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\(d y / d t=(y+1) /(t+1)\)

Short Answer

Expert verified
The general solution is |y+1| = |A(t+1)|, where A is an arbitrary constant.

Step by step solution

01

- Identify the type of differential equation

The given differential equation \(\frac{dy}{dt} = \frac{y+1}{t+1}\) is a first-order ordinary differential equation. Identify that it is separable.
02

- Rewrite the equation

Rewrite the equation to separate variables. \[\frac{dy}{y+1} = \frac{dt}{t+1}\]
03

- Integrate both sides

Integrate both sides of the equation. \[\begin{aligned} \int \frac{1}{y+1} dy &= \int \frac{1}{t+1} dt \ \ \ \text{Let } u = y+1 \text{ and } v = t+1. Then the integrals become,: \int \frac{1}{u} du &= \int \frac{1}{v} dv \end{aligned}\] \[\begin{aligned} \ln|u| &= \ln|v| + C \ \ \ \text{Subsititute back }u \text{ and }v\ \ \ \ln|y+1| &= \ln|t+1| + C \end{aligned}\]
04

- Solve for y

Step 5 - Determine the constant

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Key Concepts

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

Separable Differential Equations
A separable differential equation is a type of first-order ordinary differential equation (ODE) that can be manipulated so that all terms involving the dependent variable (y) are on one side of the equation, and all terms involving the independent variable (t) are on the other side.
In the given exercise, the equation is \(\frac{dy}{dt} = \frac{y+1}{t+1}\).
We identify that this ODE is separable, because we can rewrite it to separate the variables as follows:
  • Move y-related terms to one side: \(\frac{dy}{y+1}\).
  • Move t-related terms to the other side: \(\frac{dt}{t+1}\).
By successfully doing this, we convert the single differential equation into two integrable functions.This technique is particularly useful as it simplifies the process of solving the ODE.
When you separate variables, it paves the way to use integration on both sides, ultimately finding a solution that ties y and t together.
Integration Techniques
Integration is a fundamental part of solving separable differential equations. After separating variables, the next step involves integrating both sides of the equation to find a relation between y and t.
Here’s how to proceed with the given exercise:
  • First, we have: \(\frac{dy}{y+1} = \frac{dt}{t+1}\).
  • To integrate, use the substitution method:
    • Let \(u = y + 1\) on the left side, then \(du = dy\).
    • Let \(v = t + 1\) on the right side, then \(dv = dt\).
  • The integrals become: \( \int \frac{1}{u} \, du = \int \frac{1}{v} \, dv\).
  • Applying the integration technique, we get:
    \( \ln|u| = \ln|v| + C \).
The integration for \( \frac{1}{u} \) results in the natural logarithm, \(\ln|u|\), and similarly for \(\frac{1}{v}\).
Understanding these common integrals is crucial for tackling differential equations efficiently.
Ordinary Differential Equations
An ordinary differential equation (ODE) is an equation that contains a function of one independent variable and its derivatives. The term 'ordinary' differentiates it from partial differential equations, which may involve multiple independent variables.
In the context of this problem, the given ODE is a first-order ODE because it involves the first derivative of y with respect to t, i.e., \( \frac{dy}{dt} = \frac{y+1}{t+1}\).

To solve this type of ODE, we typically follow these steps:
  • Identify the type of ODE. In this case, it is separable.
  • Separate the variables.
  • Integrate both sides.
  • Combine like terms and solve for the dependent variable (y).
  • Determine the constant of integration (C), using any initial conditions provided.
The last step, determining the constant, is essential because the integration process results in a family of solutions, and an initial condition helps to pinpoint the exact solution.
Understanding the nature of ODEs and the standard methods to solve them forms the foundation of many applications in physics, engineering, and other scientific fields.

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Most popular questions from this chapter

\(e^{t y} d t+t e^{t y} / y d y=0\)

Equations of the form \(y=t f\left(y^{\prime}\right)+g\left(y^{\prime}\right)\) are called Lagrange equations. These equations are solved by making the substitution \(p=y^{\prime}(t)\). (a) Differentiate \(y=t f\left(y^{\prime}\right)+g\left(y^{\prime}\right)\) with respect to \(t\) to obtain $$ y^{\prime}=t f^{\prime}\left(y^{\prime}\right) y^{\prime \prime}+f\left(y^{\prime}\right)+g^{\prime}\left(y^{\prime}\right) y^{\prime \prime} . $$ (b) Substitute \(p\) into the equation to obtain $$ \begin{aligned} p &=t f^{\prime}(p) \frac{d p}{d t}+f(p)+g^{\prime}(p) \frac{d p}{d t} \\ &=f(p)+\left(t f^{\prime}(p)+g^{\prime}(p)\right) \frac{d p}{d t} \end{aligned} $$ (c) Solve this equation for \(d t / d p\) to obtain the linear equation $$ \frac{d t}{d p}=\frac{t f^{\prime}(p)+g^{\prime}(p)}{p-f(p)} $$ which is equivalent to $$ \frac{d t}{d p}+\frac{f^{\prime}(p)}{f(p)-p} t=\frac{g^{\prime}(p)}{p-f(p)} . $$ This first order linear equation can be solved for \(t\) in terms of \(p\). Then, \(t(p)\) can be used with \(y=t f(p)+g(p)\) to obtain an equation for \(y\).

\(d y / d t=\left(4 y^{2}-t^{2}\right) /(2 t y), y(1)=1\)

Suppose that a drug is added to the body at a rate \(r(t)\), and let \(y(t)\) represent the concentration of the drug in the bloodstream at time \(t\) hours. In addition, suppose that the drug is used by the body at the rate \(k y\), where \(k\) is a positive constant. Then, the net rate of change in \(y(t)\) is given by the equation \(d y / d t=r(t)-k y\). If at \(t=0\), there is no drug in the body, we determine \(y(t)\) by solving the initial value problem $$ d y / d t=r(t)-k y, \quad y(0)=0 . $$ (a) Suppose that \(r(t)=r\), where \(r\) is a positive constant. In this case, the drug is added at a constant rate. Sketch the phase line for \(d y / d t=r-k y\). Solve the IVP. Determine \(\lim _{t \rightarrow \infty} y(t)\). How does this limit correspond to the phase line? (b) Suppose that \(r(t)=1+\sin t\) and \(k=\) 1. In this case, the drug is added at a periodic rate. Solve the IVP. Determine \(\lim _{t \rightarrow \infty} y(t)\), if it exists. Describe what happens to the drug concentration over time. (c) Suppose that \(r(t)=e^{-t}\) and \(k=1\). In this case, the rate at which the drug is added decreases over time. Solve the IVP. Determine \(\lim _{t \rightarrow \infty} y(t)\), if it exists. Describe what happens to the drug concentration over time.

\(d \theta / d t=e^{2 t}+2 \theta, \theta(0)=0\)

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