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Magazine Chapter 1: Problem 13
$$ x y^{\prime}=y+\sqrt{x^{2}+y^{2}} $$
Short Answer
Expert verified
The solution is \( y = x\sinh(\ln|x| + C) \).
Step by step solution
01
Substitute for simplicity
To simplify the equation, let's set a new variable. Set \( y = ux \), which gives us \( y' = u'x + u \). This substitution helps us express the problem in terms of \( u \) and \( x \).
02
Rewrite the equation
Substitute \( y = ux \) and \( y' = u'x + u \) into the original differential equation: \[ x(u'x + u) = ux + \sqrt{x^2 + (ux)^2} \]. This equation can be manipulated further.
03
Simplify the equation
Divide the entire equation by \( x \) to make it simpler: \[ u'x + u = u + \frac{\sqrt{x^2 + u^2x^2}}{x} \]. Simplify inside the square root to get \( \sqrt{1 + u^2} \).
04
Cancel terms and simplify
Notice that \( ux \) cancels with the \( u \) term on the left, reducing the equation to \( xu' = \sqrt{1 + u^2} \).
05
Separate variables
Rearrange the equation to allow separation of variables: \[ u' = \frac{\sqrt{1 + u^2}}{x} \]. This can be rewritten for separation: \[ \frac{du}{\sqrt{1 + u^2}} = \frac{dx}{x} \].
06
Integrate both sides
Integrate both sides of the equation: \( \int \frac{du}{\sqrt{1 + u^2}} = \int \frac{dx}{x} \). The left-hand side integrates to \( \sinh^{-1}(u) \), and the right-hand side integrates to \( \ln|x| + C \), where \( C \) is the integration constant.
07
Solve for \( u \)
The equation from integration becomes \( \sinh^{-1}(u) = \ln|x| + C \). Solve for \( u \) to get \( u = \sinh(\ln|x| + C) \).
08
Back-substitute to find \( y \)
Recall that \( y = ux \), substitute \( u = \sinh(\ln|x|+C) \) back to get \( y = x\sinh(\ln|x|+C) \). This is the general solution of the differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
Sometimes, it can be tricky to solve a differential equation directly. To simplify things, you can use a technique called the "substitution method." This involves introducing a new variable to make the equation more manageable.
In our original problem, we made the substitution by setting \( y = ux \). This means we're expressing \( y \) in terms of a product of two other variables, \( u \) and \( x \). This substitution simplifies the relationship between \( x \), \( y' \), and \( y \).
In our original problem, we made the substitution by setting \( y = ux \). This means we're expressing \( y \) in terms of a product of two other variables, \( u \) and \( x \). This substitution simplifies the relationship between \( x \), \( y' \), and \( y \).
- This gets us into a form that is easier to work with.
- It allows us to express the derivative \( y' \) in terms of the new variable \( u \), making complex equations more straightforward.
Variable Separation
After substitution, the next step is often "variable separation." This is a way to manipulate the equation such that each variable appears on only one side of the equation.
Our differential equation simplifies to \( xu' = \sqrt{1 + u^2} \) with substitution. To separate the variables, we rewrite this as \( u' = \frac{\sqrt{1 + u^2}}{x} \) and then rearrange to get \( \frac{du}{\sqrt{1 + u^2}} = \frac{dx}{x} \).
Our differential equation simplifies to \( xu' = \sqrt{1 + u^2} \) with substitution. To separate the variables, we rewrite this as \( u' = \frac{\sqrt{1 + u^2}}{x} \) and then rearrange to get \( \frac{du}{\sqrt{1 + u^2}} = \frac{dx}{x} \).
- This separation allows us to handle each variable separately during the integration process.
- The goal is to turn the differential equation into integrable parts for both \( u \) and \( x \).
Integration Techniques
Once variables have been separated, we move on to integrating both sides of the equated expression. In our problem, the separated equation looks like \( \int \frac{du}{\sqrt{1 + u^2}} = \int \frac{dx}{x} \).
Integration techniques help solve this by evaluating integrals:
Integration techniques help solve this by evaluating integrals:
- The integral of \( \frac{du}{\sqrt{1 + u^2}} \) is a special kind of integral that results in \( \sinh^{-1}(u) \) (also known as the inverse hyperbolic sine function).
- The integral \( \frac{dx}{x} \) is much simpler, resulting in the natural logarithm, \( \ln|x| \).
General Solution
Finding a general solution involves bringing all the pieces together after integration. We discover a relationship in the form of \( \sinh^{-1}(u) = \ln|x| + C \), where \( C \) is the constant of integration.
To express the solution in terms of the original variable \( y \), substitute back \( u = \sinh(\ln|x| + C) \). Since \( y = ux \), this gives us \( y = x \sinh(\ln|x| + C) \).
To express the solution in terms of the original variable \( y \), substitute back \( u = \sinh(\ln|x| + C) \). Since \( y = ux \), this gives us \( y = x \sinh(\ln|x| + C) \).
- This solution covers all possible values of \( y \) for any \( x \), depending on the constant \( C \).
- It's a general solution because it includes a family of solutions, each corresponding to a different initial condition.
