Chapter 2: Problem 8
Solve the given differential equations. $$ \left(x^{2}+x-2\right) \frac{d y}{d x}+3(x+1) y=x-1 $$
Short Answer
Expert verified
The solution for the given first-order linear differential equation can be written as:
\[y(x) = \frac{1}{e^{\frac{3}{2}x^2 + 3x}}\left(\int (x-1)e^{\frac{3}{2}x^2 + 3x} dx + C\right)\]
The integral in this expression may require a specific technique, a numerical method, or may be difficult to calculate algebraically.
Step by step solution
01
Identify the Differential Equation Type
We are given a first-order linear differential equation:
\[
\left(x^{2}+x-2\right) \frac{d y}{d x}+3(x+1) y=x-1
\]
This type of equation can be written in the form:
\[
P(x)\frac{dy}{dx} + Q(x)y = R(x)
\]
where P(x) = x^2 + x - 2, Q(x) = 3(x+1) and R(x) = x-1.
02
Calculate the Integrating Factor
The next step is to find the integrating factor (IF). The integrating factor for a first-order linear differential equation can be found using the formula:
\[
IF = e^{\int Q(x)dx}
\]
Let us first calculate the integral of Q(x):
\[
\int Q(x)dx = \int 3(x+1)dx = \int (3x+3)dx = \frac{3}{2}x^2 + 3x + C
\]
Now C can be removed as we are looking for an integrating factor, not the actual integral:
\[
IF = e^{\int Q(x)dx} = e^{\frac{3}{2}x^2 + 3x}
\]
03
Multiply the Given Differential Equation by the Integrating Factor
Now, we multiply both sides of the differential equation by the integrating factor:
\[
\begin{aligned} e^{\frac{3}{2}x^2 + 3x} \left[(x^{2}+x-2) \frac{d y}{d x}+3(x+1) y\right] &= (x-1)e^{\frac{3}{2}x^2 + 3x} \\ \Rightarrow \frac{d}{dx}\left(ye^{\frac{3}{2}x^2 + 3x}\right) &= (x-1)e^{\frac{3}{2}x^2 + 3x} \end{aligned}
\]
We can see that the left side of the differential equation now represents the derivative of the product of y(x) and the integrating factor.
04
Integrate Both Sides of the Equation
Integrate both sides of the equation with respect to x:
\[
\int \frac{d}{dx}\left(ye^{\frac{3}{2}x^2 + 3x}\right) dx= \int (x-1)e^{\frac{3}{2}x^2 + 3x} dx
\]
The left side simplifies to:
\[
ye^{\frac{3}{2}x^2 + 3x} = \int (x-1)e^{\frac{3}{2}x^2 + 3x} dx + C
\]
The right side of the equation may require a numerical method or a specific substitution to calculate the integral, which might be difficult to perform.
05
Solve for y(x)
Finally, solve for y(x) by dividing both sides of the equation by the integrating factor:
\[
y(x) = \frac{1}{e^{\frac{3}{2}x^2 + 3x}}\left(\int (x-1)e^{\frac{3}{2}x^2 + 3x} dx + C\right)
\]
The integral in this expression may require a specific technique, a numerical method, or may be difficult to calculate algebraically. If the problem does not require an explicit solution, the equation above represents y(x) in terms of an integral expression.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
The integrating factor is a crucial concept when solving first-order linear differential equations. It is essentially a function that, when multiplied by the original differential equation, transforms it into an integrable form. The mystique of the integrating factor lies in its ability to simplify what might otherwise appear as a complex equation.
To derive the integrating factor, we use the standard form of a first-order linear differential equation:
To derive the integrating factor, we use the standard form of a first-order linear differential equation:
- \[ \frac{dy}{dx} + P(x)y = Q(x) \]
- \[ IF = e^{\int P(x)dx} \]
Differential Equations
Differential equations are mathematical equations involving functions and their derivatives. They appear prominently in fields such as physics, engineering, and economics due to their ability to model changing systems. The equations depict relationships where change, observed through the derivative, plays a key role.
In this context, we encounter a type known as first-order linear differential equations, which are of the form:
In this context, we encounter a type known as first-order linear differential equations, which are of the form:
- \[ P(x) \frac{dy}{dx} + Q(x)y = R(x) \]
Integration Techniques
Integration acts as the reverse operation of differentiation. It's a core technique used to solve differential equations. Particularly, it allows us to find the function given its rate of change.
When dealing with a differential equation, after applying the integrating factor, the equation is rearranged so that it becomes integrable. At this point, techniques such as substitution or even numerical methods might be utilized, depending on the complexity of the functions involved. To ensure you're prepared to tackle most integrals, familiarize yourself with:
When dealing with a differential equation, after applying the integrating factor, the equation is rearranged so that it becomes integrable. At this point, techniques such as substitution or even numerical methods might be utilized, depending on the complexity of the functions involved. To ensure you're prepared to tackle most integrals, familiarize yourself with:
- The power, exponential, and trigonometric integrals
- Substitution method
- Integration by parts
Solving Linear Differential Equations
Solving linear differential equations requires understanding and applying systematic techniques. The primary focus is transforming the problem into a form that is straightforward to solve.
For first-order linear differential equations, the method typically involves:
For first-order linear differential equations, the method typically involves:
- Identifying the type of the equation
- Calculating the integrating factor
- Multiplying through by the integrating factor to form a simple integrable expression
- Integrating both sides of the equation to find the function