Chapter 2: Problem 13
Solve the initial-value problems. $$ (2 x+3 y+1) d x+(4 x+6 y+1) d y=0, \quad y(-2)=2 $$
Short Answer
Expert verified
The implicit solution for the given initial value problem can be represented as:
\(y(x) = (x+1)^{\frac{4}{7}}\left(2-\int(x+1)^{-\frac{4}{7}}(4x + 6y + 1)dx\right)\)
Step by step solution
01
Write down the given ODE
We are given the following ODE:
\[
(2x + 3y + 1)dx + (4x + 6y + 1)dy = 0
\]
And the initial condition is: \(y(-2) = 2\).
02
Rewrite the given equation in a more convenient form
We can rearrange the given equation to show the relationship between the differentials \(dx\) and \(dy\):
\[
\frac{dy}{dx} = -\frac{2x + 3y + 1}{4x + 6y + 1}
\]
03
Separate variables
Now we want to separate the variables in the ODE.
\[
\frac{dy}{2x + 3y + 1} = -\frac{dx}{4x + 6y + 1}
\]
04
Integrate both sides
Next, we integrate both sides with respect to their respective variables:
\[\begin{aligned}
\int \frac{dy}{2x + 3y + 1} &= -\int \frac{dx}{4x + 6y + 1}
\end{aligned}\]
Notice that the integral on the left side depends on both x and y. To proceed, we will use the integrating factor method. Since the term \(3y\) inside the integral is multiplied with the dependent variable, we will consider the integrating factor to be a function of x, say \(\mu(x)\).
05
Determine the integrating factor
For the left side, our integrating factor \(\mu(x)\) should satisfy the following condition:
\[
\frac{d\mu}{dx} = - \frac{4}{2x+3y+1}\mu(x)
\]
Integrating both sides, we have
\[
\mu(x) = e^{\int-\frac{4}{2x+3(2)+1}dx}=e^{-\frac{4}{7x+7}dx}=e^{-\frac{4}{7}\ln(|x+1|)}=(x+1)^{-\frac{4}{7}}
\]
06
Apply integrating factor and solve the left integral
Now we multiply both sides of our separated ODE by our integrating factor \(\mu(x)\):
\[\begin{aligned}
(x+1)^{-\frac{4}{7}}\int \frac{dy}{2x + 3y + 1} &= -(x+1)^{-\frac{4}{7}}\int \frac{dx}{4x + 6y + 1}
\end{aligned}\]
Now, it's easy to solve the left side integral:
\[\begin{aligned}
(x+1)^{-\frac{4}{7}}y &= C_1-\int(x+1)^{-\frac{4}{7}}(4x + 6y + 1)dx
\end{aligned}\]
07
Solve for y and apply the initial condition
First, isolate y on the left side of the equation:
\[
y(x) = (x+1)^{\frac{4}{7}}\left(C_1-\int(x+1)^{-\frac{4}{7}}(4x + 6y + 1)dx\right)
\]
Now apply the initial condition \(y(-2) = 2\):
\[
2 = (1)^{\frac{4}{7}}\left(C_1-\int(1)^{-\frac{4}{7}}(4(-2) + 6(2) + 1)dx\right)
\]
Solve for \(C_1\):
\[
C_1 = 2
\]
Finally, we obtain the particular solution for the given initial value problem:
\[
y(x) = (x+1)^{\frac{4}{7}}\left(2-\int(x+1)^{-\frac{4}{7}}(4x + 6y + 1)dx\right)
\]
Due to the complexity of the integral, it may not be possible to find an explicit solution for the function y(x). However, the above equation represents an implicit solution for the given initial value problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations that involve functions of one variable and their derivatives. They are widely used to model various real-world phenomena like physics, engineering, and biology. An ODE contains only one independent variable, and the derivatives describe how the function changes with respect to this variable.
The example given in the original exercise represents an ordinary differential equation:
The example given in the original exercise represents an ordinary differential equation:
- Consists of expressions involving the function and its derivatives: (2x + 3y + 1)dx + (4x + 6y + 1)dy = 0.
- Typically requires an initial condition to ensure a unique solution, like y(-2) = 2 in our problem.
Integrating Factor Method
The integrating factor method is a powerful technique used to solve differential equations that aren't easily separable. This method involves finding a function, called the integrating factor, that simplifies the equation to allow integration. Instead of solving the equation directly, you convert it to a form where integration is straightforward.
In the given exercise, we used:
In the given exercise, we used:
- The integrating factor \(\mu(x)\) determined as \((x+1)^{-\frac{4}{7}}\), which simplifies the integration.
- Applying the integrating factor ensures terms can be integrated separately, helping in solving the equation involving both x and y.
Separation of Variables
Separation of Variables is a simple yet effective method for solving differential equations, especially ODEs. It involves rearranging the terms of the equation to isolate variables on opposite sides of the equation, hence 'separating' them. Once separated, integration can be performed on both sides to solve for the variables.
During the exercise:
During the exercise:
- We transformed \(\frac{dy}{dx} = -\frac{2x + 3y + 1}{4x + 6y + 1}\) so that all terms involving y and constants go to one side and terms involving x to the other: \(\frac{dy}{2x + 3y + 1} = -\frac{dx}{4x + 6y + 1}\).
- After separation, each side is ready for integration, paving the way to progress toward a solution.
Implicit Solutions
An implicit solution expresses a solution in terms of a relationship between variables, rather than as an explicit formula like y = f(x). Sometimes, due to complexity, an ODE can't be solved for an explicit solution, but an implicit solution can still describe the behavior of the function.
In the exercise:
In the exercise:
- The relation \(y(x) = (x+1)^{\frac{4}{7}}\left(2-\int(x+1)^{\-rac{4}{7}}(4x + 6y + 1)dx\right)\) represents the implicit solution.
- This equation implies a relationship between x and y that must satisfy the original differential equation.