Math Problem Statement

Solve the initial value problem: y' = 2cos(x/3) + 4cos(3x), with y(π/2) = 5. Do not include 'y =' in your answer.

Solution

To solve the initial value problem for the differential equation:

y=2cos(x3)+4cos(3x),y' = 2 \cos\left(\frac{x}{3}\right) + 4 \cos(3x), with the initial condition y(π2)=5y\left(\frac{\pi}{2}\right) = 5.

Step 1: Integrate yy' to find y(x)y(x)

Since y=dydxy' = \frac{dy}{dx}, we can integrate both sides with respect to xx to find y(x)y(x):

y(x)=(2cos(x3)+4cos(3x))dx.y(x) = \int \left(2 \cos\left(\frac{x}{3}\right) + 4 \cos(3x)\right) \, dx.

Step 2: Integrate each term separately

  1. First term: 2cos(x3)dx\int 2 \cos\left(\frac{x}{3}\right) \, dx

    • Let u=x3u = \frac{x}{3}, then du=13dxdu = \frac{1}{3} dx or dx=3dudx = 3 \, du.
    • Substitute into the integral: 2cos(x3)dx=2cos(u)3du=6cos(u)du=6sin(u).\int 2 \cos\left(\frac{x}{3}\right) \, dx = 2 \int \cos(u) \cdot 3 \, du = 6 \int \cos(u) \, du = 6 \sin(u).
    • Substitute u=x3u = \frac{x}{3} back: =6sin(x3).= 6 \sin\left(\frac{x}{3}\right).
  2. Second term: 4cos(3x)dx\int 4 \cos(3x) \, dx

    • Let v=3xv = 3x, then dv=3dxdv = 3 \, dx or dx=dv3dx = \frac{dv}{3}.
    • Substitute into the integral: 4cos(3x)dx=4cos(v)dv3=43cos(v)dv=43sin(v).\int 4 \cos(3x) \, dx = 4 \int \cos(v) \cdot \frac{dv}{3} = \frac{4}{3} \int \cos(v) \, dv = \frac{4}{3} \sin(v).
    • Substitute v=3xv = 3x back: =43sin(3x).= \frac{4}{3} \sin(3x).

Step 3: Combine results

So, y(x)=6sin(x3)+43sin(3x)+C,y(x) = 6 \sin\left(\frac{x}{3}\right) + \frac{4}{3} \sin(3x) + C, where CC is the constant of integration.

Step 4: Apply the initial condition y(π2)=5y\left(\frac{\pi}{2}\right) = 5

Substitute x=π2x = \frac{\pi}{2} and y=5y = 5: 5=6sin(π6)+43sin(3π2)+C.5 = 6 \sin\left(\frac{\pi}{6}\right) + \frac{4}{3} \sin\left(\frac{3\pi}{2}\right) + C.

Calculate each term:

  • sin(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, so 6sin(π6)=612=36 \sin\left(\frac{\pi}{6}\right) = 6 \cdot \frac{1}{2} = 3.
  • sin(3π2)=1\sin\left(\frac{3\pi}{2}\right) = -1, so 43sin(3π2)=43(1)=43\frac{4}{3} \sin\left(\frac{3\pi}{2}\right) = \frac{4}{3} \cdot (-1) = -\frac{4}{3}.

Thus, 5=343+C.5 = 3 - \frac{4}{3} + C.

Simplify: 5=9343+C=53+C.5 = \frac{9}{3} - \frac{4}{3} + C = \frac{5}{3} + C.

So, C=553=15353=103.C = 5 - \frac{5}{3} = \frac{15}{3} - \frac{5}{3} = \frac{10}{3}.

Final Solution

Substitute C=103C = \frac{10}{3} into the solution: y(x)=6sin(x3)+43sin(3x)+103.y(x) = 6 \sin\left(\frac{x}{3}\right) + \frac{4}{3} \sin(3x) + \frac{10}{3}.


Let me know if you need further explanation on any steps!

Related Questions

  1. How would the solution change if the initial condition were different?
  2. Can this problem be solved using a numerical method? If so, which one?
  3. How do the solutions behave as xx \to \infty in this equation?
  4. What if the cosine terms were replaced by sine terms in the original differential equation?
  5. How can we verify this solution by differentiating?

Tip

Remember that integrating terms with trigonometric functions often requires a substitution to simplify the calculation.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Integration
Initial Value Problems
Trigonometric Functions

Formulas

y(x) = ∫(2cos(x/3) + 4cos(3x)) dx
Initial Condition: y(π/2) = 5

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

College Calculus