Chapter 8: Problem 51
In Problems, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=\frac{y+1}{x-1}, \text { with } y(2)=5 $$
Short Answer
Expert verified
The solution is \( y = 6x - 7 \).
Step by step solution
01
Recognize the Form of the Differential Equation
This differential equation is of the form \( \frac{dy}{dx} = \frac{y+1}{x-1} \), which is a separable differential equation. This means the variables can be separated on different sides of the equation.
02
Separate the Variables
Rearrange the equation to separate the variables:\[\frac{dy}{y+1} = \frac{dx}{x-1}\]Here, we have moved all the terms involving \( y \) to one side and those involving \( x \) to the other side.
03
Integrate Both Sides
Integrate both sides of the equation:\[\int \frac{1}{y+1} \, dy = \int \frac{1}{x-1} \, dx\]This results in the natural logarithm functions:\[\ln|y+1| = \ln|x-1| + C\]
04
Solve for y
Exponentiate both sides to solve for \( y \):\[y+1 = C(x-1)\]Thus,\[y = C(x-1) - 1\]
05
Use the Initial Condition to Solve for C
Plug in the initial condition \( y(2) = 5 \) into the equation to find \( C \):\[5 = C(2-1) - 1\]Solve for \( C \):\[5 = C - 1 \rightarrow C = 6\]
06
Write the Particular Solution
Substitute \( C = 6 \) back into the equation for \( y \):\[y = 6(x-1) - 1\]Simplify to get the final solution:\[y = 6x - 7\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Differential Equations
Separable differential equations are a special type of differential equation which allows variables to be separated on opposite sides of the equation. This usually involves moving all terms containing the dependent variable (say, \( y \)) to one side of the equation, and terms containing the independent variable (say, \( x \)) to the other. Here's a process that makes these equations manageable:
- Identify the form of the equation to see if it's separable, like \( \frac{dy}{dx} = g(y)h(x) \).
- Rearrange the terms so that all \( y \)-dependent terms are on one side, and all \( x \)-dependent terms are on the other.
- Find the integral of both sides separately, which eventually helps to solve for \( y \).
Initial Conditions
Initial conditions are the values supplied to help determine a specific solution to a differential equation. They specify the state of the function at a particular point, which aids in finding an exact rather than a general solution.In this case, the initial condition provided was \( y(2)=5 \). This tells us the exact value the function \( y \) takes when \( x = 2 \). By substituting this pair of \( x \) and \( y \) values into the solution, you can solve for any constants, such as \( C \) in earlier steps. This process:
- Narrows down your range from all possible functions to the one that actually fits this condition.
- Makes sure your general solution aligns with real-world or specific given scenarios.
Integration Methods
Integration is a fundamental tool in solving separable differential equations. Once the equation has been separated into terms involving \( y \) and \( x \), we integrate both sides separately. Here, since each part was in a form that corresponds to natural logarithms:
- For \( \int \frac{1}{y+1} \, dy \), this equals \( \ln|y+1| + C_1 \).
- For \( \int \frac{1}{x-1} \, dx \), this equals \( \ln|x-1| + C_2 \).