Math Problem Statement

Find the particular solution of the differential equation that satisfies the initial condition: 2x y' - y = x^3 - x with y(4) = 5.

Solution

The given differential equation is:

2xyy=x3x2x y' - y = x^3 - x

with the initial condition:

y(4)=5.y(4) = 5.

Steps to solve:

  1. Rearrange the equation: Divide both sides by 2x2x to isolate yy':

    yy2x=x3x2x.y' - \frac{y}{2x} = \frac{x^3 - x}{2x}.

    Simplifying the right-hand side:

    yy2x=x2212.y' - \frac{y}{2x} = \frac{x^2}{2} - \frac{1}{2}.

  2. Recognize the form: This is a first-order linear differential equation of the form:

    y+P(x)y=Q(x),y' + P(x) y = Q(x),

    where P(x)=12xP(x) = -\frac{1}{2x} and Q(x)=x2212Q(x) = \frac{x^2}{2} - \frac{1}{2}.

  3. Find the integrating factor: The integrating factor μ(x)\mu(x) is given by:

    μ(x)=eP(x)dx=e12xdx=e12lnx=x12=1x.\mu(x) = e^{\int P(x) dx} = e^{\int -\frac{1}{2x} dx} = e^{-\frac{1}{2} \ln |x|} = |x|^{-\frac{1}{2}} = \frac{1}{\sqrt{x}}.

  4. Multiply through by the integrating factor:

    1xy12x3/2y=(x2212)1x.\frac{1}{\sqrt{x}} y' - \frac{1}{2x^{3/2}} y = \left( \frac{x^2}{2} - \frac{1}{2} \right) \frac{1}{\sqrt{x}}.

    This simplifies to:

    ddx(yx)=x3/2212x.\frac{d}{dx} \left( \frac{y}{\sqrt{x}} \right) = \frac{x^{3/2}}{2} - \frac{1}{2\sqrt{x}}.

  5. Integrate both sides:

    yx=(x3/2212x)dx.\frac{y}{\sqrt{x}} = \int \left( \frac{x^{3/2}}{2} - \frac{1}{2\sqrt{x}} \right) dx.

    Compute the integrals:

    x3/22dx=x5/25,12xdx=x.\int \frac{x^{3/2}}{2} dx = \frac{x^{5/2}}{5}, \quad \int \frac{1}{2\sqrt{x}} dx = \sqrt{x}.

    So,

    yx=x5/25x+C,\frac{y}{\sqrt{x}} = \frac{x^{5/2}}{5} - \sqrt{x} + C,

    where CC is the constant of integration.

  6. Solve for yy:

    Multiply through by x\sqrt{x}:

    y=x35x+Cx.y = \frac{x^3}{5} - x + C \sqrt{x}.

  7. Use the initial condition y(4)=5y(4) = 5:

    Substitute x=4x = 4 and y=5y = 5:

    5=4354+C4.5 = \frac{4^3}{5} - 4 + C \sqrt{4}.

    Simplifying:

    5=6454+2C,5 = \frac{64}{5} - 4 + 2C,

    5=12.84+2C,5 = 12.8 - 4 + 2C,

    5=8.8+2C,5 = 8.8 + 2C,

    2C=58.8=3.8,2C = 5 - 8.8 = -3.8,

    C=1.9.C = -1.9.

  8. Final solution:

    The particular solution is:

    y=x35x1.9x.y = \frac{x^3}{5} - x - 1.9 \sqrt{x}.


Do you need any further explanation or steps clarified?

Here are some related questions to expand the topic:

  1. What are the general steps for solving first-order linear differential equations?
  2. How is the integrating factor derived in solving linear differential equations?
  3. What are the methods for verifying solutions of differential equations?
  4. How does the initial condition influence the solution of a differential equation?
  5. Can you explain the difference between homogeneous and non-homogeneous differential equations?

Tip: Always verify a differential equation solution by substituting it back into the original equation.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
First-Order Linear Differential Equations
Integrating Factors

Formulas

y' - (y/2x) = (x^3 - x)/2x
Integrating Factor: e^(∫P(x) dx) = e^(-1/2 ln|x|) = 1/√x

Theorems

First-Order Linear Differential Equation Theorem

Suitable Grade Level

College Level / Advanced High School