result typically includes a constant of integration \( C \). The initial condition allows us to solve for \( C \) and find the unique solution that satisfies the condition. In our example, after determining \( C \), we find that \( y = -e^{\frac{x^2}{2}} \), which satisfies \( y(0) = -1 \).
The process of applying initial conditions involves:
- Solving the differential equation to find the general solution.
- Using the initial condition to calculate the exact value of any unknown constants.
Separable Differential Equation
A separable differential equation is a type of differential equation where variables can be separated on different sides of the equation. For the equation \( y' = xy \), you can rearrange it as \( \frac{dy}{y} = x \, dx \), effectively separating the variables \( x \) and \( y \).
The method of solving these equations is called "separation of variables," and it usually involves the following steps:
The method of solving these equations is called "separation of variables," and it usually involves the following steps:
- Identify that the equation is separable.
- Rearrange or manipulate the equation so that all \( y \)-dependent terms are on one side and all \( x \)-dependent terms are on the other.
- Integrate each side separately. This gives you two expressions, one in terms of \( y \) and one in terms of \( x \).
Constants of Integration
In the process of solving differential equations, especially those involving integration, we encounter the "constant of integration" \( C \). This constant arises because the integral of a function is not unique; it can differ by an arbitrary constant.
Whenever we integrate a function, we add a constant of integration because there are infinitely many antiderivatives for a given function differing only by a constant. For instance, when we solved the integral \( \int \frac{1}{y} \, dy = \int x \, dx \), the result was \( \ln |y| = \frac{x^2}{2} + C \). Here, \( C \) represents the constant of integration.
To find the specific value of \( C \), we use initial conditions. In our example with the initial condition \( y(0) = -1 \), substituting into the equation \( -1 = A e^0 \) gives \( A = -1 \). This ensures that \( y = -e^{\frac{x^2}{2}} \) satisfies both the equation and the initial condition.
Whenever we integrate a function, we add a constant of integration because there are infinitely many antiderivatives for a given function differing only by a constant. For instance, when we solved the integral \( \int \frac{1}{y} \, dy = \int x \, dx \), the result was \( \ln |y| = \frac{x^2}{2} + C \). Here, \( C \) represents the constant of integration.
To find the specific value of \( C \), we use initial conditions. In our example with the initial condition \( y(0) = -1 \), substituting into the equation \( -1 = A e^0 \) gives \( A = -1 \). This ensures that \( y = -e^{\frac{x^2}{2}} \) satisfies both the equation and the initial condition.