Problem 50
Question
If \(\frac{d y}{d x}=y+3>0\) and \(y(0)=2\), then \(y(\ln 2)\) is equal to: [2011] (a) 5 (b) 13 (c) \(-2\) (d) 7
Step-by-Step Solution
Verified Answer
The value of \( y(\ln 2) \) is 7.
1Step 1: Identify the Differential Equation
You are given a first-order linear differential equation \( \frac{d y}{d x} = y + 3 \). This expresses that the rate of change of \( y \) with respect to \( x \) is given by \( y + 3 \).
2Step 2: Separate Variables
To solve this differential equation, we start by rewriting it in a form that allows us to separate the variables \( y \) and \( x \). This can be done by isolating \( \frac{d y}{d x} \) on one side:\[ \frac{d y}{d x} - y = 3. \]
3Step 3: Solve the Homogeneous Equation
Solve the homogeneous equation \( \frac{d y}{d x} - y = 0 \), which is obtained by ignoring the constant on the right side. The equation is separable and the solutions are of the form:\[ y = Ce^{x}. \]
4Step 4: Particular Solution of Non-homogeneous Equation
To find a particular solution to the non-homogeneous equation \( \frac{d y}{d x} = y + 3 \), consider \( y_p = A \), a constant solution. Substituting into the equation, we get \( 0 = A - (-3) \), yielding \( A = -3 \).
5Step 5: General Solution
The general solution combines the homogeneous and particular solutions: \[ y = Ce^{x} - 3. \]
6Step 6: Apply Initial Condition
We use the initial condition \( y(0) = 2 \) to find \( C \). Substituting into the general solution:\[ 2 = Ce^{0} - 3 \] \[ 2 = C - 3 \]\[ C = 5. \]
7Step 7: Write Complete Solution
Substitute \( C = 5 \) into the general solution:\[ y = 5e^{x} - 3. \]
8Step 8: Evaluate at Specific Point
Find \( y(\ln 2) \) by substituting \( x = \ln 2 \) into the complete solution:\[ y(\ln 2) = 5e^{\ln 2} - 3 = 5 \times 2 - 3 = 10 - 3 \]\[ y(\ln 2) = 7. \]
Key Concepts
First-order linear differential equationSeparation of variablesHomogeneous equationNon-homogeneous equationInitial condition in differential equations
First-order linear differential equation
A first-order linear differential equation has the general form \( \frac{d y}{d x} + P(x)y = Q(x) \).These equations involve derivatives of the function \( y \) that are first-order, meaning they contain terms like \( \frac{d y}{d x} \).To solve a first-order linear differential equation, our main task is to find the function \( y \).The solution involves a combination of methods like separation of variables and finding integrating factors, depending on the structure of the equation.
In the given exercise, we identified \( \frac{d y}{d x} = y + 3 \) as a linear equation because it follows the pattern \( \frac{d y}{d x} + (-1)y = 3 \).Understanding the linearity of the problem helps guide us to the appropriate solving method.
In the given exercise, we identified \( \frac{d y}{d x} = y + 3 \) as a linear equation because it follows the pattern \( \frac{d y}{d x} + (-1)y = 3 \).Understanding the linearity of the problem helps guide us to the appropriate solving method.
- Co-efficient of \( y \) in the equation is constant.
- Non-linear characteristics are absent.
- The equation can be transformed for various solution techniques.
Separation of variables
Separation of variables is a technique used to solve differential equations.It involves rearranging the equation so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other.This technique works particularly well with separable differential equations.
In the exercise, we performed a transformation from\[ \frac{d y}{d x} = y + 3 \] to \[ \frac{d y}{d x} - y = 3, \] effectively separating \( y \) and \( x \) terms.While this example does not consider the full separation directly,we are moving toward it by attempting to solve the homogeneous part,which often facilitates the use of integration.
In the exercise, we performed a transformation from\[ \frac{d y}{d x} = y + 3 \] to \[ \frac{d y}{d x} - y = 3, \] effectively separating \( y \) and \( x \) terms.While this example does not consider the full separation directly,we are moving toward it by attempting to solve the homogeneous part,which often facilitates the use of integration.
