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First-order ODEs are equations involving the first derivative of a function. They come in various forms, including linear, nonlinear, and autonomous. Understanding these types helps in choosing the right solving method, like separation of variables.

These equations model real-world phenomena like population growth and radioactive decay. By applying initial conditions, we can find particular solutions that describe specific scenarios. This connects math to practical applications in science and engineering.

First-Order Ordinary Differential Equations (ODEs)

Classification of first-order ODEs

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  • First-order ODEs involve the first derivative of the dependent variable with respect to the independent variable
    • General form: dydx=f(x,y)\frac{dy}{dx} = f(x, y) where yy is the dependent variable and xx is the independent variable
  • Linear first-order ODEs have the form: dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)
    • P(x)P(x) and Q(x)Q(x) are functions of the independent variable xx only, not involving the dependent variable yy
    • Example: dydx+3x2y=sin(x)\frac{dy}{dx} + 3x^2y = \sin(x) is a
  • Nonlinear first-order ODEs cannot be written in the linear form due to the presence of nonlinear terms involving yy or its derivatives
    • Example: dydx=y2+x\frac{dy}{dx} = y^2 + x is a because of the y2y^2 term
  • Autonomous first-order ODEs have the form: dydx=f(y)\frac{dy}{dx} = f(y) where the right-hand side depends only on the dependent variable yy
    • Example: dydx=y(1y)\frac{dy}{dx} = y(1-y) is an (logistic equation)

Separation of variables technique

  • Separable first-order ODEs have the form: dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y) where g(x)g(x) is a function of the independent variable xx and h(y)h(y) is a function of the dependent variable yy
    • Example: dydx=x2y3\frac{dy}{dx} = x^2y^3 is a with g(x)=x2g(x) = x^2 and h(y)=y3h(y) = y^3
  • Separation of variables technique involves the following steps:
    1. Rearrange the equation to separate variables: dyh(y)=g(x)dx\frac{dy}{h(y)} = g(x)dx
    2. Integrate both sides: dyh(y)=g(x)dx\int \frac{dy}{h(y)} = \int g(x)dx
    3. Solve for yy as a function of xx by evaluating the integrals and applying algebraic manipulations
  • Example: dydx=x2y3\frac{dy}{dx} = x^2y^3
    • Rearrange: dyy3=x2dx\frac{dy}{y^3} = x^2dx
    • Integrate: dyy3=x2dx\int \frac{dy}{y^3} = \int x^2dx
    • Solve: 12y2=x33+C-\frac{1}{2y^2} = \frac{x^3}{3} + C, where CC is an arbitrary constant

Particular solutions with initial conditions

  • of a contains an arbitrary constant CC that represents a family of solutions
  • specifies the value of the dependent variable at a specific point, such as y(x0)=y0y(x_0) = y_0, where x0x_0 and y0y_0 are known values
    • Example: y(0)=1y(0) = 1 means the value of yy is 1 when x=0x = 0
  • Substitute the initial condition into the general solution to determine the value of CC
  • Replace CC in the general solution with the determined value to obtain the that satisfies the initial condition
  • Example: Given the general solution y=Ce2xy = Ce^{2x} and the initial condition y(0)=3y(0) = 3, find the particular solution
    • Substitute: 3=Ce2(0)=C3 = Ce^{2(0)} = C
    • Replace: y=3e2xy = 3e^{2x} is the particular solution

Applications of First-Order ODEs

Exponential growth and decay modeling

  • and decay problems can be modeled using first-order ODEs of the form: dydt=ky\frac{dy}{dt} = ky, where yy is the quantity undergoing growth or decay and kk is the growth or
    • k>0k > 0 for exponential growth (population growth, compound interest)
    • k<0k < 0 for (radioactive decay, cooling of objects)
  • Solution to the exponential growth/decay ODE: y(t)=y0ekty(t) = y_0e^{kt}, where y0y_0 is the initial value at t=0t = 0
  • (t1/2t_{1/2}) is the time required for the quantity to reduce to half its initial value in exponential decay
    • For exponential decay: t1/2=ln(2)kt_{1/2} = \frac{\ln(2)}{|k|}, where k|k| is the absolute value of the decay constant
    • Example: If a radioactive substance decays with a half-life of 10 days, its decay constant is k=ln(2)10k = -\frac{\ln(2)}{10}
  • (tdt_d) is the time required for the quantity to double its initial value in exponential growth
    • For exponential growth: td=ln(2)kt_d = \frac{\ln(2)}{k}, where kk is the growth constant
    • Example: If a population grows with a doubling time of 5 years, its growth constant is k=ln(2)5k = \frac{\ln(2)}{5}
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© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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