Math Problem Statement

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Solve the initial value problem cos(x)y' + sin(x)y = 2cos^3(x)sin(x) - 1 with the condition y(π/4) = 3√2.

Solution

We are given the initial value problem:

cos(x)y+sin(x)y=2cos3(x)sin(x)1\cos(x)y' + \sin(x)y = 2\cos^3(x)\sin(x) - 1

with the initial condition y(π4)=32y\left(\frac{\pi}{4}\right) = 3\sqrt{2}.

Step 1: Rewrite the equation

We can rearrange the equation to isolate yy': y+sin(x)cos(x)y=2cos3(x)sin(x)1cos(x)y' + \frac{\sin(x)}{\cos(x)}y = \frac{2\cos^3(x)\sin(x) - 1}{\cos(x)} This becomes: y+tan(x)y=2cos2(x)sin(x)sec(x)y' + \tan(x)y = 2\cos^2(x)\sin(x) - \sec(x)

Step 2: Solve the homogeneous part

First, we solve the homogeneous part of the equation: y+tan(x)y=0y' + \tan(x)y = 0 This is a separable differential equation: dyy=tan(x)dx\frac{dy}{y} = -\tan(x) dx Integrating both sides: lny=lnsec(x)+C\ln|y| = \ln|\sec(x)| + C Thus, the solution to the homogeneous part is: yh=Csec(x)y_h = C\sec(x)

Step 3: Solve the non-homogeneous part (particular solution)

Now, we find the particular solution using variation of parameters. Let yp=v(x)sec(x)y_p = v(x)\sec(x), where v(x)v(x) is a function to be determined.

Taking the derivative: yp=v(x)sec(x)+v(x)sec(x)tan(x)y_p' = v'(x)\sec(x) + v(x)\sec(x)\tan(x) Substitute into the original equation: (v(x)sec(x)+v(x)sec(x)tan(x))+tan(x)v(x)sec(x)=2cos2(x)sin(x)sec(x)(v'(x)\sec(x) + v(x)\sec(x)\tan(x)) + \tan(x)v(x)\sec(x) = 2\cos^2(x)\sin(x) - \sec(x) Simplifying: v(x)sec(x)=2cos2(x)sin(x)sec(x)v'(x)\sec(x) = 2\cos^2(x)\sin(x) - \sec(x) Thus: v(x)=2cos3(x)sin(x)1v'(x) = 2\cos^3(x)\sin(x) - 1

Step 4: Integrate to find v(x)v(x)

To find v(x)v(x), we integrate both sides: v(x)=(2cos3(x)sin(x)1)dxv(x) = \int (2\cos^3(x)\sin(x) - 1) dx We can break this integral into two parts: v(x)=2cos3(x)sin(x)dx1dxv(x) = \int 2\cos^3(x)\sin(x) dx - \int 1 dx

The first integral can be solved using substitution. Let u=cos(x)u = \cos(x), then du=sin(x)dxdu = -\sin(x) dx: 2cos3(x)sin(x)dx=2u3du=u42+C=cos4(x)2+C\int 2\cos^3(x)\sin(x) dx = -\int 2u^3 du = -\frac{u^4}{2} + C = -\frac{\cos^4(x)}{2} + C Thus: v(x)=cos4(x)2x+Cv(x) = -\frac{\cos^4(x)}{2} - x + C

Step 5: General solution

Now, substitute v(x)v(x) back into yp=v(x)sec(x)y_p = v(x)\sec(x): yp=(cos4(x)2x+C)sec(x)y_p = \left(-\frac{\cos^4(x)}{2} - x + C\right)\sec(x) Simplifying the general solution: y=Csec(x)+(cos4(x)2x)sec(x)y = C\sec(x) + \left(-\frac{\cos^4(x)}{2} - x\right)\sec(x)

Step 6: Apply the initial condition

We are given the initial condition y(π4)=32y\left(\frac{\pi}{4}\right) = 3\sqrt{2}. Substitute x=π4x = \frac{\pi}{4} and solve for CC.

At x=π4x = \frac{\pi}{4}: y(π4)=C2+((12)42π4)2y\left(\frac{\pi}{4}\right) = C\sqrt{2} + \left(-\frac{\left(\frac{1}{\sqrt{2}}\right)^4}{2} - \frac{\pi}{4}\right)\sqrt{2} Now solve for CC.

Would you like further details on how to complete the calculation for CC or expand on any particular part of the solution?


Relative Questions:

  1. How is the variation of parameters method generally applied?
  2. What is the significance of the homogeneous solution in differential equations?
  3. How do we handle integrals involving powers of trigonometric functions like cos(x)\cos(x)?
  4. Can initial conditions be applied to any type of solution, homogeneous or particular?
  5. Why do we need both a particular and homogeneous solution for non-homogeneous ODEs?

Tip: For non-homogeneous linear equations, always start by solving the homogeneous part before finding the particular solution.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Trigonometric Identities
Initial Value Problems
Separable Equations
Variation of Parameters

Formulas

y' + P(x)y = Q(x)
Separable Equation: dy/y = f(x)dx
Integration by Substitution

Theorems

Fundamental Theorem of Calculus
Existence and Uniqueness Theorem for ODEs

Suitable Grade Level

University Level (Differential Equations)