Problem 17 - Advanced Engineering Mathematics

Question

Solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=y\left(x y^{3}-1\right) $$

Step-by-Step Solution

Answer

Use the substitution \( v = xy \), derive \( \frac{dv}{dx} \, \), and reintegrate to solve.

Step 1: Identify a Suitable Substitution

The given differential equation is \( \frac{dy}{dx} = y(xy^3 - 1) \). Notice that the term \( xy^3 \) suggests a substitution involving \( y \). We can try the substitution \( v = y^n \) for some \( n \) or look for a simpler form. Let us try \( v = xy \). This suggests differentiating \( v = xy \) to link \( dy/dx \) and \( dv/dx \).

Step 2: Differentiate Substitution Equation

Using the substitution \( v = xy \), differentiate it with respect to \( x \):\[ \frac{dv}{dx} = y + x \frac{dy}{dx} \]Plugging \( \frac{dy}{dx} = y(xy^3 - 1) \) into it gives:\[ \frac{dv}{dx} = y + x(y(xy^3 - 1)) = y + x(y^4x - y) = y + x^2y^4 - xy \].

Step 3: Simplify and Solve for \( \frac{dv}{dx} \)

Simplify the expression for \( \frac{dv}{dx} \):\[ \frac{dv}{dx} = y - xy + x^2y^4 = y(1 - x) + x^2y^4 \]By substituting \( v = xy \), \( y = \frac{v}{x} \), rewrite the equation: \[ \frac{dv}{dx} = \frac{v}{x}(1 - x) + x^2\left(\frac{v}{x}\right)^4 \].

Step 4: Simplify and Integrate

Substituting \( y = \frac{v}{x} \), we now have:\[ \frac{dv}{dx} = \frac{v}{x}(1 - x) + \frac{v^4}{x^2} \], resulting in:\[ \frac{dv}{dx} = \frac{v(1 - x) + v^4x^2}{x^2} \]Now separate variables and integrate accordingly. This task might need simplification before integrating.

Step 5: Solve and Back Substitute

After carrying out the integration, solve for \( v \) and substitute back \( v = xy \). This would provide an implicit or explicit solution to the original differential equation. Adjust constants as per initial conditions if provided. The specific integration is complex, so the exact steps depend on the reduction achieved from simplification.

Key Concepts

Substitution Method

The substitution method is a powerful algebraic technique used to simplify differential equations by transforming them into a more familiar form. In our original exercise, the equation given is:
\( \frac{dy}{dx} = y(xy^3 - 1) \).
Here, a substitution helps in reducing the complexity of the problem.
To apply substitution effectively, look for repetitive or recursive expressions. In this case, noticing the expression \( xy^3 \) indicates a potential transformation can be applied by substituting with a product of variables. By letting \( v = xy \), we cleverly simplify the right-hand side terms.
After substituting, we differentiate \( v = xy \) with respect to \( x \). This step is crucial as it allows us to express \( \frac{dy}{dx} \) in terms of \( \frac{dv}{dx} \), connecting the substitution back to the original variables. The substitution method essentially swaps an originally difficult differential equation for one that is easier to integrate.

Integration Techniques

Once you've applied a substitution to a differential equation, integrating the resulting expression is the next step. In our case, after substituting \( y = \frac{v}{x} \) and transforming dependencies, we arrive at:
\[ \frac{dv}{dx} = \frac{v(1-x) + v^4}{x^2} \]
We must then handle integrating this equation. Integration techniques vary based on the equation form, involving different strategies such as separation of variables, partial fractions, or even trigonometric identities. Here, the separation of variables is primarily used. We manipulate the equation to isolate \( dv \) on one side and \( dx \) on the other, enabling integration over both sides.
Simplifying expressions before integration can also help to ease this process. Check if the terms can be factored or rearranged on either side; this often reveals simpler forms for calculating integrals.
The importance of integration techniques lies in their ability to transition from a differential form to a functional form, allowing for the determination of solutions, whether implicit or explicit.

Implicit and Explicit Solutions

Differential equations can yield solutions in implicit or explicit forms. In this exercise, once substitution and integration are complete, we must back-solve for the original variables.
An explicit solution is one where the dependent variable is isolated on one side of the equation, such as \( y = f(x) \). These are usually more straightforward and easy to interpret, allowing direct calculation of outputs for given inputs.
However, an implicit solution involves the dependent variable mingled with other terms, represented in equations like \( F(x, y) = 0 \). They convey relations between variables without necessarily separating them. Often, implicit forms are more common after solving differential equations, especially when algebraic manipulation or simplification doesn't lead to a clean separation.
In practice, even if your solution remains implicit, utilizing initial conditions or further algebraic methods can sometimes enable or simplify transitioning to an explicit form, assuring a more direct understanding of the variable relationships inherent to the system described by the differential equation.