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The given differential equation is:

\[
2x y' - y = x^3 - x
\]

with the initial condition:

\[
y(4) = 5.
\]

### Steps to solve:

1. **Rearrange the equation:**
   Divide both sides by \( 2x \) to isolate \( y' \):

   \[
   y' - \frac{y}{2x} = \frac{x^3 - x}{2x}.
   \]

   Simplifying the right-hand side:

   \[
   y' - \frac{y}{2x} = \frac{x^2}{2} - \frac{1}{2}.
   \]

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

   \[
   y' + P(x) y = Q(x),
   \]

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

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

   \[
   \mu(x) = e^{\int P(x) dx} = e^{\int -\frac{1}{2x} dx} = e^{-\frac{1}{2} \ln |x|} = |x|^{-\frac{1}{2}} = \frac{1}{\sqrt{x}}.
   \]

4. **Multiply through by the integrating factor:**

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

   This simplifies to:

   \[
   \frac{d}{dx} \left( \frac{y}{\sqrt{x}} \right) = \frac{x^{3/2}}{2} - \frac{1}{2\sqrt{x}}.
   \]

5. **Integrate both sides:**

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

   Compute the integrals:

   \[
   \int \frac{x^{3/2}}{2} dx = \frac{x^{5/2}}{5}, \quad \int \frac{1}{2\sqrt{x}} dx = \sqrt{x}.
   \]

   So,

   \[
   \frac{y}{\sqrt{x}} = \frac{x^{5/2}}{5} - \sqrt{x} + C,
   \]

   where \( C \) is the constant of integration.

6. **Solve for \( y \):**

   Multiply through by \( \sqrt{x} \):

   \[
   y = \frac{x^3}{5} - x + C \sqrt{x}.
   \]

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

   Substitute \( x = 4 \) and \( y = 5 \):

   \[
   5 = \frac{4^3}{5} - 4 + C \sqrt{4}.
   \]

   Simplifying:

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

   \[
   5 = 12.8 - 4 + 2C,
   \]

   \[
   5 = 8.8 + 2C,
   \]

   \[
   2C = 5 - 8.8 = -3.8,
   \]

   \[
   C = -1.9.
   \]

8. **Final solution:**

   The particular solution is:

   \[
   y = \frac{x^3}{5} - x - 1.9 \sqrt{x}.
   \]

---

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.

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

This problem involves solving the differential equation 2x y' - y = x^3 - x using integrating factors, with an initial condition y(4) = 5. Learn the step-by-step process to find the particular solution, including key formulas and methods for solving linear differential equations.

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