Differential Equations Study Guide

Handout 7A: Verifying Solutions, Slope Fields, etc.

1. Differential Equation Overview

  • Consider the differential equation given by:
    dydx=2xy\frac{dy}{dx} = 2xy
  • Conditions that apply:
    • y > 0
    • \frac{dy}{dx} > 0 when x > 0
a. Particular Solution Tangent Line
  • Let y=g(x)y = g(x) be a particular solution.
  • To find the equation of the tangent line at the point (1, 3):
    • The slope at this point is obtained from the differential equation:
    • At x=1x = 1, substitute into the differential equation:
      dydx=2(1)(g(1))=2⋅3=6\frac{dy}{dx} = 2(1)(g(1)) = 2 \cdot 3 = 6
    • Therefore, the slope mm of the tangent line at (1, 3) is 6.
    • Equation of tangent line using the point-slope form:
    • Point-slope form: y−y1=m(x−x1)y - y_1 = m(x - x_1)
    • Thus, y−3=6(x−1)y - 3 = 6(x - 1)
    • Rearranging gives:
      y=6x−3y = 6x - 3
  • Use this equation to estimate the value of g(0.9)g(0.9):
    • Substitute x=0.9x = 0.9 into the tangent line equation:
    • g(0.9)≈6(0.9)−3=5.4−3=2.4g(0.9) \approx 6(0.9) - 3 = 5.4 - 3 = 2.4
b. Second Derivative Expression
  • Finding the second derivative:
    • Start from the differential equation:
    • dydx=2xy\frac{dy}{dx} = 2xy
    • Differentiate both sides:
    • d2ydx2=2y+2xdydx\frac{d^2y}{dx^2} = 2y + 2x\frac{dy}{dx}
    • Substitute dydx\frac{dy}{dx} back into the equation:
    • At x=1x = 1, where g(1)=3g(1) = 3, thus compute:
      d2ydx2=2(3)+2(1)(6)=6+12=18\frac{d^2y}{dx^2} = 2(3) + 2(1)(6) = 6 + 12 = 18
  • Comparison with estimate from part (a):
    • The estimate g(0.9)=2.4g(0.9) = 2.4 is less than the actual g(1)=3g(1) = 3 evaluated.
    • Conclusion: our previous estimate is less than the exact value because the differential equation confirms that dydx\frac{dy}{dx} is increasing.
c. Verification of Particular Solution
  • To verify that y=5ex2y = 5e^{x^2} satisfies the differential equation:
    • Compute the derivative:
    • dydx=5ex2⋅2x=10xex2\frac{dy}{dx} = 5e^{x^2} \cdot 2x = 10xe^{x^2}
    • Check:
    • Substitute yy into the original differential equation:
      • 2xy=2x(5ex2)=10xex22xy = 2x(5e^{x^2}) = 10xe^{x^2}
    • Since both sides match, verification is complete.
d. Estimate Comparison for Another Point
  • It is known that for all x≥1x \geq 1:
    • \frac{d^2y}{dx^2} > 0
  • This suggests that the function is concave up in this interval.
  • Reflection on estimate from part (a) when evaluated at g(1.2)g(1.2) indicates the potential for it being less than the actual value due to the concavity of the function.

2. Additional Example

a. Verification of Solution
  • Consider the solution y=3x−1y = 3x - 1 to the differential equation:
    dydx=1+y\frac{dy}{dx} = 1 + y
  • Calculate the derivative:
    • dydx=3\frac{dy}{dx} = 3
  • Check against the differential equation:
    • Right-hand side calculation:
    • 1+(3x−1)=3x1 + (3x - 1) = 3x
    • Hence, both sides are equal when evaluated.

3. Estimation Comparison for Initial Condition

a. Tangent Line Estimate
  • Considering the differential equation with initial condition y(0)=−1y(0) = -1:
    • dydx=y−x\frac{dy}{dx} = y - x
    • At point x=0x = 0, we have:
    • y′(0)=−1−0=−1y'(0) = -1 - 0 = -1
    • Equation of the tangent line at this point is given by:
    • Point-slope: y−(−1)=−1(x−0)y - (-1) = -1(x - 0)
      • Thus,
        y=−x−1y = -x - 1
  • Use this equation to estimate the value of y(0.5)y(0.5):
    • Substitute x=0.5x = 0.5:
    • y(0.5)=−0.5−1=−1.5y(0.5) = -0.5 - 1 = -1.5
b. Expression for Second Derivative
  • Obtain the expression for the second derivative:
  • From dydx=y−x\frac{dy}{dx} = y - x:
    • Differentiate:
    • d2ydx2=dydx−1=y−x−1\frac{d^2y}{dx^2} = \frac{dy}{dx} - 1 = y - x - 1
  • Use previous result for evaluation of the over/under estimate at y(0.5)y(0.5):
    • Compare the tangent estimate y(0.5)=−1.5y(0.5) = -1.5 with the actual computed value, determining if it surpasses or falls short of the expected reality based on the concavity as established in earlier sections.