Find the general solution of each of the following differential equations. \(y^{\prime}+y=e^{x}\)

Short Answer

Expert verified
The general solution is \( y = \frac{1}{2} e^x + Ce^{-x} \).

Step by step solution

01

Identify the form of the differential equation

The differential equation given is a first-order linear differential equation of the form \(y' + P(x)y = Q(x)\). Here, \(P(x) = 1\) and \(Q(x) = e^{x}\).
02

Find the integrating factor

The integrating factor \( \mu(x) \) is calculated using the formula \( \mu(x) = e^{\int P(x) dx} \).Plugging in \(P(x) = 1\), we get \( \mu(x) = e^{\int 1 dx} = e^x \).
03

Multiply the differential equation by the integrating factor

Multiply both sides of the differential equation by the integrating factor \(e^x\):\( e^x y' + e^x y = e^x e^x \). This simplifies to \( e^x y' + e^x y = e^{2x} \).
04

Simplify the left-hand side

Notice that the left-hand side \( e^x y' + e^x y \) can be written as the derivative of \( y \cdot e^x\):\( \frac{d}{dx}(y e^x) = e^{2x} \).
05

Integrate both sides

Integrate both sides with respect to x:\( \int \frac{d}{dx}(y e^x)dx = \int e^{2x} dx \).This simplifies to \( y e^x = \int e^{2x} dx \).
06

Perform the integration

Integrate \( \int e^{2x} dx \):\( \int e^{2x} dx = \frac{1}{2} e^{2x} + C \), where C is the constant of integration.So, \( y e^x = \frac{1}{2} e^{2x} + C \).
07

Solve for y

Divide both sides by \( e^x\) to solve for y:\( y = \frac{1}{2} e^x + Ce^{-x} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor
First-order linear differential equations often require special techniques to solve. One of these methods is using an integrating factor. An integrating factor is a function that simplifies the equation, making it easier to solve.

For a differential equation of the form:
\( y' + P(x)y = Q(x) \),
the integrating factor, denoted as \(\bf{\textmu(x)} \), is found using the formula:
\[ \textmu(x) = e^{\int P(x) \, dx} \]
Here, we see that our given differential equation is \( y' + y = e^x \). Identifying \(P(x) = 1 \) and \(Q(x) = e^x \), we calculate the integrating factor:
\[ \textmu(x) = e^{\int 1 \, dx} = e^{x} \]
This function \(e^x\) will make our differential equation easier to work with.
General Solution
After calculating the integrating factor, the next step in solving the differential equation is to use it to find the general solution.

We do this by multiplying the entire differential equation by the integrating factor \(e^x\):

\( e^x y' + e^x y = e^{x} e^x \)
Simplifies to: \( e^x y' + e^x y = e^{2x} \)
Left-hand side can be written as the derivative: \( \frac{d}{dx}(y e^x) = e^{2x} \)

Now, we integrate both sides with respect to x:
\[ \int \frac{d}{dx}(y e^x)dx = \int e^{2x} dx \]
Which simplifies to: \( y e^x = \int e^{2x} dx \)

Performing the integration, we get:
\[ \int e^{2x} dx = \frac{1}{2}e^{2x} + C \]
Thus, the solution in its intermediate form is: \( y e^x = \frac{1}{2} e^{2x} + C \)

Finally, we solve for y:
\( y = \frac{1}{2} e^x + Ce^{-x} \)
This represents the general solution of the differential equation.
Differential Equation Integration
Integration is a fundamental technique to solve differential equations. After transforming the equation using the integrating factor, we often encounter integrals we need to evaluate:

Consider the equation after applying the integrating factor \(e^x\):
\( \frac{d}{dx}(y e^x) = e^{2x} \)
This tells us that the left-hand side is the derivative of \( y e^x \), so to eliminate the derivative, we integrate both sides:

\[ \int \frac{d}{dx}(y e^x) dx = \int e^{2x} dx \]

The integral on the left-hand side becomes \( y e^x \), while the integral on the right side is performed as follows:
\[ \int e^{2x} dx = \frac{1}{2}e^{2x} + C \]

Thus resulting in:
\( y e^x = \frac{1}{2} e^{2x} + C \)
Finally, to solve for y, we divide both sides by \(e^x\), simplifying to:
\(y = \frac{1}{2} e^x + Ce^{-x} \)
This demonstrates how integrating both sides facilitates finding the solution to the differential equation. Integration helps turn the transformed equation into a clearer, explicit solution for y.

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Most popular questions from this chapter

An object of mass \(m\) falls from rest under gravity subject to an air resistance proportional to its velocity. Taking the \(y\) axis as positive down, show that the differential equation of motion is \(m(d v / d t)=m g-k v\), where \(k\) is a positive constant. Find \(v\) as a function of \(t\), and find the limiting value of \(v\) as \(t\) tends to infinity; this limit is called the terminal relocity. Can you find the terminal velocity directly from the differential equation without solving it? Hint: What is \(d v / d t\) after \(v\) has reached an essentially constant value? Consider the following specific examples of this problem. (a) A person drops from an airplane with a parachute. Find a reasonable value of \(k\). (b) In the Millikan oil drop experiment to measure the charge of an electron, tiny electrically charged drops of oil fall through air under gravity or rise under the combination of gravity and an electric field. Measurements can be made only after they have reached terminal velocity. Find a formula for the time required for a drop starting at rest to reach \(99 \%\) of its terminal velocity.

Find a particular solution satisfying the given conditions. \(x y^{\prime}-y=x^{2}, \quad y=6\) when \(x=2\)

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(y^{\prime}=\frac{2 x y^{2}+x}{x^{2} y-y}\) \(y=0\) when \(x=\sqrt{2}\)

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime \prime}+2 y^{\prime \prime}+2 y^{\prime}=0\)

Find the general solution of each of the following differential equations. \(y^{\prime} \cos x+y=\cos ^{2} x\)

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