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Find the solution of the following differential equations: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=2\) (b) \(\frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=t\) (c) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=\mathrm{e}^{-4 t}\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+t x=-2 t\)

Short Answer

Expert verified
Solutions are (a) \(x(t) = \frac{2}{3} + Ce^{-3t}\); (b) \(x(t) = -\frac{1}{4} te^{-4t} - \frac{1}{16} e^{-4t} + Ce^{4t}\); (c) \(x(t) = -\frac{1}{2} e^{-4t} + Ce^{-2t}\); (d) \(x(t) = -2 + Ce^{-\frac{t^2}{2}}\).

Step by step solution

01

Differential Equation (a)

The equation \( \frac{\mathrm{d} x}{\mathrm{~d} t} + 3x = 2 \) is a first-order linear differential equation. The general solution formula is given by \( x(t) = e^{-\int P(t) \, dt} \left( \int Q(t) e^{\int P(t) \, dt} \, dt + C \right) \), where \( P(t) = 3 \) and \( Q(t) = 2 \).
02

Integrating Factor for (a)

The integrating factor \( \mu(t) \) is determined by \( e^{\int P(t) \, dt} = e^{\int 3 \, dt} = e^{3t} \).
03

Solve (a)

Substitute into the general solution formula: \( x(t) = e^{-3t} \left( \int 2 e^{3t} \, dt + C \right) = e^{-3t} \left( \frac{2}{3} e^{3t} + C \right) \).Simplifying gives: \( x(t) = \frac{2}{3} + Ce^{-3t} \).
04

Differential Equation (b)

The equation \( \frac{\mathrm{d} x}{\mathrm{~d} t} - 4x = t \) is another first-order linear differential equation. Here, \( P(t) = -4 \) and \( Q(t) = t \).
05

Integrating Factor for (b)

Calculate the integrating factor: \( \mu(t) = e^{\int -4 \, dt} = e^{-4t} \).
06

Solve (b)

Using the formula: \( x(t) = e^{4t} \left( \int t e^{-4t} \, dt + C \right) \).The integral \( \int t e^{-4t} \, dt \) is solved using integration by parts: Let \( u = t \), \( dv = e^{-4t} dt \). After integration, \( x(t) = -\frac{1}{4} t e^{-4t} - \frac{1}{16} e^{-4t} + Ce^{4t} \).
07

Differential Equation (c)

The equation \( \frac{\mathrm{d} x}{\mathrm{~d} t} + 2x = e^{-4t} \) with \( P(t) = 2 \) and \( Q(t) = e^{-4t} \).
08

Integrating Factor for (c)

The integrating factor is \( e^{\int 2 \, dt} = e^{2t} \).
09

Solve (c)

Using the formula: \( x(t) = e^{-2t} \left( \int e^{-4t} e^{2t} \, dt + C \right) = e^{-2t} \left( \int e^{-2t} \, dt + C \right) \). The integral simplifies to \( x(t) = -\frac{1}{2} e^{-4t} + Ce^{-2t} \).
10

Differential Equation (d)

The equation \( \frac{\mathrm{d} x}{\mathrm{~d} t} + tx = -2t \) with \( P(t) = t \) and \( Q(t) = -2t \).
11

Integrating Factor for (d)

Calculate the integrating factor: \( e^{\int t \, dt} = e^{\frac{t^2}{2}} \).
12

Solve (d)

Substitute into the formula: \( x(t) = e^{-\frac{t^2}{2}} \left( \int -2t e^{\frac{t^2}{2}} \, dt + C \right) \). The integral simplifies to \( x(t) = -2 + Ce^{-\frac{t^2}{2}} \).

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Key Concepts

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

First-order linear differential equations
First-order linear differential equations are the most basic type of differential equations you will encounter. These equations are of the form \( \frac{\mathrm{d} x}{\mathrm{~d} t} + P(t)x = Q(t) \). They involve only the first derivative of the unknown function, in this case, \( x(t) \). These equations often appear in modeling simple systems like electrical circuits or population dynamics.
To solve such equations, we need a systematic method that involves transforming the equation into a more manageable form. This transformation is achieved using a mathematical tool known as the "integrating factor." Once we apply the integrating factor, the differential equation becomes easier to integrate directly, allowing us to find the solution to the original problem.
Basic steps include:
  • Identify \(P(t)\) and \(Q(t)\).
  • Compute the integrating factor \(\mu(t)\).
  • Multiply the entire equation by this integrating factor.
  • Integrate both sides to find the solution.
Understanding this method is crucial, as it's a foundation for solving more complex differential equations that you might encounter later on in your studies.
Integrating factor
The integrating factor is a key component in solving first-order linear differential equations. It's a function, typically expressed as \( \mu(t) = e^{\int P(t) \, dt} \), which we multiply with the entire differential equation to simplify it.
Once we multiply the differential equation by the integrating factor, the left side of the equation typically becomes the derivative of a product of the integrating factor and the unknown function. This transformation simplifies the problem because it translates into a more direct integration process.
To use the integrating factor effectively:
  • Compute \( \mu(t)\) based on the identified \(P(t)\).
  • Multiply both sides of the differential equation by \( \mu(t)\).
  • Recognize the product rule derivative form, which allows for easy integration.
This technique is extraordinarily valuable because it converts a potentially complicated problem into a straightforward integration, making the path to the solution much clearer.
Integration by parts
Integration by parts is another essential technique used, especially in situations where the integrals are too complex to solve directly. It extends the concept of the product rule for differentiation to integration. The formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Here, \(u\) and \(dv\) are functions of \(t\). You choose them based on ease of integration and differentiation.
Follow these steps for integration by parts:
  • Identify parts of the integrand to use as \(u\) and \(dv\).
  • Differentiating \(u\) gives \(du\), and integrating \(dv\) gives \(v\).
  • Substitute into the integration by parts formula.
  • Evaluate the resulting integrals.
This method can be particularly useful when solving integrals that appear as part of solving linear differential equations, as demonstrated in equations like \( \int t e^{-4t} \, dt \). Mastery of this technique significantly broadens the range of differential equations you can solve.

