Chapter 7: Problem 136
Obtain the general solution for: \(\quad(\mathrm{dy} / \mathrm{d} \mathrm{x})=\mathrm{e}^{\mathrm{x}}+\mathrm{y}\)
Short Answer
Expert verified
The general solution to the given first-order linear ODE is: \( y(x) = e^x (x + C) \).
Step by step solution
01
Identify the type of ODE and its components
Our given ODE is: \( \frac{dy}{dx} = e^x + y \)
We rearrange it to get the standard linear ODE form: \( \frac{dy}{dx} - y = e^x \)
Now, we clearly see that \(P(x) = -1\) and \(Q(x) = e^x\) .
02
Compute the integrating factor (IF)
The integrating factor is given by the exponential of the integral of \(P(x)\), i.e.,
IF = \( e^{\int{P(x) dx}} = e^{\int{-1 dx}} = e^{-x} \)
03
Multiply the original ODE by the integrating factor
Insert the integrating factor into the equation by multiplying each term:
\( e^{-x} \frac{dy}{dx} - e^{-x} y = e^{x}e^{-x} \)
which simplifies to:
\( e^{-x} \frac{dy}{dx} - e^{-x} y = 1 \)
04
Integrate both sides of the equation
Now, the left-hand side of the equation will be the exact derivative of \(y \cdot e^{-x}\).
So, we can rewrite the equation as:
\( \frac{d}{dx}(y e^{-x}) = 1 \)
Now, integrate both sides with respect to x:
\( \int{\frac{d}{dx}(y e^{-x}) dx} = \int{1 dx} \)
Which gives:
\( y e^{-x} = x + C \), where C is the constant of integration.
05
Solve for the general solution
Finally, isolate 'y' to find the general solution:
\( y = e^x (x + C) \)
So, the general solution to the given first-order linear ODE is:
\( y(x) = e^x (x + C) \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
To solve certain types of first-order linear differential equations, we use a method that involves something called the 'integrating factor'. This method transforms the equation into a form where it becomes easier to integrate.
The integrating factor itself is typically a function derived from the rearranged standard linear equation of the format \[ \frac{dy}{dx} + P(x)y = Q(x) \].
The integrating factor itself is typically a function derived from the rearranged standard linear equation of the format \[ \frac{dy}{dx} + P(x)y = Q(x) \].
- First, identify the coefficient \(P(x)\) in front of \(y\).
- Second, calculate the integrating factor which is \(e^{\int P(x) \, dx}\).
- In our problem, \(P(x)\) equals \(-1\), so the integrating factor is \(e^{-x}\).
General Solution
Finding the general solution to a first-order linear differential equation involves a systematic approach using an integrating factor. Once the equation is rearranged appropriately, and the integrating factor is applied, integrating the transformed equation is the next step.
- After rewiring, the left side becomes the derivative \( \frac{d}{dx}(y \cdot e^{-x}) \).
- We integrate both sides with respect to \(x\): \[ \int{\frac{d}{dx}(y \cdot e^{-x}) \, dx} = \int{1 \, dx} \].
- This results in: \[ y \cdot e^{-x} = x + C \].
- Finally, solve for \(y\) to get the general solution: \[ y = e^{x} (x + C) \].
Ordinary Differential Equation
An Ordinary Differential Equation (ODE) is an equation containing a function of one independent variable and its derivatives. Such equations are used to model real-life scenarios where change is continuous, like population growth or cooling of a body.
In our example, the differential equation \(\frac{dy}{dx} = e^x + y\) includes:
In our example, the differential equation \(\frac{dy}{dx} = e^x + y\) includes:
- The function \(y(x)\), which we want to solve for.
- Its derivative \(\frac{dy}{dx}\).
- The terms \(e^x\) and \(y\), representing how \(y\) changes with respect to \(x\).
Separable Variables
The concept of 'separable variables' applies to a subclass of differential equations where the variables can be relocated on separate sides of the equation. This is a step towards solving the differential equation by integration.
Unfortunately, not all first-order linear differential equations are separable, but recognizing when you can separate variables offers a straightforward solution method.
For an equation to be separable, it should be expressible in the form: \[ \frac{dy}{dx} = g(y)h(x) \].
Unfortunately, not all first-order linear differential equations are separable, but recognizing when you can separate variables offers a straightforward solution method.
For an equation to be separable, it should be expressible in the form: \[ \frac{dy}{dx} = g(y)h(x) \].
- Rearrange terms to form: \[ \frac{dy}{g(y)} = h(x) \, dx \].
- Integrate each side: \[ \int \frac{1}{g(y)} \, dy = \int h(x) \, dx \].