Problem 19
Question
Put \(x y=v\), i.e., \(y+x \frac{d y}{d x}=\frac{d v}{d x}\)
Step-by-Step Solution
Verified Answer
The substitution works to simplify the problem.
1Step 1: Recognize the Substitution
The problem suggests using the substitution \(v = xy\). This means that \(y = \frac{v}{x}\). The idea is to express \(y\) and its derivative \(\frac{dy}{dx}\) in terms of \(v\).
2Step 2: Differentiate the Substitution
To find \(\frac{d v}{d x}\), differentiate \(v = xy\) with respect to \(x\). The differentiation of the product gives us \(\frac{dv}{dx} = y + x \frac{dy}{dx}\). This matches the left-hand side of the original equation, confirming that the substitution is correct.
Key Concepts
DifferentiationProduct RuleSubstitution in Differential Equations
Differentiation
Differentiation is a fundamental tool in calculus used to calculate the rate at which a function changes at any given point. Imagine tracking how a car's speed changes instantaneously; this is what differentiation does for mathematical functions.
When we differentiate a function, we are looking for its derivative. The derivative of a function at a particular point provides the slope of the tangent to the function at that point. It's like finding out how steep a hill is at a certain location. To differentiate a function, you follow specific rules, such as the power rule, product rule, and chain rule, to systematically find these derivatives.
In our problem, we differentiate the equation using the substitution method. That means we first replaced a combination of variables with a single variable, which makes it easier to handle. Differentiation in this context helps us determine the relationship between these variables by finding how they change with respect to an independent variable, like time or space.
When we differentiate a function, we are looking for its derivative. The derivative of a function at a particular point provides the slope of the tangent to the function at that point. It's like finding out how steep a hill is at a certain location. To differentiate a function, you follow specific rules, such as the power rule, product rule, and chain rule, to systematically find these derivatives.
In our problem, we differentiate the equation using the substitution method. That means we first replaced a combination of variables with a single variable, which makes it easier to handle. Differentiation in this context helps us determine the relationship between these variables by finding how they change with respect to an independent variable, like time or space.
Product Rule
The product rule is a technique used in differentiation when you have to find the derivative of the product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), where both are differentiable, the derivative of their product is given by this formula:
In the exercise, the product rule comes into play when differentiating the expression \( v = xy \) with respect to \( x \). Here, \( x \) is one function, and \( y \) is another function of \( x \). Applying the product rule, you get \( y + x \frac{dy}{dx} \), which is key to solving the problem as it forms part of the substitution equation.
- \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \)
In the exercise, the product rule comes into play when differentiating the expression \( v = xy \) with respect to \( x \). Here, \( x \) is one function, and \( y \) is another function of \( x \). Applying the product rule, you get \( y + x \frac{dy}{dx} \), which is key to solving the problem as it forms part of the substitution equation.
Substitution in Differential Equations
Substitution in differential equations is a method used to simplify and solve complex equations by introducing a new variable. This technique is particularly useful when equations have complicated terms or patterns. It's akin to using a code name for a long, complex name to make a speech easier to follow.
In the given exercise, the substitution \( v = xy \) helps reduce the complexity of the differential equation. By expressing \( y \) in terms of \( v \) and \( x \), you transform the equation into a simpler form that's easier to handle mathematically.
This substitution allows us to shift the problem from a difficult equation to one that is more straightforward, taking advantage of already known differentiation methods. After substituting, we can solve for \( \frac{dv}{dx} \) using the relationship \( y + x \frac{dy}{dx} \), making substitution a powerful and effective problem-solving tool.
In the given exercise, the substitution \( v = xy \) helps reduce the complexity of the differential equation. By expressing \( y \) in terms of \( v \) and \( x \), you transform the equation into a simpler form that's easier to handle mathematically.
This substitution allows us to shift the problem from a difficult equation to one that is more straightforward, taking advantage of already known differentiation methods. After substituting, we can solve for \( \frac{dv}{dx} \) using the relationship \( y + x \frac{dy}{dx} \), making substitution a powerful and effective problem-solving tool.
Other exercises in this chapter
Problem 16
\(d y-\sin x \sin y d x=0\)
View solution Problem 18
We have \(\frac{d y}{d x}=\sin \frac{x-y}{2}-\sin \frac{x+y}{2}\) \(=-2 \cos \frac{x}{2} \sin \frac{y}{2}\)
View solution Problem 26
\(\frac{y^{\prime \prime \prime}}{y^{\prime \prime}}=8\) or \(\log y^{\prime \prime}=8 x+c\)
View solution Problem 28
\(\frac{d y}{d x}=\frac{y}{x}\left[\log \frac{y}{x}+1\right]\)
View solution