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Chapter 1: Problem 1

Find all solutions of the following equations: (a) \(y^{\prime}-2 y=1\) (b) \(y^{\prime}+y=e^{x}\) (c) \(y^{\prime}-2 y=x^{2}+x\) (d) \(3 y^{\prime}+y=2 e^{-x}\) (e) \(y^{\prime}+3 y=e^{i x}\)

Short Answer

Expert verified
(a) \( y = -\frac{1}{2} + Ce^{2x} \); (b) \( y = \frac{1}{2}e^x + Ce^{-x} \); (c) Solve \( e^{-2x}y = \int(x^2+x)e^{-2x}dx + C \); (d) Integrate and solve \( e^{x/3}y \); (e) Integrate and solve \( e^{3x}y \).

Step by step solution

01

Identify the Type of Equation

Each of the given equations is a first-order linear differential equation. The standard form is \( y'+P(x)y=Q(x) \). We can solve these equations using the integrating factor method.
02

Solve Equation (a)

For the equation \( y^{\prime} - 2y = 1 \), identify \( P(x) = -2 \) and \( Q(x) = 1 \). The integrating factor \( \mu(x) = e^{\int P(x)\,dx} = e^{-2x} \). Multiply the entire equation by \( e^{-2x} \) to get \( e^{-2x}y^{\prime} - 2e^{-2x}y = e^{-2x} \). Integrating both sides gives \( e^{-2x}y = -\frac{1}{2}e^{-2x} + C \), solve for \( y \): \( y = -\frac{1}{2} + Ce^{2x} \).
03

Solve Equation (b)

For \( y^{\prime} + y = e^{x} \), identify \( P(x) = 1 \) and \( Q(x) = e^x \). The integrating factor is \( \mu(x) = e^{\int 1\,dx} = e^x \). Multiply through by \( e^x \) gives \( e^xy^{\prime} + e^xy = e^{2x} \). Integrate to find \( e^xy = \frac{1}{2}e^{2x} + C \), solve for \( y \): \( y = \frac{1}{2}e^{x} + Ce^{-x} \).
04

Solve Equation (c)

For \( y^{\prime} - 2y = x^2 + x \), find \( P(x) = -2 \), \( Q(x) = x^2 + x \). The integrating factor is \( \mu(x) = e^{-2x} \). Multiply: \( e^{-2x}y^{\prime} - 2e^{-2x}y = e^{-2x}(x^2 + x) \). Integrate the right side term by term to obtain \( e^{-2x}y = \int (x^2+x)e^{-2x}dx + C \). Solve for \( y \) to get the general solution.
05

Solve Equation (d)

Rearrange \( 3y^{\prime} + y = 2e^{-x} \) to \( y^{\prime} + \frac{1}{3}y = \frac{2}{3}e^{-x} \) where \( P(x) = \frac{1}{3} \) and \( Q(x) = \frac{2}{3}e^{-x} \). The integrating factor is \( \mu(x) = e^{x/3} \). Multiply through by \( e^{x/3} \) and integrate to find \( e^{x/3}y = \int \frac{2}{3}e^{-2x/3}dx + C \). Solve for \( y \).
06

Solve Equation (e)

For \( y^{\prime} + 3y = e^{ix} \), identify \( P(x) = 3 \) and \( Q(x) = e^{ix} \). The integrating factor is \( \mu(x) = e^{3x} \). Multiply the equation by \( e^{3x} \) and integrate to find \( e^{3x}y = \int e^{(i-3)x}dx + C \). Solve for \( y \) to get the solution.

