Problem 19 - Advanced Engineering Mathematics

Question

Each \(D E\) in Problems \(15-22\) is a Bernoulli equation. In Problems 15-20, solve the given differential equation by using an appropriate substitution. $$ t^{2} \frac{d y}{d t}+y^{2}=t y $$

Step-by-Step Solution

Answer

The solution for the given Bernoulli differential equation is \( y = \frac{1}{\sqrt{2(\ln|t| + \frac{1}{t} + C)}} \).

Step 1: Identify Type of Differential Equation

The given equation is of the form \( t^2 \frac{dy}{dt} + y^2 = ty \). This is a Bernoulli differential equation because it has the form \( \frac{dy}{dt} + P(t)y = Q(t)y^n \). In this equation, \( n = 2 \), \( P(t) = 0 \), and \( Q(t) = \frac{1}{t^2} \).

Step 2: Apply Appropriate Substitution

For Bernoulli equations, we use the substitution \( v = y^{1-n} = y^{-1} \). Differentiate both sides with respect to \( t \): \( \frac{dv}{dt} = -y^{-2}\frac{dy}{dt} \). Using this substitution, the original equation can be transformed into a linear differential equation in terms of \( v \).

Step 3: Rewrite the Differential Equation

Substitute \( y = \frac{1}{v} \) and \( \frac{dy}{dt} = -v^{-2} \frac{dv}{dt} \) into the original equation. This transforms it into: \[ t^2 \left(-v^{-2} \frac{dv}{dt}\right) + \left(\frac{1}{v}\right)^2 = t\left(\frac{1}{v}\right) \] which simplifies to \[ -t^2 \frac{dv}{dt} + \frac{1}{v} = \frac{t}{v} \].

Step 4: Simplify and Solve for v

Rearrange and simplify the transformed equation: \[ -t^2 \frac{dv}{dt} = \frac{t}{v} - \frac{1}{v} \] which simplifies further to \[ -t^2 \frac{dv}{dt} = \frac{t-1}{v} \]. Multiply both sides by \( v \) to clear the fraction: \[ -t^2 v \frac{dv}{dt} = t - 1 \].

Step 5: Separation of Variables

Separate variables such that \( -v \frac{dv}{dt} \) is on one side and \( \frac{t-1}{t^2} \) is on the other: \[ v dv = \frac{t-1}{t^2} dt \]. Integrate both sides: \[ \int v dv = \int \frac{t-1}{t^2} dt \].

Step 6: Integrate Both Sides

Integrate: \( \frac{1}{2}v^2 = \int \left(\frac{1}{t} - \frac{1}{t^2}\right) dt = \ln|t| + \frac{1}{t} + C \).

Step 7: Back-substitute for y

Recall the substitution \( v = \frac{1}{y} \), thus \( v^2 = \frac{1}{y^2} \). Solve for \( y \): \[ \frac{1}{2} \frac{1}{y^2} = \ln|t| + \frac{1}{t} + C \]. Rearrange to find \( y \): \[ y^2 = \frac{1}{2(\ln|t| + \frac{1}{t} + C)} \]. Therefore, \( y = \frac{1}{\sqrt{2(\ln|t| + \frac{1}{t} + C)}} \).

Key Concepts

Differential Equations

Differential equations are fundamental in understanding how varying quantities relate to one another. Often appearing in the form of equations involving derivatives, they describe how a particular function changes over time or space. For instance, the equation \( t^2 \frac{dy}{dt} + y^2 = ty \) represents a relationship between time \( t \) and a function \( y(t) \). This equation is not just an ordinary differential equation—it is specifically identified as a Bernoulli differential equation, which often deals with the product of a function and its derivative raised to a power.

• The key to solving differential equations is identifying the type: linear, separable, or in this case, Bernoulli.
• Bernoulli equations take the form \( \frac{dy}{dt} + P(t)y = Q(t)y^n \).
• Recognizing this allows us to apply appropriate transformations or substitution methods to simplify the problem.

Understanding these distinctions lays the groundwork for developing the right strategies to solve different forms of differential equations.

Substitution Methods

Substitution methods are powerful techniques used to simplify complicated differential equations. By replacing parts of the equation with new variables, you can transform the equation into a more familiar or manageable form. In the context of the Bernoulli differential equation, the substitution \( v = y^{1-n} \) is pivotal.This particular substitution is designed to make the equation linear, enabling easier manipulation and resolution:

• Start with the substitution \( v = y^{-1} \), corresponding to the case where \( n = 2 \).
• Differentiating both sides gives \( \frac{dv}{dt} = -y^{-2} \frac{dy}{dt} \).
• Substitute back into the original equation to convert it into a linear form with respect to \( v \).

Such a transformation allows for the separation of variables, a helpful technique that paves the way towards finding a solution.

Linear Differential Equation

Once transformed, our Bernoulli differential equation becomes a linear differential equation. Linear differential equations represent a class of problems where the unknown function and its derivatives appear in a linear form.The steps involved include:

• Rewrite in terms of the new variable by substituting \( y = \frac{1}{v} \) and its derivative accordingly.
• Simplify the equation to clear any fractions and make it solvable.
• After simplification, employ techniques like separation of variables and integration to find\( v \).

Eventually, solving for \( v \) will guide us back to the original function \( y \) through back-substitution. This loop from nonlinear to linear back to the solution of \( y \) illustrates the elegance of using linear transformations in differential equations.