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. $$y^{\prime}(x) \sqrt{y+4}=\sqrt{x+1}, y(3)=5$$

Short answer

Short Answer: The solution to the initial value problem is given in implicit form by $$2\sqrt{y+4} = \frac{2}{3}(x+1)^{\frac{3}{2}} + 2\sqrt{9} - \frac{16}{3}\sqrt{2}$$. Using graphing software like Desmos, GeoGebra, or Wolfram Alpha, we can plot this implicit solution, making sure to indicate the function representing the IVP solution and verifying it satisfies the initial condition $$y(3) = 5$$.

Step-by-step solution

Separate variables

To separate the variables, we will divide both sides of the equation by $$\sqrt{y+4}$$ and multiply by dx: $$\frac{y^\prime(x)}{\sqrt{y(x)+4}} = \frac{\sqrt{x+1}}{1}$$ Now, we have: $$\frac{dy}{\sqrt{y+4}} = \sqrt{x+1}dx$$

Integrate both sides

Integrate both sides of the equation with respect to their respective variables: $$\int \frac{1}{\sqrt{y+4}} dy = \int \sqrt{x+1} dx$$ Let's perform the integration of these functions: $$2\sqrt{y+4} = \frac{2}{3}(x+1)^{\frac{3}{2}} + C$$

Apply the initial condition

We are given the initial condition $$y(3) = 5$$. Let's substitute this into the equation to determine the constant of integration C: $$2\sqrt{5+4} = \frac{2}{3}(3+1)^{\frac{3}{2}} + C$$ Solving for C, we find that: $$C = 2\sqrt{9} - \frac{16}{3}\sqrt{2}$$

Write the implicit solution

Now, substitute the value of C back into the equation and rewrite it in implicit form: $$2\sqrt{y+4} = \frac{2}{3}(x+1)^{\frac{3}{2}} + 2\sqrt{9} - \frac{16}{3}\sqrt{2}$$ This is the implicit solution to the initial value problem.

Plot the solution using graphing software

Use a graphing software like Desmos, GeoGebra, or Wolfram Alpha to plot the implicit solution: $$2\sqrt{y+4} = \frac{2}{3}(x+1)^{\frac{3}{2}} + 2\sqrt{9} - \frac{16}{3}\sqrt{2}$$ Remember to indicate which function corresponds to the IVP solution on the graph, and ensure that it satisfies the initial condition $$y(3) = 5$$.

Key concepts

Separating Variables

When solving differential equations, separating variables is a technique used to isolate the function variables on one side and the independent variable on the other side. This method applies to equations where the rate of change of the function can be expressed as the product of a function of the independent variable and a function of the dependent variable.

For the initial value problem given, \(y^\prime(x) \sqrt{y+4}=\sqrt{x+1}, y(3)=5\), the separation was achieved by dividing both sides by \(\sqrt{y+4}\) and multiplying by \(dx\), the differential of the independent variable. This manipulation led to an equation with all \(y\)-terms and differentials \(dy\) on one side and all \(x\)-terms and differentials \(dx\) on the other, \(\frac{dy}{\sqrt{y+4}} = \sqrt{x+1}dx\). The ease of integrating each side separately is the main advantage of this technique.

By using this strategy, you ensure the integration process becomes straightforward, allowing you to move to the next step of finding the solution to the differential equation.

Integrating Factors

In some differential equations, you cannot straightforwardly separate variables. In such cases, an approach involving integrating factors can be helpful. An integrating factor is a function that we multiply by the differential equation to facilitate simplification. Usually, it is a function of the independent variable only and is derived from the equation itself.

Although the initial value problem in this exercise did not require the use of an integrating factor since the variables were easily separated, it's crucial to be aware of this technique for more complex equations. Typically, you'd employ integrating factors in linear first-order differential equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)\).

The integrating factor, usually denoted by \(\mu(x)\), makes the left side of the modified equation a product of a derivative, transforming it into a form where a direct integration is possible. This clever method transforms a potentially complex equation into something much more manageable.

Implicit Solutions

An implicit solution to a differential equation expresses the relationship between the independent variable \(x\) and the dependent variable \(y\) without explicitly solving for \(y\) as a function of \(x\). This form can be beneficial when finding an explicit solution is too challenging or when the explicit solution involves complex expressions that are difficult to work with.

In the given exercise, after integrating both sides and applying the initial condition, \(y(3) = 5\), the solution is expressed implicitly. The constant of integration \(C\) is determined to satisfy the initial condition, resulting in the implicit solution \(2\sqrt{y+4} = \frac{2}{3}(x+1)^{\frac{3}{2}} + 2\sqrt{9} - \frac{16}{3}\sqrt{2}\).

Although the solution might describe more than one function, indicating which one corresponds to the initial value problem is important, especially when graphing. Implicit solutions also retain the authenticity of the main function's behavior when the explicit form might imply restrictions not present in the original equation.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are powerful tools in expressing the rates at which quantities change, often used in physics, engineering, economics, and biology to model complex systems and phenomena.

The initial value problem presented is an example of an ordinary differential equation (ODE) because it involves one independent variable. An ODE is classified by its order, determined by the highest derivative present in the equation. In this exercise, the equation \(y^\prime(x) \sqrt{y+4}=\sqrt{x+1}\) is a first-order ODE since the highest derivative is the first derivative of \(y\).

Solving an ODE typically involves finding a function that satisfies the given equation and adheres to certain initial or boundary conditions. The process might involve techniques such as separating variables, integrating factors, or more sophisticated methods when dealing with higher-order or partial differential equations. The goal is a solution that provides insight into the behavior and prediction of the system described by the differential equation.