Chapter 10: Problem 24
Find the solution of the following initial-value problems: (a) \((x-1) \frac{\mathrm{d} x}{\mathrm{~d} t}+t+1=0, \quad x(0)=2\) (b) \((2 x+t) \frac{\mathrm{d} x}{\mathrm{~d} t}+x-t=0, \quad x(0)=-1\) (c) \(\left(2-x t^{2}\right) \frac{\mathrm{d} x}{\mathrm{~d} t}-x^{2} t=0, \quad x(1)=2\) (d) \(\cos t \frac{\mathrm{d} x}{\mathrm{~d} t}-x \sin t+1=0, \quad x(0)=2\)
Short Answer
Step by step solution
Solve Part (a)
Solve Part (b)
Solve Part (c)
Solve Part (d)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problem
Consider an example where we solve the initial-value problem \((x-1) \frac{\mathrm{d} x}{\mathrm{~d} t} + t + 1=0, \ x(0)=2\). Here, the initial condition \(x(0) = 2\) is imperative since it allows us to solve the constant of integration that emerges when we integrate to find the general solution. Without such a condition, our solution would be incomplete, leaving an arbitrary constant, \(C\), resulting in an entire family of potential solutions, instead of a specific one.
Separation of Variables
For instance, if we consider the equation \((x-1) \frac{\mathrm{d} x}{\mathrm{~d} t} = -(t + 1)\), we can rearrange to \(\frac{\mathrm{d} x}{x-1} = -(t + 1) \mathrm{d} t\). This format allows us to integrate each side individually.
- The left side integrates with respect to \(x\),
- while the right side integrates with respect to \(t\).
Integration Factor
The integration factor, \(\mu(t)\), is derived by exponentiating the integral of the function \(P(t)\). Considering the equation \(\cos t \frac{\mathrm{d} x}{\mathrm{~d} t} - x \sin t + 1=0\), once it is written in the form \(\frac{\mathrm{d} x}{\mathrm{d} t} + \tan t x = \cos t\), the integration factor is \( \mu(t) = e^{\int \tan t \, \mathrm{d} t} = \sec t\).
- This integration factor simplifies and allows the exact integration of the linear transformed equation.
- The process leads to finding an expression involving \(x\).
First-Order Linear Differential Equations
Linear differential equations may be homogeneous (when \(Q(t) = 0\)) or non-homogeneous (when \(Q(t) eq 0\)). Solving these equations often involves an integration factor, which simplifies them into integrable forms.
- This method consists of finding \(\mu(t) = e^{\int P(t) \, \mathrm{d} t}\) which helps transform the equation.
- After multiplying through by \(\mu(t)\), the equation becomes integrable straightforwardly in terms of \(x\).