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Most popular questions from this chapter

The end of a chain, coiled near the edge of a horizontal surface, falls over the edge. If the friction between the chain and the horizontal surface is negligible and the chain is inextensible then, when a length \(x\) of chain has fallen, the equation of motion is $$ \frac{\mathrm{d}}{\mathrm{d} t}(m x v)=m g x $$ where \(m\) is the mass per unit length of the chain, \(g\) is gravitational acceleration and \(v\) is the velocity of the falling length of the chain. If the mass per unit length of the chain is constant show that this equation can be expressed as $$ x v \frac{\mathrm{d} v}{\mathrm{~d} x}+v^{2}=g x $$ and, by putting \(y=v^{2}\), show that \(\left.v=\sqrt{(} 2 g x / 3\right)\).

A chemical \(A\) is formed by an irreversible reaction from chemicals \(B\) and \(C\). Assuming that the amounts of \(B\) and \(C\) are adequate to sustain the reaction, the amount of \(A\) formed at time \(t\) is governed by the differential equation $$ \frac{\mathrm{d} A}{\mathrm{~d} t}=K(1-\alpha A)^{7} $$ If no \(A\) is present at time \(t=0\), find an expression for the amount of \(A\) present at time \(t\).

State the order of each of the following differential equations and name the dependent and independent variables. Classify each equation as linear homogeneous, linear nonhomogeneous or nonlinear differential equations. (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=0\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=0\) (c) \(\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}+x=0\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=t^{2}\) (e) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=\cos t+\mathrm{e}^{t}\)

For each of the following differential equations write down the differential operator \(\mathrm{L}\) that would enable the equation to be expressed as \(\mathrm{L}[x(t)]=0\) : (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=f(t) x\) (b) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+(\sin t) \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4 t^{2} x=0\) (c) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+(\sin t) \frac{\mathrm{d} x}{\mathrm{~d} t}=(t+\cos t) x\) (d) \((\sin t) \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\cos t}{t} x\) (e) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{b x}{t}\) (f) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=x \mathrm{t} \mathrm{e}^{t^{2}}\) (g) \(\frac{\mathrm{d}}{\mathrm{d} t}\left(t^{2} \frac{\mathrm{d} x}{\mathrm{~d} t}\right)=t \frac{\mathrm{d}}{\mathrm{d} t}(x t)\) (h) \(\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{t} \frac{\mathrm{d}}{\mathrm{d} t}\left(t^{2} x\right)\right]=x t\)

Which of the following sets of functions are linearly dependent and which are linearly independent? (a) \(\\{\sin t+2 \cos t, \sin t-2 \cos t, 2 \sin t+\cos t,\), \(2 \sin t-\cos t\\}\) (b) \(\\{\sin t, \cos t, \sin 2 t, \cos 2 t, \sin 3 t, \cos 3 t\\}\) (c) \(\left\\{1+2 t, 2 t-3 t^{2}, 3 t^{2}+4 t^{3}, 4 t^{3}-5 t^{4}\right\\}\) (d) \(\left\\{1+2 t, 2 t-3 t^{2}, 3 t^{2}+4 t^{3}, 4 t^{3}\right\\}\) (e) \(\left\\{1,1+2 t, 2 t-3 t^{2}, 3 t^{2}+4 t^{3}, 4 t^{3}\right\\}\) (f) \(\\{\ln a, \ln b, \ln a b\\}\) (g) \(\left\\{\mathrm{e}^{s}, \mathrm{e}^{t}, \mathrm{e}^{s+t}\right\\}\) (h) \(\left\\{\mathrm{e}^{t}, \mathrm{e}^{2 t}-\mathrm{e}^{t}, \mathrm{e}^{3 t}-\mathrm{e}^{2 t}, \mathrm{e}^{2 t}\right\\}\) (i) \(\\{f(t), f(t)-g(t), f(t)+g(t)\\}\) (j) \(\left\\{1-2 t^{2}, t-3 t^{3}, 2 t^{2}-4 t^{4}, 3 t^{3}-5 t^{5}\right\\}\) (k) \(\left\\{1,1+t, 1+t+t^{2}, 1+t+t^{2}+t^{3}\right\\}\)

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