Chapter 2: Problem 19
Solve the initial value problem and find the interval of validity of the solution. $$ y^{\prime}=\frac{2 x}{1+2 y}, \quad y(2)=0 $$
Short Answer
Expert verified
The function y(x) is:
$$ y(x) = \frac{1}{2}(e^{x^2 + 4} - 1) $$
The interval of validity for this solution is:
$$ x \in (-\infty, \infty) $$
Step by step solution
01
Rewrite as a separable equation
First, we will rewrite the given differential equation as a separable equation. We can do this by multiplying both sides by \((1+2y)\) and dividing by \(2x\).
$$ (1+2y)y^{\prime} = 2x $$
02
Integrate both sides
Now that the equation is separable, we can do the integration on both sides. To integrate, we will write the equation as:
$$ (1+2y) \frac{dy}{dx} = 2x $$
Now we divide both sides by \((1+2y)\) and multiply by \(dx\):
$$ \frac{dy}{(1+2y)} = \frac{2x dx} $$
Integrate both sides with respect to their variables:
$$ \int \frac{1}{1+2y} dy = \int x dx $$
On the left side, use substitution: \(u = 1+2y\), \(du = 2 dy\):
$$ \frac{1}{2}\int \frac{1}{u} du = \int x dx $$
Now, integrate both sides:
$$ \frac{1}{2} (\ln |u| + C_1) = \frac{1}{2}x^2 + C_2 $$
Substitute back for \(u\):
$$ \frac{1}{2} (\ln |1+2y| + C_1) = \frac{1}{2}x^2 + C_2 $$
Combine both constants on the left side:
$$ \ln |1+2y| + C_3 = x^2 + C_4 $$
03
Solve for y using the initial condition
Now, we will solve for y using the initial condition \(y(2)=0\). Substitute \(x = 2\) and \(y = 0\) into the equation:
$$ \ln |1+2(0)| + C_3 = (2)^2 + C_4 $$
Simplifying, we get:
$$ \ln 1 + C_3 = 4 + C_4 $$
Since \(\ln 1 = 0\), this tells us \(C_3 = -4 + C_4\). We substitute this back into the equation:
$$ \ln |1+2y| - 4 + C_4 = x^2 + C_4 $$
This simplifies to:
$$ \ln |1+2y| = x^2 + 4 $$
Now, we isolate y by exponentiating both sides:
$$ |1+2y| = e^{x^2 + 4} $$
Since \(1+2y\) is always positive in our domain, we can write:
$$ 1+2y = e^{x^2 + 4} $$
Finally, solve for y:
$$ y(x) = \frac{1}{2}(e^{x^2 + 4} - 1) $$
04
Determine the interval of validity
Now, we will determine the interval of validity for the solution. As the exponential function is valid for all real number inputs, the function exists for all x. Moreover, the function itself is continuous, so the interval of validity is all real numbers:
$$ x \in (-\infty, \infty) $$
So the solution to the initial value problem is:
$$ y(x) = \frac{1}{2}(e^{x^2 + 4} - 1), \quad x \in (-\infty, \infty) $$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Equations
A separable equation is a type of differential equation in which the variables can be separated on either side of the equation. This makes the equation easier to solve through integration. In the given initial value problem, we are dealing with the differential equation \( y' = \frac{2x}{1+2y} \). To make it separable, we rewrite it as \((1+2y) y' = 2x\).
By multiplying and dividing appropriately, we have all terms involving \(y\) on one side and terms involving \(x\) on the other. The idea is to isolate \(dy\) with its associated function and \(dx\) with its function, as in \(\frac{dy}{(1+2y)} = 2x \, dx\).
The separated form prepares the equation for integration, allowing a direct integration of each side with respect to its own variable.
By multiplying and dividing appropriately, we have all terms involving \(y\) on one side and terms involving \(x\) on the other. The idea is to isolate \(dy\) with its associated function and \(dx\) with its function, as in \(\frac{dy}{(1+2y)} = 2x \, dx\).
The separated form prepares the equation for integration, allowing a direct integration of each side with respect to its own variable.
Interval of Validity
The interval of validity in differential equations refers to the set of values of \(x\) for which a solution is defined and meaningful. In our problem, the equation is valid along the domain where both sides of the equation can be correctly calculated.
For the solution derived as \(y(x) = \frac{1}{2}(e^{x^2 + 4} - 1)\), we find the interval of validity by considering the behavior of the function components. The exponential expression, \(e^{x^2 + 4}\), is well defined and continuous for all real numbers. This continuity ensures that the solution is also valid for all \(x \in (-\infty, \infty)\).
This broad interval shows the robustness of exponential functions and how they influence the solution's validity on the real number line.
For the solution derived as \(y(x) = \frac{1}{2}(e^{x^2 + 4} - 1)\), we find the interval of validity by considering the behavior of the function components. The exponential expression, \(e^{x^2 + 4}\), is well defined and continuous for all real numbers. This continuity ensures that the solution is also valid for all \(x \in (-\infty, \infty)\).
This broad interval shows the robustness of exponential functions and how they influence the solution's validity on the real number line.
Differential Equations Integration
Integration of differential equations involves finding an antiderivative that satisfies the original equation. Once a differential equation is made separable, like \(\frac{dy}{(1+2y)} = 2x \, dx\), we integrate each side.
On the left side, we use substitution to simplify and solve the integral. Letting \(u = 1 + 2y\), transforms the integral into \(\frac{1}{2} \int \frac{1}{u} \, du\). Integrate to get \(\frac{1}{2} (\ln |u| + C)\).
On right side, it's a straightforward integration as \(\int x \, dx = \frac{1}{2}x^2 + C\).
These integrals give us an equation involving constants that we solve using initial conditions, such as \(y(2) = 0\), ensuring the derived particular solution satisfies the initial value problem.
On the left side, we use substitution to simplify and solve the integral. Letting \(u = 1 + 2y\), transforms the integral into \(\frac{1}{2} \int \frac{1}{u} \, du\). Integrate to get \(\frac{1}{2} (\ln |u| + C)\).
On right side, it's a straightforward integration as \(\int x \, dx = \frac{1}{2}x^2 + C\).
These integrals give us an equation involving constants that we solve using initial conditions, such as \(y(2) = 0\), ensuring the derived particular solution satisfies the initial value problem.
Exponential Functions
Exponential functions are functions of the form \(e^{f(x)}\), where \(e\) is the base of natural logarithms. They are particularly notable for their continuous and smooth nature, making them valid across the entire real line.
In the problem's solution, after solving the integration and applying initial conditions, we arrive at \(1 + 2y = e^{x^2 + 4}\). This leads us to our final solution \(y(x) = \frac{1}{2}(e^{x^2 + 4} - 1)\).
Because exponential functions grow rapidly and never touch zero, they can change significantly over a small change in \(x\). This property ensures that our solution function behaves predictably and is valid across its entire domain for real numbers, which underscores why \(x \in (-\infty, \infty)\) is valid.
In the problem's solution, after solving the integration and applying initial conditions, we arrive at \(1 + 2y = e^{x^2 + 4}\). This leads us to our final solution \(y(x) = \frac{1}{2}(e^{x^2 + 4} - 1)\).
Because exponential functions grow rapidly and never touch zero, they can change significantly over a small change in \(x\). This property ensures that our solution function behaves predictably and is valid across its entire domain for real numbers, which underscores why \(x \in (-\infty, \infty)\) is valid.
