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 function, be sure to indicate which function 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 following initial value problem and leave the solution in implicit form. Plot the solution on a graph. Initial value problem: $$y^{\prime}(x) = \frac{1+x}{2-y}, \quad y(1) = 1$$ Answer: The solution to the initial value problem is given by the implicit equation: $$2y-\frac{y^2}{2} = x + \frac{x^2}{2}$$ You can view the plot of the implicit solution using the following link: https://www.desmos.com/calculator/fctowrmz6h. The red curve on the graph represents the solution passing through the point (1,1).

Step-by-step solution

Separate Variables

To separate the variables, we need to bring all the terms involving y and its derivative to one side and all the terms involving x to the other side. The given equation is: $$y^{\prime}(x)=\frac{1+x}{2-y}$$ Let's rewrite it with the separated variables: $$y'(x)(2-y) = 1+x$$

Integrate Both Sides

Now that we have separated the variables, we can integrate both sides of the equation with respect to x: $$(2-y) \frac{dy}{dx} = 1+x$$ $$\int(2-y) dy = \int (1+x) dx$$

Solve the Integrals

We can find the antiderivatives of both sides: $$\int(2-y) dy = 2y - \frac{y^2}{2} + C_1$$ $$\int (1+x) dx = x + \frac{x^2}{2} + C_2$$ We can now equate these expressions:\\ $$2y - \frac{y^2}{2} + C_1 = x + \frac{x^2}{2} + C_2$$

Simplify and Combine Constants

We can combine the constants \(C_1\) and \(C_2\) into a single constant, say, \(C\). Thus, our equation becomes: $$2y - \frac{y^2}{2} = x + \frac{x^2}{2} + C$$

Apply the Initial Condition

Now, we need to find the constant C using the given initial condition, which is \(y(1) = 1\). Substituting this into our equation, we get: $$2(1) - \frac{1^2}{2} = 1 + \frac{1^2}{2} + C$$ Solving for C, we get: $$C = 0$$

Write the Implicit Solution

We can now write the implicit solution of the initial value problem by substituting \(C=0\): $$2y-\frac{y^2}{2} = x + \frac{x^2}{2}$$

Plot the Solution

Finally, we will plot the solution using a graphing software. Note that this is an implicit solution and can possibly describe more than one function. The graph for the given implicit solution is available here: https://www.desmos.com/calculator/fctowrmz6h The solution of the initial value problem is the curve passing through the point (1,1), which is the red curve on the graph.

Key concepts

Separation of Variables

The method of separation of variables is a powerful technique used to solve differential equations, where we can separate the variables involved, such as 'x' and 'y', on opposite sides of the equation.

For our example, the differential equation is given by \(y^\prime(x)=\frac{1+x}{2-y}\). To apply this method, we manipulate the equation to isolate the terms involving 'y' from those involving 'x', resulting in an equation where each side is a function of one variable only. This allows us to integrate each side independently, bringing us closer to a solution.

In practical terms, we'd rewrite our equation as \(y'(x)(2-y) = 1+x\), with the 'dy' on one side and 'dx' on the other after separating, like this:\[(2-y) \frac{dy}{dx} = 1+x\]. Now we're ready to integrate both sides to find a more complete form of the solution.

Implicit Solution

An implicit solution to a differential equation involves a relationship between the dependent variable 'y' and the independent variable 'x' that isn’t explicitly solved for 'y'.

In our case, after integrating and applying the initial conditions, we end up with an equation that defines 'y' in terms of 'x' and vice versa, but it doesn't give us a direct expression for 'y=' something. The implicit solution for our equation is \(2y - \frac{y^2}{2} = x + \frac{x^2}{2}\).

This form can often describe more than one function, which is important when we examine the system's behavior on a graph. Being able to interpret implicit solutions is a critical skill, as they provide insights into complex relationships that explicit solutions may not readily show.

Integrating Differential Equations

Integrating differential equations is the next step after separation of variables. It involves finding the antiderivatives of both sides of the equation to get an equation in a more recognizable function form.

From our separated equation, we integrate the left side with respect to 'y' and the right side with respect to 'x'. This yields \[\int (2-y) dy = 2y - \frac{y^2}{2} + C_1\] and \[\int (1+x) dx = x + \frac{x^2}{2} + C_2\]. These antiderivatives encapsulate all the possible solutions of the differential equation before we apply any initial conditions to find the particular solution tailored to our problem.

Applying Initial Conditions

Applying initial conditions is a crucial step in determining the specific solution to an initial value problem. Initial conditions are given values for the variables at a particular point, which we use to solve for the constants that arise from integrating the differential equation.

In our example, we're given the initial condition \(y(1) = 1\). When we substitute these values into the integrated form of the equation, we are able to solve for the constant 'C'. For our problem, substituting into \(2y - \frac{y^2}{2} = x + \frac{x^2}{2} + C\) and solving gives us \(C=0\).

By applying the initial condition correctly, we ensure that our solution curve passes through the specified point, as it must for any accurate solution of an initial value problem. Combining this constant with the integrated functions gives us the particular solution to our original differential equation.