Problem 32 - Calculus Early Transcendentals

Question

Find the general solution of the following equations. $$\frac{d y}{d x}=y\left(x^{2}+1\right), \text { where } y>0$$

Step-by-Step Solution

Answer

Answer: The general solution of the given ODE is: $$y=ke^{\frac{x^{3}}{3}+x}$$, where k is an arbitrary constant determined by initial or boundary conditions.

Step 1: Rewrite the equation in the form (1/y)dy = (x^2+1)dx

To separate the variables, we first need to rewrite the equation in the form (1/y)dy = (x^2+1)dx. We can achieve this by dividing both sides of the equation by y: $$\frac{1}{y}\frac{dy}{dx}=x^{2}+1$$

Step 2: Integrate both sides

Now we can integrate both sides of the equation with respect to the corresponding variables. For the left side, we'll integrate with respect to y, and for the right side, we'll integrate with respect to x: $$\int\frac{1}{y}dy=\int\left(x^{2}+1\right)dx$$

Step 3: Evaluate the integrals

Evaluating the left integral, we have: $$\ln|y|= \int\left(x^{2}+1\right)dx$$ And evaluating the right integral, we get: $$\ln|y|=\frac{x^{3}}{3}+x+C$$ where C is the constant of integration.

Step 4: Solve for y

To find the general solution, we need to solve for y. First, we'll take the exponent of both sides to remove the natural logarithm: $$|y|=e^{\frac{x^{3}}{3}+x+C}$$ Since y > 0, we can remove the absolute value sign and introduce a constant k (k = e^C) to get the general solution: $$y=ke^{\frac{x^{3}}{3}+x}$$

Step 5: Write the general solution

The general solution of the given ODE is: $$y=ke^{\frac{x^{3}}{3}+x}$$ where k is an arbitrary constant determined by initial or boundary conditions.

Key Concepts

Integrating Factors

One of the techniques used in solving differential equations is the use of integrating factors. An integrating factor is a function that is typically used to transform a non-exact first-order ordinary differential equation into an exact one by multiplication. This opens up a pathway to simplify the equation, making it solvable.

However, in the context of separable differential equations, such as in our exercise, the role of integrating factors is diminished, as these types of equations can be solved simply by separating variables and integrating directly. Separable differential equations are already in a form that can be integrated without further manipulation. One must remember though, integrating factors are extremely useful when dealing with more complex, non-separable first-order equations.

First-order Differential Equations

First-order differential equations are equations involving the first derivative of a function and the function itself. They often represent physical phenomena, like growth rates or motion.

In the problem provided, we have a classic first-order separable differential equation, where the solution method involves separating the variables involved - the \(y\) variables on one side and the \(x\) variables on the other. This separation allows for each side of the equation to be integrated with respect to its own variable. The step by step solution provided is a prime example of this method, starting with the separation of variables and following through with integration on both sides.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a fundamental mathematical function that helps in solving differential equations. It is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. The natural logarithm converts multiplication into addition, providing a powerful tool for solving equations that involve exponential growth or decay.

The exercise demonstrates this by taking the \(\ln\) of both sides of the equation after integrating, which then makes it possible to exponentiate both sides to solve for \(y\). This manipulation is typical when working with exponents and is key to finding the general solution to many first-order differential equations.

Integration of Functions

Integration is the process of finding the integral of a function, which can be thought of as the opposite operation of differentiation. When we integrate a function, we are essentially finding the area under the curve of that function with respect to a given variable.

In our step by step solution, integration is used to solve for \(y\) after the variables have been separated. The left side of the equation involves the integration of \(1/y\), which results in the natural logarithm of the absolute value of \(y\), while the right side involves the integration of a polynomial in \(x\). The integration process is vital for determining the relationship between \(y\) and \(x\) and for deriving the general solution of the differential equation.