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Calculus - Differential Equations (Separation of Variables)

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Differential Equations Definition: A differential equation is an equation that relates a function to its derivatives. In Grade 12 Calculus, we focus on first-order equations of the form dydx=f(x,y)\frac{dy}{dx} = f(x, y). Visually, this equation defines the slope of a tangent line at any point (x,y)(x, y) on a coordinate plane.

Separable Equations: A first-order differential equation is 'separable' if it can be expressed in the form dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y). This means the variables xx and yy can be isolated on opposite sides of the equals sign. Graphically, the slope at any point is the product of a horizontal component and a vertical component.

The Method of Separation: To solve, rewrite the equation as 1h(y)dy=g(x)dx\frac{1}{h(y)} dy = g(x) dx. We then apply the integral operator to both sides: 1h(y)dy=g(x)dx\int \frac{1}{h(y)} dy = \int g(x) dx. This transforms a derivative problem into an integration problem.

The Constant of Integration (CC): When integrating both sides, we technically get a constant on each side (C1C_1 and C2C_2), but we combine them into a single arbitrary constant CC on the side of the independent variable xx. This constant is vital as it represents a family of possible solution curves rather than a single line.

General vs. Particular Solutions: The 'General Solution' contains the constant CC and represents an infinite family of curves. A 'Particular Solution' is found when an initial condition (x0,y0)(x_0, y_0) is provided, allowing us to solve for a specific value of CC. Visually, the general solution is like a set of parallel-ish paths, while the particular solution is the specific path passing through a given dot.

Slope Fields (Direction Fields): A slope field is a visual representation of a differential equation. It consists of short line segments at various points (x,y)(x, y) where the gradient of each segment is dydx\frac{dy}{dx}. If you follow the 'flow' of these segments, you can sketch the shape of the solution curves even without solving the equation analytically.

Exponential Growth and Decay: A common application is the differential equation dydx=ky\frac{dy}{dx} = ky. When separated and solved, this yields y=Aekxy = Ae^{kx}. Visually, if k>0k > 0, the graph shows an upward-curving growth; if k<0k < 0, it shows a downward curve that asymptotically approaches the xx-axis.

📐Formulae

dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y) (Standard separable form)

1h(y)dy=g(x)dx\int \frac{1}{h(y)} dy = \int g(x) dx (Integral form)

y=f(x)+Cy = f(x) + C (General solution representation)

dydt=ky    y=y0ekt\frac{dy}{dt} = ky \implies y = y_0 e^{kt} (Exponential growth/decay model)

1xdx=lnx+C\int \frac{1}{x} dx = \ln|x| + C (Commonly used integral)

eudu=eu+C\int e^{u} du = e^{u} + C (Commonly used integral)

💡Examples

Problem 1:

Solve the differential equation dydx=2xy\frac{dy}{dx} = \frac{2x}{y} given the initial condition y(0)=3y(0) = 3.

Solution:

  1. Separate the variables: ydy=2xdxy \, dy = 2x \, dx
  2. Integrate both sides: ydy=2xdx\int y \, dy = \int 2x \, dx
  3. Perform the integration: 12y2=x2+C\frac{1}{2}y^2 = x^2 + C
  4. Multiply by 2 to simplify: y2=2x2+2Cy^2 = 2x^2 + 2C. Let 2C=K2C = K: y2=2x2+Ky^2 = 2x^2 + K
  5. Apply the initial condition (0,3)(0, 3): (3)2=2(0)2+K    9=K(3)^2 = 2(0)^2 + K \implies 9 = K
  6. Write the particular solution: y2=2x2+9y^2 = 2x^2 + 9 or y=2x2+9y = \sqrt{2x^2 + 9} (taking the positive root as y(0)=3y(0)=3).

Explanation:

We first isolate yy on the left and xx on the right. After integrating both sides, we include the constant CC. By substituting the point (0,3)(0, 3), we determine the specific curve that passes through that point.

Problem 2:

Find the general solution for dydx=3y\frac{dy}{dx} = 3y.

Solution:

  1. Separate the variables: 1ydy=3dx\frac{1}{y} dy = 3 \, dx
  2. Integrate both sides: 1ydy=3dx\int \frac{1}{y} dy = \int 3 \, dx
  3. Calculate integrals: lny=3x+C\ln|y| = 3x + C
  4. Exponentiate both sides: elny=e3x+Ce^{\ln|y|} = e^{3x + C}
  5. Simplify: y=e3xeC|y| = e^{3x} \cdot e^{C}
  6. Let A=±eCA = \pm e^{C}: y=Ae3xy = Ae^{3x}

Explanation:

This is a classic growth equation. The constant CC moves from the exponent to become a coefficient AA through the laws of indices, representing the initial value of yy at x=0x=0.