Question

For the following problems, find the general solution to the differential equation.\(y^{\prime}=3 x+e^{x}\)

Short answer

The general solution is \( y = \frac{3x^2}{2} + e^x + C \).

Step-by-step solution

Identify the Differential Equation

The differential equation given is \( y' = 3x + e^x \). This is a first-order ordinary differential equation where you need to find \( y \) in terms of \( x \).

Integrate the Right Side

To find the general solution, integrate the right side of the equation with respect to \( x \):\[ \int (3x + e^x) \, dx \]

Integrate Each Term Separately

Separate the integral into two parts and integrate each component:\[ \int 3x \, dx + \int e^x \, dx \]

Find the Integral of Each Term

Calculate each integral:- For \( \int 3x \, dx \), use the power rule to get: \[ \frac{3x^2}{2} + C_1 \]- For \( \int e^x \, dx \), the result is \( e^x \): \[ e^x + C_2 \]

Combine the Solutions and Simplify

Combine the results from Step 4:\[ y = \frac{3x^2}{2} + e^x + C \]where \( C = C_1 + C_2 \) is the constant of integration.

Key concepts

Integration

Integration is a fundamental concept in calculus, specifically essential in solving differential equations.
When you integrate a function, you are essentially finding the original function from its derivative. This process is the opposite of differentiation.

In the problem given, the solution involves integrating the right-hand side of the differential equation, which is given by the expression \(3x + e^{x}\).

• Integration of a linear term \(3x\) involves using the power rule, where you increase the power of \(x\) by one and divide by the new power. This gives us \(\frac{3x^2}{2}\).
• For \(e^x\), integration is straightforward since the integral of \(e^x\) is \(e^x\). This is due to the unique property of the exponential function, where its derivative and integral are the same.

The purpose of integration in this context is to help us transition from the derivative of the solution back to finding what function \(y\) can possibly match this derivative. Therefore, integration is crucial to finding the general solution in a differential equation.

General Solution

A general solution in the context of differential equations is a solution that contains arbitrary constants, representing a family of solutions.
This makes it adaptable to satisfy certain initial conditions when provided.

For our given differential equation \(y^{\prime}=3x+e^{x}\), after integrating the right side step by step, we obtained:

• \( y = \frac{3x^2}{2} + e^x + C \)

Here, \(C\) is the constant of integration that arises from indefinite integration.
This constant is critical because without it, the solution wouldn't be complete, as it allows the solution to cover an entire set of possible values for particular solutions.

The concept of a general solution is fundamental because it underlines the inherent flexibility of solutions in mathematics, showcasing how a differential equation can describe multiple scenarios through its constants.

First-Order Differential Equation

First-order differential equations involve the first derivative of a function but no higher derivatives.
This means that in these equations, you are essentially trying to find a function, let's say \(y\), whose rate of change is given.

In the problem provided, \( y' = 3x + e^x \) is a first-order ordinary differential equation.

• The equation is termed 'first-order' because the highest derivative present is the first derivative \(y'\).
• These types of equations often model real-life situations where you need to understand how a quantity changes over time or space.

Solving them usually involves integration, as you seek to jump from the derivative \(y'\) back to the original function \(y\).

Being adept at these equations is crucial in fields like physics and engineering, where systems change continuously, and understanding these changes is key. The first-order differential equation in this problem sets the stage for how mathematicians and scientists can deduce the behavior and solutions of various systems.