Problem 19 - Advanced Engineering Mathematics

Question

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 is \( y(t) = \frac{t}{\ln |t| + C} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is written as \( t^2 \frac{dy}{dt} + y^2 = ty \). We notice that this is a first-order differential equation with variable coefficients and a non-linear term \( y^2 \), indicating it could be Bernoulli's type.

Step 2: Rewrite in Standard Bernoulli's Form

A Bernoulli's differential equation is of the form \( y' + P(t)y = Q(t)y^n \). Rewrite the equation: \( t^2 \frac{dy}{dt} = ty - y^2 \). Rearranging gives \( \frac{dy}{dt} + \left(-\frac{1}{t}\right)y = \frac{1}{t^2}y^2 \). It matches the Bernoulli's form with \( n = 2 \), \( P(t) = -\frac{1}{t} \), and \( Q(t) = \frac{1}{t^2} \).

Step 3: Apply the Substitution

In Bernoulli's equation, use the substitution \( v = y^{1-n} = y^{-1} \), so \( y = \frac{1}{v} \). Differentiate this: \( \frac{dy}{dt} = -\frac{1}{v^2} \cdot \frac{dv}{dt} \). Substitute \( y \) and \( \frac{dy}{dt} \) into the equation.

Step 4: Transform the Equation

By substituting, the original equation becomes \( t^2 \left(-\frac{1}{v^2} \frac{dv}{dt}\right) + \frac{1}{v^2} = \frac{t}{v} \). Multiplying through by \( v^2 \) to clear fractions gives: \( -t^2 \frac{dv}{dt} + 1 = tv \). Simplify to get \( -t^2 \frac{dv}{dt} = tv - 1 \).

Step 5: Solve the Linear Differential Equation for v

Rearrange to \( \frac{dv}{dt} + \frac{v}{t} = \frac{1}{t^2} \). This is a first-order linear differential equation in standard form. To solve, use the integrating factor \( \mu(t) = e^{\int \frac{1}{t} dt} = t \). Multiply throughout by \( t \) resulting in \( t \frac{dv}{dt} + v = \frac{1}{t} \).

Step 6: Integrate the Transformed Equation

The equation can now be written as \( \frac{d}{dt}(tv) = \frac{1}{t} \). Integrate both sides with respect to \( t \), obtaining \( tv = \ln|t| + C \), where \( C \) is the integration constant.

Step 7: Back-Substitute to Find y(t)

Solve for \( v \) to get \( v = \frac{\ln|t| + C}{t} \). Recall \( y = \frac{1}{v} \), therefore, \( y(t) = \frac{t}{\ln|t| + C} \). This is the solution to the original differential equation.

Key Concepts

first-order differential equations

A first-order differential equation involves derivatives of a function with respect to one variable and has the first derivative as its highest-order derivative. In this context, we study how a function changes as its variables change.
For example, our exercise includes the equation \( t^2 \frac{dy}{dt} + y^2 = ty \). Here, \( \frac{dy}{dt} \) is the first derivative of \( y \) with respect to \( t \), showing it's a first-order differential equation.

• These equations are crucial because they model real-world phenomena such as physics, biology, and economics.
• The challenge often lies in finding a solution that satisfies the equation under specific conditions.

By finding such solutions, we can better understand and predict behaviors of systems described by these equations.

non-linear differential equations

Non-linear differential equations are equations in which the dependent variable or its derivatives appear to a power or are multiplied by each other. Such equations are more complex compared to linear ones.
In our problem, the term \( y^2 \) makes the equation non-linear, since the variable \( y \) is squared.

• Non-linear differential equations are prevalent in natural sciences as they model phenomena such as weather patterns, populations dynamics, and electrical circuits.
• Solutions to non-linear equations can exhibit complex behavior that is not seen in linear equations, like chaos or multiple solutions.

Understanding non-linear differential equations is essential for advancing in various scientific fields as they reflect more intricate systems than their linear counterparts.

variable coefficient differential equations

Variable coefficient differential equations have coefficients that are functions of the independent variable, rather than constants. This feature adds another layer of complexity to solving them.
In the given exercise, coefficients like \( t^2 \) and \( \frac{1}{t} \) are functions of the independent variable \( t \).

• These equations require specialized techniques, such as substitutions or transformations, to solve.
• Recognizing the form of the equation aids in choosing the appropriate method, like using integrating factors or transformation via substitution as in our example.

Solving variable coefficient equations is significant in engineering and physics where systems' dynamics depend significantly on time or space.