Chapter 1: Problem 17
$$x^{3} y^{\prime}+3 x^{2} y=5 \sinh 10 x$$
Short Answer
Expert verified
The solution is \( y = -\frac{1}{2x^2} \cosh(10x) + \frac{C}{x^3} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \(x^{3} y^{\prime} + 3x^{2} y = 5 \sinh (10x)\). This is a first-order linear differential equation in standard form: \( y' + P(x)y = Q(x) \). To solve it, identify \(P(x)\) and \(Q(x)\). Initially, divide every term by \(x^3\) to simplify.
02
Simplify the Differential Equation
Simplify the equation by dividing all terms by \(x^3\): \(y' + \frac{3}{x}y = \frac{5}{x^3}\sinh(10x)\). Here, \( P(x) = \frac{3}{x} \) and \( Q(x) = \frac{5}{x^3}\sinh(10x) \).
03
Find the Integrating Factor
The integrating factor \(\mu(x)\) is calculated using \(\mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3\ln |x|} = x^3\).
04
Multiply by the Integrating Factor
Multiply the entire equation by the integrating factor \(x^3\): \(x^3 y' + 3x^2 y = 5 \sinh(10x)\). Notice that it's identical to the original equation, confirming correctness.
05
Rewrite as an Exact Differential
Observe and confirm that \((x^3 y)' = x^3 y' + 3x^2 y\). Therefore, the left-hand side becomes an exact derivative: \((x^3 y)' = 5 \sinh(10x)\).
06
Integrate Both Sides
Integrate both sides with respect to \(x\): \(\int (x^3 y)' \, dx = \int 5 \sinh(10x) \, dx\). The left side integrates to \(x^3 y\) and the right side becomes \(-\frac{1}{2}\cosh(10x) + C\), where \(C\) is the integration constant.
07
Solve for y
Solve for \(y\): \(x^3 y = -\frac{1}{2}x \cosh(10x) + C \Rightarrow y = -\frac{1}{2x^2} \cosh(10x) + \frac{C}{x^3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order linear differential equation
A **First-order linear differential equation** is a type of differential equation that involves the first derivative of the unknown function, usually denoted as \( y' \), and the function itself, \( y \). The general form is
In our problem, dividing the original equation by \( x^3 \) puts it into this standard recognizable form.
- \( y' + P(x)y = Q(x) \)
- They only involve the first derivative \( y' \) of the function \( y(x) \).
- Both \( P(x) \) and \( Q(x) \) are functions solely of \( x \). This linearity with respect to \( y \) and its derivatives is what qualifies them as "linear."
In our problem, dividing the original equation by \( x^3 \) puts it into this standard recognizable form.
Integrating factor
Solving **first-order linear differential equations** often requires an integrating factor, which is a function that simplifies the solution process. The integrating factor \( \mu(x) \) is calculated from the function \( P(x) \) found in the differential equation:
Once we perform this operation, solving the equation becomes a matter of simple integration over both sides, leading directly to the solution in terms of \( y \).
- For example, if \( y' + \frac{3}{x}y = \frac{5}{x^3} \sinh(10x) \), then \( P(x) = \frac{3}{x} \).
- The integrating factor is then \( \mu(x) = e^{\int P(x) \, dx} = e^{3\ln |x|} = x^3 \).
Once we perform this operation, solving the equation becomes a matter of simple integration over both sides, leading directly to the solution in terms of \( y \).
Hyperbolic functions
You might have noticed the term involving **hyperbolic functions** in the equation: \( 5 \sinh(10x) \). Hyperbolic functions, such as \( \sinh \) and \( \cosh \), relate closely to exponential functions:
When integrating hyperbolic functions, they often translate smoothly into their exponential form, making them relatively straightforward to handle in calculus. For instance, integrating \( 5 \sinh(10x) \), we find the result \(-\frac{1}{2}\cosh(10x) + C\), which is easier to manage when expressed using \( \cosh \).
Understanding these functions' relationships and properties can greatly simplify working with complex differential equations like this one.
- The hyperbolic sine function is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
- The hyperbolic cosine function is similarly defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
When integrating hyperbolic functions, they often translate smoothly into their exponential form, making them relatively straightforward to handle in calculus. For instance, integrating \( 5 \sinh(10x) \), we find the result \(-\frac{1}{2}\cosh(10x) + C\), which is easier to manage when expressed using \( \cosh \).
Understanding these functions' relationships and properties can greatly simplify working with complex differential equations like this one.