Problem 9
Question
Solve the given differential equation by using an appropriate substitution. $$ -y d x+(x+\sqrt{x y}) d y=0 $$
Step-by-Step Solution
Verified Answer
The solutions are \(y = 0\) or \(y = \frac{4}{x}\).
1Step 1: Identify the substitution
The given differential equation is \(-y \, dx + (x + \sqrt{x y}) \, dy = 0\). To simplify this, notice the term \(\sqrt{x y}\). Using the substitution \(v = \sqrt{x y}\), implies \(y = v^2 / x\). By letting \(y = v^2/x\), differentiating gives \(dy = \frac{2v}{x} \frac{dx}{dx} - \frac{v^2}{x^2} dx\).
2Step 2: Apply the substitution
Substitute \(y = \frac{v^2}{x}\) and the corresponding differential into the equation:\(-\frac{v^2}{x} \, dx + (x + v) \left(\frac{2v}{x} dx - \frac{v^2}{x^2} dx \right) = 0\).
3Step 3: Simplify and rearrange
Substitute into the original equation resulting in:\[-v^2 dx + 2v \, dx = 0\].Simplify the equation, resulting in:\[v(v-2) \, dx = 0\].
4Step 4: Solve the simplified equation
Since the equation \(v(v-2) \, dx = 0\) is satisfied for all \(dx\), it indicates that either \(v = 0\) or \(v = 2\).
5Step 5: Back-substitute to original variables
Recalling that the substitution was \(v = \sqrt{x y} \), substituting back gives:1. \(\sqrt{x y} = 0\) implies \(xy = 0\), hence \(y = 0\) when \(x eq 0\).2. \(\sqrt{x y} = 2\) implies \(xy = 4\), solving for \(y\) gives \(y = \frac{4}{x}\).
6Step 6: General solution
The general solution to the differential equation is either \(y = 0\) or \(y = \frac{4}{x}\).
Key Concepts
Substitution MethodGeneral SolutionSeparable Equations
Substitution Method
The substitution method is a powerful tool used in solving differential equations. It involves replacing one variable with another that simplifies the equation, making it easier to solve.
To use the substitution method effectively, you first identify an appropriate substitution that simplifies the problematic term. In the example given, the term \( \sqrt{x y} \) suggests a substitution because it is not easily separable. By letting \( v = \sqrt{x y} \), the equation becomes more manageable and can be rewritten in terms of \( v \). This substitution turns derivatives and expressions into forms that might be familiar from standard solved problems.
Once the substitution is made, the differential equation is solved in terms of the new variable. After solving, it's essential to substitute back to the original variable to find the final solution to the given differential equation. This method breaks down complex problems into simpler parts that are easier to tackle.
To use the substitution method effectively, you first identify an appropriate substitution that simplifies the problematic term. In the example given, the term \( \sqrt{x y} \) suggests a substitution because it is not easily separable. By letting \( v = \sqrt{x y} \), the equation becomes more manageable and can be rewritten in terms of \( v \). This substitution turns derivatives and expressions into forms that might be familiar from standard solved problems.
Once the substitution is made, the differential equation is solved in terms of the new variable. After solving, it's essential to substitute back to the original variable to find the final solution to the given differential equation. This method breaks down complex problems into simpler parts that are easier to tackle.
General Solution
The general solution of a differential equation involves finding an expression that describes all possible solutions.
General solutions often depend on constants, termed 'arbitrary constants', to cover the whole family of solutions. In the provided exercise, you arrive at a general solution after back-substituting the variable \( v \) into the original terms. This shows all potential functions \( y \) that satisfy the original differential equation.
For example, the solutions \( y = 0 \) and \( y = \frac{4}{x} \) are both derived from the values \( v = 0 \) and \( v = 2 \) obtained in the simplified solution. These expressions represent the family of solutions that the equation can yield, providing flexibility in application across different contexts or boundary conditions. It demonstrates that different initial or boundary conditions might lead to different specific solutions within this general framework.
General solutions often depend on constants, termed 'arbitrary constants', to cover the whole family of solutions. In the provided exercise, you arrive at a general solution after back-substituting the variable \( v \) into the original terms. This shows all potential functions \( y \) that satisfy the original differential equation.
For example, the solutions \( y = 0 \) and \( y = \frac{4}{x} \) are both derived from the values \( v = 0 \) and \( v = 2 \) obtained in the simplified solution. These expressions represent the family of solutions that the equation can yield, providing flexibility in application across different contexts or boundary conditions. It demonstrates that different initial or boundary conditions might lead to different specific solutions within this general framework.
Separable Equations
A separable equation is a type of differential equation where variables can be separated on opposite sides of the equation, allowing it to be solved by simple integration.
In separable equations, the goal is to manipulate the equation into a form where all terms containing one variable are on one side, and all terms containing the other variable are on the opposite side. This separation allows you to integrate each side independently with respect to its variable.
Even though the original exercise does not directly feature a separable equation, the process of using substitution and simplification results in expressions similar to separable forms. Once the relationship between the variables is clearer and in simplified forms, integration becomes straightforward. Understanding separable equations helps solve differential equations not initially in separable form by transforming them into one.
In separable equations, the goal is to manipulate the equation into a form where all terms containing one variable are on one side, and all terms containing the other variable are on the opposite side. This separation allows you to integrate each side independently with respect to its variable.
Even though the original exercise does not directly feature a separable equation, the process of using substitution and simplification results in expressions similar to separable forms. Once the relationship between the variables is clearer and in simplified forms, integration becomes straightforward. Understanding separable equations helps solve differential equations not initially in separable form by transforming them into one.
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