Question

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y^{\prime}(x)=\frac{1+x}{2-y}, y(1)=1$$

Short answer

Question: Solve the initial value problem (IVP) $$y^{\prime}(x)=\frac{1+x}{2-y}, y(1)=1$$ and find an implicit solution. Use graphing software to plot the solution, and indicate the curve corresponding to the solution of the IVP. Answer: The implicit solution for the given IVP is $$ 2y-\frac{1}{2}y^2 = x+\frac{1}{2}x^2.$$ Plot the implicit solution using graphing software and include the point (1, 1) to indicate the curve corresponding to the solution of the IVP.

Step-by-step solution

Separate variables

We start by separating the variables x and y by dividing both sides of the equation by \((1+x)\) and multiplying both sides by \((2-y)\). This gives us: $$ (2-y)y^{\prime} = 1+x$$ Now we can integrate both sides of the equation with respect to \(x\).

Integrate both sides

To integrate both sides, we want to eliminate \(y^{\prime}\) (i.e., replace it with \(\frac{dy}{dx}\)). So, we need to rewrite our equation as: $$ (2-y)\frac{dy}{dx} = 1+x $$ Integrating both sides with respect to \(x\), we get: $$\int (2-y)dy = \int (1+x)dx$$

Integrate and find the implicit solution

Now we are ready to integrate both sides of the equation: On the left side, we have: $$\int (2-y)dy = 2\int dy - \int y dy = 2y-\frac{1}{2}y^2 + C_1.$$ On the right side, we have: $$\int (1+x)dx = x+\frac{1}{2}x^2+C_2.$$ Put them together: $$ 2y-\frac{1}{2}y^2+C_1 = x+\frac{1}{2}x^2+C_2.$$ We can simplify by combining \(C_1\) and \(C_2\) into a single constant \(C\): $$ 2y-\frac{1}{2}y^2 = x+\frac{1}{2}x^2+C.$$ The resulting equation is the implicit solution for the given IVP.

Apply the initial condition

Now, let's use the initial condition to find the constant \(C\). We are given \(y(1) = 1\). Plugging this into our implicit solution, we get: $$2(1) - \frac{1}{2}(1)^2 = (1) + \frac{1}{2}(1)^2 + C$$ Simplifying, we obtain: $$1 = 1+C$$ Thus, the constant \(C=0\). This gives us the final implicit solution: $$ 2y-\frac{1}{2}y^2 = x+\frac{1}{2}x^2.$$

Plot the solution

Using graphing software (such as WolframAlpha, Desmos, or GeoGebra), we can plot the implicit solution: $$2y-\frac{1}{2}y^2 = x+\frac{1}{2}x^2$$ Making sure to plot the point \((1, 1)\) to indicate the curve corresponding to the solution of the initial value problem.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in differential equations involves finding a solution to a differential equation that also satisfies a specific condition at a given point, known as the initial condition. For instance, in our exercise, the differential equation \(y'(x) = \frac{1+x}{2-y}\) comes with the condition \(y(1) = 1\).

This means that not only do we need a function \(y(x)\) that satisfies the differential equation, but it must also satisfy this initial condition when \(x = 1\) — specifically \(y(1) = 1\).

• A solution that satisfies the initial condition is not just any arbitrary function but a specific one that crosses or touches a given point on the graph that represents \(y(1) = 1\).
• IVPs are common in real-world applications, such as in physics for motion prediction, chemistry for reaction rates, and other sciences that involve change or growth modeling.

It is crucial to apply the initial condition after finding a general solution to ensure the solution is accurate and valid for the problem.

Implicit Solution

An implicit solution is a form of solution for a differential equation where the dependent variable is not explicitly isolated on one side of the equation. It often remains embedded within an equation involving both the independent and dependent variables.

For the problem \(2y-\frac{1}{2}y^2 = x+\frac{1}{2}x^2\), the solution is implicit because \(y\) is not clearly solved for in terms of \(x\) alone. In many cases, it’s acceptable and often necessary to leave solutions in this form, especially when solving for \(y\) explicitly is difficult or impossible.

• This form of solution can describe multiple curves or regions on a graph, particularly because it often represents a family of curves rather than a single line.
• One must graph these implicit equations to determine which part of the solution curve fits the given initial conditions.

Implicit solutions are powerful tools in mathematics because they can represent complex relationships that simple explicit solutions cannot capture.

Variable Separation

Variable separation is a method used to solve differential equations by rearranging terms to isolate the variables of the differential equation on opposite sides of the equation.

In our problem, this involves setting up the equation \((2-y)y' = 1+x\). To achieve separation, we manipulate it to \((2-y)\frac{dy}{dx} = 1+x\), which allows us to integrate with respect to \(x\).

• This method works effectively for first-order differential equations and is quite straightforward as it breaks down a complex equation into manageable parts.
• The goal is to get all terms containing \(y\) on one side and all terms containing \(x\) on the other.
• Once separation is achieved, each side can be integrated independently to solve for the unknowns.

This technique is highly applicable for differential equations in various scientific fields where rates of change are expressed using separable functions.

Integration

Integration is a fundamental process in solving differential equations, allowing us to find functions from their derivatives. In the given problem, after applying variable separation, we have \(\int (2-y) \; dy = \int (1+x) \; dx\).

This means we need to integrate both sides independently:

• On the left, the integration of \((2-y)\) with respect to \(y\) results in \(2y - \frac{1}{2}y^2 + C_1\).
• On the right, the integration of \((1+x)\) with respect to \(x\) yields \(x + \frac{1}{2}x^2 + C_2\).

Combining these, we get our implicit solution:
\(2y - \frac{1}{2}y^2 = x + \frac{1}{2}x^2 + C\).

Integration is sometimes described metaphorically as "undoing differentiation," and it plays a crucial role in recovering original functions from their rates of change or gradients.

Graphing Solutions

Graphing solutions is a key step in fully understanding the behavior of a differential equation, especially when working with implicit solutions.

For the implicit solution \(2y - \frac{1}{2}y^2 = x + \frac{1}{2}x^2\), graphing is essential to identify which curve corresponds to our initial value problem

• By plotting this equation, often with the aid of graphing software, we can visualize the overall behavior of the solution.
• It’s crucial to ensure that the specific point \((1, 1)\) lies on the curve, as this confirms the solution meets the initial condition \(y(1) = 1\).

Graphing not only verifies our solution but also provides a visual representation, making it easier to comprehend the dynamics involved. It highlights the unique curve that matches the initial conditions among potentially many others.