- Allows transformation of complex equations into simpler forms.
- Particularly useful for equations that can be integrated directly after separation.
- Often requires manipulating equations into suitable forms for straightforward integration.
Homogeneous equation
A homogeneous equation is a special kind of differential equation where every term is a function of the dependent variable and its derivatives.In its standard form, it looks like \( M(x,y)dx + N(x,y)dy = 0 \), where both \( M \) and \( N \) are homogenous functions.
In linear terms, it is defined as \( \frac{d y}{d x} - y = 0 \),as we saw when isolating the terms of our differential equation.
Solving the homogeneous version involves finding solutions to\[ \frac{d y}{d x} = y \],which results in functions like \( y = Ce^{x} \), illustrating the pattern of exponential solutions typical of homogeneous equations.
In linear terms, it is defined as \( \frac{d y}{d x} - y = 0 \),as we saw when isolating the terms of our differential equation.
Solving the homogeneous version involves finding solutions to\[ \frac{d y}{d x} = y \],which results in functions like \( y = Ce^{x} \), illustrating the pattern of exponential solutions typical of homogeneous equations.
- Assumes the absence of any non-variable functions on the right side of the equation.
- Helps in constructing a general solution framework for more complex equations.
- Fundamental to identifying part of the solution to linear differential equations.
Non-homogeneous equation
Non-homogeneous equations differ from their homogeneous counterparts by having an additional term, usually a function or constant that does not depend on the dependent variable or its derivatives.For example, in our scenario of \( \frac{d y}{d x} = y + 3 \),the term "+3" marks it as non-homogeneous.
To solve, we locate a particular solution, which often involves substituting constants or varied functions.Here, we proposed \( y_p = A \), and found a constant solution \( A = -3 \) to satisfy the non-homogeneous aspect.This aligns with the non-homogeneous part of the solution,combining with the homogeneous counterpart to form a complete general solution.
To solve, we locate a particular solution, which often involves substituting constants or varied functions.Here, we proposed \( y_p = A \), and found a constant solution \( A = -3 \) to satisfy the non-homogeneous aspect.This aligns with the non-homogeneous part of the solution,combining with the homogeneous counterpart to form a complete general solution.
- Includes a constant or functional term added to the homogeneous equation.
- Requires a special approach to find particular solutions.
- Critical for constructing a complete solution by adding particular solutions to a homogeneous framework.
Initial condition in differential equations
Initial conditions are used in the context of differential equations to find specific solutions from a general solution.They specify the value of the solution and its derivatives at certain points, thereby narrowing down the infinite set of possibilities.
For our particular problem, the given initial condition is \( y(0) = 2 \).This translates to substituting \( x = 0 \) and \( y = 2 \) in the general solution \( y = Ce^{x} - 3 \).Using this, we can determine the value of \( C \) as 5, leading to a specific solution \( y = 5e^{x} - 3 \).
For our particular problem, the given initial condition is \( y(0) = 2 \).This translates to substituting \( x = 0 \) and \( y = 2 \) in the general solution \( y = Ce^{x} - 3 \).Using this, we can determine the value of \( C \) as 5, leading to a specific solution \( y = 5e^{x} - 3 \).
- Essential for identifying particular solutions among a broad family of solutions.
- Simplifies and specifies the form of a general solution for specific scenarios.
- Allows verification and applicability of solutions to real-world problems.
Other exercises in this chapter
Problem 47
Let \(y(x)\) be a solution of \(\frac{(2+\sin x)}{(1+y)} \frac{d y}{d x}=\cos x\). If \(y(0)=2\), then \(y\left(\frac{\pi}{2}\right)\) equals (a) \(\frac{5}{2}\
View solution Problem 48
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View solution Problem 52
The normal to a curve at \(P(x, y)\) meets the \(x\)-axis at \(G\). If the distance of \(G\) from the origin is twice the abscissa of \(P\), then the curve is a
View solution Problem 53
If \(x \frac{d y}{d x}=y(\log y-\log x+1)\), then the solution of the equation is \(\quad[\mathbf{2 0 0 5}]\) (a) \(y \log \left(\frac{x}{y}\right)=c x\) (b) \(
View solution