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Key Concepts

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

Integrating Factor Method
The integrating factor method is a reliable technique for solving first-order linear differential equations. It simplifies the process by transforming a differential equation into an easily integrable form. The method starts by rewriting the given differential equation in standard form:
  • Identify the standard form: \( y' + P(x)y = Q(x) \)
  • Determine the integrating factor, \( \mu(x) = e^{\int P(x)\,dx} \)
  • Multiply the entire differential equation by \( \mu(x) \)
  • The left side now becomes the derivative of the product: \( (\mu(x)y)' \)
This method transforms the differential equation into a recognizable format whose left side is an exact derivative, making it straightforward to integrate. The integrating factor typically appears as an exponential function, linking the rate of change of one variable to another through integration.
First-Order Differential Equation
A first-order differential equation is a type of equation that involves the derivatives of a function concerning one variable. In essence, it predicts how a function behaves by understanding its rate of change. These equations take the form:
  • \( y' = f(x, y) \)
where \( y' \) represents the derivative of the function \( y \) with respect to \( x \). First-order linear differential equations specifically have the structure:
  • \( y' + P(x)y = Q(x) \)
with \( P(x) \) and \( Q(x) \) being known functions of \( x \). Understanding this structure is key to applying the integrating factor method, as it allows you to set up the equation for integration. The primary goal is to find the function \( y(x) \) that satisfies the equation, providing insights into various phenomena modeled by differential equations.
Solving Differential Equations
Solving differential equations involves identifying a function that satisfies the relationship outlined by the equation's derivatives. The process typically follows these steps:
  • Start with the initial equation in its standard form \( y' + P(x)y = Q(x) \).
  • Find the integrating factor \( \mu(x) = e^{\int P(x)\,dx} \).
  • Multiply the whole equation by \( \mu(x) \) to simplify it into a form that is straightforward to integrate.
  • Integrate both sides to find a general solution containing an arbitrary constant \( C \).
  • Solve for \( y \) in terms of \( x \) and \( C \).
This method allows you to determine a family of solutions for the differential equation. Each solution corresponds to different initial conditions or parameters, represented by the constant \( C \). These solutions could represent real-world phenomena, such as growth and decay processes, oscillations, and much more, making this approach vital in both mathematics and applied sciences.

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Most popular questions from this chapter

4\. Consider the equation $$ L y^{\prime}+R y=E \sin \omega x $$ where \(L, R, E, \omega\) are positive constants. (a) Compute the solution \(\phi\) satisfying \(\phi(0)=0\). (b) Show that this solution may be written in the form $$ \phi(x)=\frac{E \omega L}{R^{2}+\omega^{2} L^{2}} e^{-(R / L) x}+\frac{E}{\sqrt{R^{2}+\omega^{2} L^{2}}} \sin (\omega x-\alpha) $$ where \(\alpha\) is the angle satisfying $$ \cos \alpha=\frac{\mathrm{R}}{\sqrt{R^{2}+\omega^{2} L^{2}}}, \quad \sin \alpha=\frac{\omega L}{\sqrt{R^{2}+\omega^{2} L^{2}}} $$ (c) Sketch the graph of the solution given in (b).

Consider the equation \(y^{\prime}+(\cos x) y=e^{-\sin x}\). (a) Find the solution \(\phi\) which satisfies \(\phi(\pi)=\pi\). (b) Show that any solution \(\phi\) has the property that $$ \phi(\pi k)-\phi(0)=\pi k $$ where \(k\) is any integer.

Verify that the following are solutions of the differential equations given: (a) \(\phi(x)=e^{-\sin x}\), for \(y^{\prime}+(\cos x) y=0\) (b) \(\phi(x)=\sin x-1\), for \(y^{\prime}+(\cos x) y=\sin x \cos x\) (c) \(\phi(x)=1\), for \(y^{\prime \prime}-y^{\prime}=0\) (d) \(\phi(x)=e^{x}\), for \(y^{\prime \prime}-y^{\prime}=0\) (e) \(\phi(x)=c_{1}+c_{2} e^{x}\), for \(y^{\prime \prime}-y^{\prime}=0,\left(c_{1}, c_{2}\right.\) any constants) (f) \(\phi(x)=\sin 2 x\), for \(y^{\prime \prime}+4 y=0\) (g) \(\phi(x)=e^{2 i x}\), for \(y^{\prime \prime}+4 y=0\) (h) \(\phi(x)=c_{1} \cos k x+c_{2} \sin k x\), for \(y^{\prime \prime}+k^{2} y=0,\left(k\right.\) a positive constant, and \(c_{1}, c_{2}\) any constants)

Consider the equation \(y^{\prime}=k y\) on \(-\infty0\) then the magnitude of every non-trivial (not identically zero) solution tends to \(\infty\) as \(x \rightarrow \infty\). (d) What can you say about the magnitudes of the solutions if \(\operatorname{Re} k=0\) ?

5\. The equation $$ y^{\prime}+\alpha(x) y=\beta(x) y^{k}, \quad(k \text { constant) } $$ is called Bernoulli's equation. (a) Show that the formal substitution \(z=y^{1-k}\) transforms this into the linear equation $$ z^{\prime}+(1-k) \alpha(x) z=(1-k) \beta(x) $$ (b) Find all solutions of \(y^{\prime}-2 x y=x y^{2}\).
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