Question

Find the general solution to the differential equations. $$ y^{\prime}=y\left(x^{2}+1\right) $$

Short answer

The general solution is \( y = C_1 e^{\frac{x^3}{3} + x} \) where \( C_1 \) is a nonzero constant.

Step-by-step solution

Identify the Equation Type

The given differential equation is \( y' = y(x^2 + 1) \). This is a first-order differential equation that can be solved using separation of variables because the equation can be written in the form \( \frac{dy}{dx} = f(x)g(y) \).

Rearrange and Separate Variables

Separate the variables by dividing both sides by \( y \) and multiplying both sides by \( dx \), giving \( \frac{1}{y} dy = (x^2 + 1) dx \).

Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{1}{y} \, dy = \int (x^2 + 1) \, dx \).

Solve Integrals

The integral of \( \frac{1}{y} \, dy \) is \( \ln|y| \) and the integral of \( (x^2 + 1) \, dx \) is \( \frac{x^3}{3} + x \). So we have: \( \ln|y| = \frac{x^3}{3} + x + C \), where \( C \) is the constant of integration.

Solve for y

To solve for \( y \), exponentiate both sides to get rid of the natural logarithm: \( |y| = e^{\frac{x^3}{3} + x + C} \). This can be rewritten as \( |y| = e^C e^{\frac{x^3}{3} + x} \).

Express General Solution

Since \( y \) can be positive or negative, eliminate the absolute value by including a constant \( \pm e^C \) which we denote as \( C_1 \), giving the general solution: \( y = C_1 e^{\frac{x^3}{3} + x} \), where \( C_1 \) is a nonzero constant.

Key concepts

First-order Differential Equation

A first-order differential equation refers to an equation that involves the first derivative of a function. In the given exercise, the differential equation is \( y' = y(x^2 + 1) \). This means that the rate of change of \( y \) with respect to \( x \) is dependent on both \( y \) and the expression \( (x^2 + 1) \).
First-order differential equations often appear in various scientific fields, including physics and biology, describing how a quantity changes over time or space.

• First-order indicates that the highest derivative in the equation is the first derivative.
• These equations can be linear, where terms either don't multiply or take derivatives of each other, or non-linear, like our example here.

Recognizing how these equations are structured allows us to apply suitable methods for obtaining their solutions.

Separation of Variables

Separation of variables is a method used to solve differential equations, which involves rearranging the equation so that each variable and its differentials are separated on opposite sides of the equation. For the equation \( y' = y(x^2 + 1) \), you can separate variables by dividing both sides by \( y \) and multiplying by \( dx \), resulting in \( \frac{1}{y} dy = (x^2 + 1) dx \).
This allows us to integrate each side with respect to its own variable, a crucial step towards finding the solution.

• The core idea of this method is to isolate all terms involving one variable on one side of the equation and all terms involving the other variable on the other side.
• This technique is especially helpful for simple first-order and sometimes higher-order differential equations.

Successfully separating the variables sets the stage for the integration process, which will lead us to the general solution.

Integration

Integration is the process of finding an integral, which is the reverse operation of differentiation. In the context of differential equations, integration is used to solve the rearranged equation after variables are separated. For our problem, after separation, we integrate both sides: \( \int \frac{1}{y} \, dy = \int (x^2 + 1) \, dx \).
This yields the integrals \( \ln|y| \) and \( \frac{x^3}{3} + x \), respectively.

• The integral \( \int \frac{1}{y} \, dy \) is a common result that equals \( \ln|y| \), involving the natural logarithm function.
• The integral \( \int (x^2 + 1) \, dx \) results in the polynomial \( \frac{x^3}{3} + x \).
• After completing the integration, we solve for the variable \( y \) by exponentiating both sides to remove the logarithm.

Integration not only helps in solving differential equations but also plays a key role in various branches of calculus and applied mathematics.