Study Notes on Solving a Linear Differential Equation

Differential Equation Analysis and Solutions

Introduction to the Differential Equation

  • We are given the differential equation in the form:
    \frac{dy}{dx} + 2xy = x^2
    where x > 0 .

Part (a): Determine the Integrating Factor

  • To solve this first-order linear differential equation, we first determine the integrating factor, \mu(x) .

  • The standard form of a linear differential equation is
    \frac{dy}{dx} + P(x)y = Q(x)
    where

    • P(x) = 2x
    • Q(x) = x^2

  • The integrating factor is calculated using the formula:
    \mu(x) = e^{\int P(x) \, dx}

Step to Calculate the Integrating Factor:

  • Calculate the integral of P(x):
    \int P(x) \, dx = \int 2x \, dx = x^2 + C
  • Now plugging this into the integrating factor formula:
    \mu(x) = e^{x^2}

Part (b): Solve the Differential Equation

  • We multiply through the differential equation by the integrating factor:
    e^{x^2} \frac{dy}{dx} + e^{x^2}(2xy) = e^{x^2}(x^2)

  • The left side of this equation can be rewritten as the derivative of a product:
    \frac{d}{dx}(e^{x^2}y) = e^{x^2}x^2

  • We will now integrate both sides:
    \int \frac{d}{dx}(e^{x^2}y) \, dx = \int e^{x^2}x^2 \, dx

Solving the Left Side:

  • The result on the left side simplifies to:
    e^{x^2}y = \int e^{x^2}x^2 \, dx + C
    where C is the integration constant.

Solving the Right Side:

  • The integral \int e^{x^2}x^2 \, dx can be tricky, typically it is related to the error function, but we recognize it as an integral that does not have a simple form.

Conclusion for Part (b):

  • We leave the solution in terms of the integral, giving us:
    y = e^{-x^2}(\int e^{x^2}x^2 \, dx + C)

Part (c): Determine the Particular Solution with Initial Condition

  • We are given the initial condition:
    y(1) = 0 .
  • We will now substitute x = 1 into our solution from Part (b):
    0 = e^{-1^2}(\int e^{1^2}1^2 \, dx + C)

Calculate the Integral:

  • Evaluate \int e^{1} \, dx which simplifies the left-hand side.
  • Remember that we must solve for C to fulfill the initial condition, ultimately leading us to find:
    C = -\int e \ dx
  • Substituting C back into our general solution allows us to express the final particular solution for the differential equation with the initial condition. This yields the complete solution tailored to the initial condition.