Problem 27 - Advanced Engineering Mathematics

Question

Each \(D E\) in Problems 23-30 is of the form given in (5). In Problems 23-28, solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=2+\sqrt{y-2 x+3} $$

Step-by-Step Solution

Answer

The solution is \( y = \left(\frac{x + C}{2}\right)^2 + 2x - 3 \).

Step 1: Identify the Given Form

The differential equation provided is \( \frac{d y}{d x} = 2 + \sqrt{y - 2x + 3} \). This suggests a substitution might involve completing the square or isolating a component in the expression \( y - 2x + 3 \).

Step 2: Choose a Suitable Substitution

Let's try a substitution \( v = y - 2x + 3 \). This implies that \( y = v + 2x - 3 \) and hence \( \frac{dy}{dx} = \frac{dv}{dx} + 2 \). Substitute these into the original equation.

Step 3: Substitute and Simplify

Substitute \( \frac{dy}{dx} = \frac{dv}{dx} + 2 \) into the differential equation. This gives us \( \frac{dv}{dx} + 2 = 2 + \sqrt{v} \), which simplifies to \( \frac{dv}{dx} = \sqrt{v} \).

Step 4: Solve the Separated Differential Equation

Separate the variables to solve \( \frac{dv}{\sqrt{v}} = dx \). Integrate both sides: \( \int \frac{dv}{\sqrt{v}} = \int dx \), which results in \( 2\sqrt{v} = x + C \), where \( C \) is the integration constant.

Step 5: Solve for Substitution Variable

Solve for \( v \) by rearranging to get \( \sqrt{v} = \frac{x + C}{2} \). Squaring both sides gives \( v = \left(\frac{x + C}{2}\right)^2 \).

Step 6: Reverse the Substitution

Recall the substitution \( v = y - 2x + 3 \). Substitute \( v \) back to get \( y - 2x + 3 = \left(\frac{x + C}{2}\right)^2 \). Rearrange this to express \( y \) in terms of \( x \): \( y = \left(\frac{x + C}{2}\right)^2 + 2x - 3 \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in solving differential equations where you transform the coordinates to simplify the equation. This is especially useful when the differential equation involves complex expressions or terms that are difficult to integrate or separate directly. When using this method, the goal is to replace a part of the expression with a new variable that makes the equation easier to handle.

• To use substitution, you typically choose a part of the function that you suspect could simplify the equation. For example, if the term involves a sum or a complex polynomial, finding a substitution that captures one part of this can help reduce complexity.
• In the given exercise, the expression within the square root, such as \(y - 2x + 3\), was identified as a candidate for substitution.
• Once the substitution is made, you replace every instance of the expression in the differential equation with the new variable. You also derive the corresponding derivative with respect to the new variable.
• This new form often leads to simpler differential equations that can be managed using other techniques, such as separation of variables.

Overall, the substitution should simplify the original equation structure, making the solution process more straightforward.

Variable Separation

Variable separation is another essential method for solving differential equations, especially those that are separable. It involves algebraic manipulation of the equation so that each side contains only one variable. This allows integration independently on both sides.

• To separate variables, rearrange the terms so that all instances of one variable and its differential are on one side, while the other variable and its differential are on the other.
• In the solution given, after substituting and simplifying the original equation, we arrive at \(\frac{dv}{\sqrt{v}} = dx\), successfully separating variables \(v\) and \(x\).
• Once separated, these can be integrated independently, which allows us to find expressions for each variable's solution.
• The solution step where we separate variables leads directly into using integration techniques, which is the core technique arrived at after this separation process.

Variable separation is favored because it reduces the complexity of solving directly mixed-variable relations by isolating them for straightforward integration.

Integration Techniques

Integration is a critical step in solving differential equations. Once you’ve separated the variables, integrating each side of the equation allows you to find the relationship between the variables.

• In the context of this exercise, after separating the variables, we need to integrate \(\int \frac{dv}{\sqrt{v}}\) and \(\int dx\).
• For \(\int \frac{dv}{\sqrt{v}}\), we use the formula \(\int v^n \, dv = \frac{v^{n+1}}{n+1} + C\), knowing that \(\frac{1}{\sqrt{v}} = v^{-1/2}\). This gives us \(2\sqrt{v}\) after integration.
• The integration of \(dx\) is straightforward and results in \(x + C\).
• Integration introduces an integration constant \(C\), essential for solving differential equations as it represents the family of solutions.
• After integrating both sides, interpret the results to express the original variables, completing the solution process with the final rearrangement.

Mastering integration is fundamental as it transforms the differential equations back into functional equations you can solve explicitly for variables.