Question

Solve the initial value problem and graph the solution. $$ \left(3 y^{2}+4 y\right) y^{\prime}+2 x+\cos x=0, \quad y(0)=1 $$

Short answer

Answer: This is a non-linear first-order ordinary differential equation (ODE). Challenges in finding an explicit solution include determining the constant of integration (which may not be directly obtainable from the initial condition) and dealing with undefined terms, making it difficult to isolate $$y(x)$$ in the equation proress.

Step-by-step solution

Rewrite the ODE as a separable equation

Divide both sides of the ODE by $$x\sqrt{y(x)}$$ to isolate terms involving $$x$$ on the right side and terms involving $$y$$ on the left side. We get: $$\frac{y'(x)}{x\sqrt{y(x)}} - \frac{y(x)}{x\sqrt{y(x)}} = 1$$.

Integrate both sides

To obtain an implicit solution, integrate both sides with respect to $$x$$: $$\int \frac{y'(x)}{x\sqrt{y(x)}} dx - \int \frac{y(x)}{x\sqrt{y(x)}} dx = \int 1 dx$$. Now let $$u = y(x)$$ and $$v = x$$, then $$du = y'(x) dx$$ and $$dv = dx$$. The integrals become: $$\int \frac{1}{v\sqrt{u}} du - \int \frac{u}{v\sqrt{u}} dv = \int dv$$. We can compute these integrals: $$\int \frac{1}{v\sqrt{u}} du = 2\sqrt{u} + C_1$$, $$\int \frac{u}{v\sqrt{u}} dv = -\frac{u}{v} + C_2$$. So the implicit solution is: $$2\sqrt{y(x)} - \frac{y(x)}{x} = x + C$$, where $$C = C_1 - C_2$$.

Apply the initial condition

Now we apply the initial condition $$y(0) = 4$$ to find the value of $$C$$: $$2\sqrt{4} - \frac{4}{0} = 0 + C$$. Note that the term $$\frac{4}{0}$$ is undefined. As $$x=0$$ is not in the domain of the solution, we can't find the value of $$C$$ by directly plugging in the initial condition. This means we can't go further to find the explicit solution. Despite this, it's still possible to graph the implicit solution.

Graph the solution

We will graph the implicit solution $$2\sqrt{y(x)} - \frac{y(x)}{x} = x + C$$ with the understanding that the value of $$C$$ cannot be found directly from the initial condition. The graph will show the general behavior of the function, but it will not show the specific curve that starts at the point $$x=0$$, $$y=4$$ due to the undefined term in the equation. Instead, the graph will show a family of curves for different values of $$C$$. Observations from the graph can be helpful in understanding the behavior of the solution, but further analysis is needed to obtain a specific explicit solution.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve a function and its derivatives. These equations describe how one variable changes in relation to another. In the initial value problem provided, we are given a differential equation and an initial condition, which is a value of the function at a particular point. Here, the equation is \(y^{\prime}-y=x y^{1 / 2}\) with \( y(0)=4 \), indicating that when \(x=0\), \(y\) is equal to 4.
ODEs can help model various real-world phenomena such as motion, heat transfer, or population dynamics, as they account for the rate of change.
By solving ODEs, we find the function that explains these phenomena. The solution can be explicit, giving an exact function \(y=f(x)\), or implicit, which involves an equation with \(y\) and \(x\) that cannot be easily separated.

Separable Equations

Separable equations are a special class of ordinary differential equations. They can be written in a form that allows the variables to be separated on opposite sides of the equation. In essence, this means that all terms involving \(y\) can be grouped on one side, and all terms involving \(x\) on the other side.
In our problem, the ODE is initially \(y^{\prime}-y=x y^{1 / 2}\).
To make it separable, we divide through by \(x\sqrt{y(x)}\), resulting in the rearranged form \(\frac{y'(x)}{x \sqrt{y(x)}} - \frac{y(x)}{x \sqrt{y(x)}} = 1\).

• This format enables us to integrate each side with respect to its respective variable, a necessary step in solving separable differential equations.

Implicit Solutions

An implicit solution of a differential equation is a relation between \(x\) and \(y\) that is not directly solved for \(y\). Instead of expressing \(y\) strictly as a function of \(x\), it remains within an equation compared with \(x\).
In our solution process, after integrating both sides of the separable equation, we arrive at the implicit solution:
\(2\sqrt{y(x)} - \frac{y(x)}{x} = x + C\).

• This expression defines a relationship between \(y\) and \(x\), although not immediately solved for \(y\).
• Implicit solutions are common when integrals cannot be expressed in terms of elementary functions.
• These solutions are valid and can describe the same relationship between variables as explicit solutions, sometimes offering a broader perspective.

Integration Techniques

Integration is a mathematical tool fundamental for solving separable differential equations. By integrating terms with respect to their corresponding variable, we can solve for a function that satisfies the differential equation. In this example, we encounter integrals like \(\int \frac{1}{v\sqrt{u}} du\) and \(\int \frac{u}{v\sqrt{u}} dv\).
These integrals might involve substitution or comparison methods, which are useful techniques to simplify complex integrals.

• In our example, substitution \(u=y(x)\) and \(v=x\) aids in simplifying terms, making the integrals more manageable.
• Once integrated, the functions should be compiled into a solution form, such as the implicit format we derived.

Effective application of these techniques can transform differential equations into understandable solutions, essential for addressing many problems involving change.