Understanding and Solving the Ordinary Differential Equation ( frac{dy}{dt} y - 2y - 3y ): A Comprehensive Guide
" "This article discusses how to solve the ordinary differential equation ( frac{dy}{dt} y - 2y - 3y ). We will explore the process of identifying that the equation is separable, decomposing the equation into partial fractions, and integrating to find the general solution.
" "Introduction to Separable Equations
" "A separable ordinary differential equation (ODE) is one that can be written in the form ( frac{dy}{dt} g(t)h(y) ). In this equation, the variables ( t ) and ( y ) can be separated such that the equation can be expressed as ( frac{dy}{h(y)} g(t)dt ). This property allows us to integrate both sides of the equation to find the general solution.
" "Revisiting the Given Equation
" "Consider the given ordinary differential equation:
" "[ frac{dy}{dt} y - 2y - 3y ]
" "This equation simplifies to:
" "[ frac{dy}{dt} -4y ]
" "Here, the equation is separable because the right-hand side is a function of ( y ) alone, and the left-hand side is a function of ( t ).
" "Decomposing and Integrating Using Partial Fractions
" "For the given equation, it is more complex as written. However, let's consider a similar structure for exposition, where the equation is of the form:
" "[ frac{dy}{dt} frac{1}{y - 2y - 3y} ]
" "This can be simplified to:
" "[ frac{dy}{dt} frac{1}{y(1-2-3)} frac{1}{y(-4)} -frac{1}{4y} ]
" "Which is indeed separable. We can then rewrite it as:
" "[ frac{dy}{y} -frac{1}{4} dt ]
" "Integrating both sides with respect to ( t ), we get:
" "[ int frac{dy}{y} -frac{1}{4} int dt ]
" "This yields:
" "[ ln|y| -frac{t}{4} C ]
" "where ( C ) is the constant of integration. Exponentiating both sides, we obtain:
" "[ |y| e^{-t/4 C} e^C e^{-t/4} ]
" "Letting ( e^C K ), where ( K ) is a positive constant, we have the general solution:
" "[ y(t) Ke^{-t/4} ]
" "Handling More Complex Forms
" "For more complex forms, such as ( frac{dy}{dt} frac{1}{y - 2y - 3y} frac{1}{y(1-2-3)} frac{1}{y(-4)} -frac{1}{4y} ), we need to decompose the expression using partial fractions.
" "Let's decompose ( frac{1}{y(1-2-3)} -frac{1}{4y} ) into partial fractions:
" "[ frac{1}{y(1-2-3)} frac{1}{y(-4y)} -frac{1}{4y} ]
" "which simplifies to:
" "[ frac{1}{y-2y-3y} frac{A}{y} frac{B}{y-2} frac{C}{y-3} ]
" "Here, we have:
" "[ frac{1}{y(1-2-3)} frac{A}{y} frac{B}{y-2} frac{C}{y-3} ]
" "To find ( A ), ( B ), and ( C ), we multiply through by ( y(1-2-3) ) to get:
" "[ 1 A(y-2)(y-3) B(y)(y-3) C(y)(y-2) ]
" "Evaluating at ( y 0 ), ( y 2 ), and ( y 3 ), we get:
" "[ y 0: 1 A(-2)(-3) Rightarrow A frac{1}{6} ]
" "[ y 2: 1 B(2)(-1) Rightarrow B -frac{1}{2} ]
" "[ y 3: 1 C(3)(1) Rightarrow C frac{1}{3} ]
" "So, the partial fraction decomposition is:
" "[ frac{1}{y-2y-3y} frac{1/6}{y} frac{-1/2}{y-2} frac{1/3}{y-3} ]
" "Integrating both sides:
" "[ t frac{1}{6} ln(y) - frac{1}{2} ln(y-2) frac{1}{3} ln(y-3) C ]
" "This simplifies to:
" "[ t frac{1}{6} ln(y) - frac{1}{2} ln(y-2) frac{1}{3} ln(y-3) C ]
" "Exponentiating both sides:
" "[ e^{6t} e^{frac{1}{6} ln(y) - frac{1}{2} ln(y-2) frac{1}{3} ln(y-3) C} ]
" "[ e^{6t} Ke^{frac{1}{6} ln(y) - frac{1}{2} ln(y-2) frac{1}{3} ln(y-3)} ]
" "[ e^{6t} Kleft(frac{y}{(y-2)^2(y-3)}right) ]
" "Where ( K e^C ).
" "Conclusion and Applications
" "Solving such differential equations is crucial in various fields of science and engineering, including physics, electrical engineering, and control systems. The methods discussed here can be applied to more complex equations by decomposing them into partial fractions and integrating step-by-step.
" "For further exploration, consider studying more complex differential equations and their applications in real-world scenarios.