Math Problem Statement

solve each differential equation by variation of parameters. 1. y0 + y = sec x

Solution

To solve the differential equation y+y=secxy' + y = \sec x using the method of variation of parameters, follow these steps:

Step 1: Solve the homogeneous equation

The homogeneous version of the equation is:

y+y=0y' + y = 0

This is a first-order linear differential equation. We can solve it as follows:

dydx+y=0\frac{dy}{dx} + y = 0

This can be rewritten as:

dydx=y\frac{dy}{dx} = -y

We can solve this by separation of variables:

dyy=dx\frac{dy}{y} = -dx

Integrating both sides:

1ydy=1dx\int \frac{1}{y} \, dy = \int -1 \, dx

lny=x+C\ln |y| = -x + C

Exponentiating both sides:

y=C1exy = C_1 e^{-x}

So, the general solution to the homogeneous equation is:

yh=C1exy_h = C_1 e^{-x}

Step 2: Set up the particular solution using variation of parameters

Now, to solve the non-homogeneous equation y+y=secxy' + y = \sec x, we use the method of variation of parameters. Assume the particular solution has the form:

yp=v(x)exy_p = v(x) e^{-x}

where v(x)v(x) is a function to be determined. We substitute this into the non-homogeneous equation. First, we need to compute the derivative of ypy_p:

yp=v(x)exv(x)exy_p' = v'(x) e^{-x} - v(x) e^{-x}

Now, substitute ypy_p and ypy_p' into the original differential equation y+y=secxy' + y = \sec x:

(v(x)exv(x)ex)+v(x)ex=secx(v'(x) e^{-x} - v(x) e^{-x}) + v(x) e^{-x} = \sec x

Simplifying:

v(x)ex=secxv'(x) e^{-x} = \sec x

Now, solve for v(x)v'(x):

v(x)=secxexv'(x) = \sec x e^{x}

Step 3: Integrate to find v(x)v(x)

To find v(x)v(x), we integrate both sides:

v(x)=secxexdxv(x) = \int \sec x e^{x} \, dx

This integral is challenging, but it can be solved by integration by parts or using a table of integrals. The solution is:

v(x)=exsecxdxv(x) = e^x \int \sec x \, dx

The integral of secx\sec x is lnsecx+tanx\ln | \sec x + \tan x |, so:

v(x)=exlnsecx+tanx+C2v(x) = e^x \ln | \sec x + \tan x | + C_2

Step 4: Write the general solution

Now that we have v(x)v(x), we can write the particular solution ypy_p:

yp=v(x)ex=(exlnsecx+tanx+C2)exy_p = v(x) e^{-x} = \left( e^x \ln | \sec x + \tan x | + C_2 \right) e^{-x}

Simplifying:

yp=lnsecx+tanx+C2exy_p = \ln | \sec x + \tan x | + C_2 e^{-x}

Thus, the general solution to the non-homogeneous differential equation is:

y=yh+yp=C1ex+lnsecx+tanx+C2exy = y_h + y_p = C_1 e^{-x} + \ln | \sec x + \tan x | + C_2 e^{-x}

Since C1C_1 and C2C_2 are arbitrary constants, we can combine them into a single constant CC:

y=Cex+lnsecx+tanxy = C e^{-x} + \ln | \sec x + \tan x |

Final Solution:

y=Cex+lnsecx+tanxy = C e^{-x} + \ln | \sec x + \tan x |


Would you like further details on any step, or have any additional questions? Here are some related questions you might find useful:

  1. How do you handle integrals involving secx\sec x in variation of parameters?
  2. Can you apply the method of variation of parameters to second-order linear equations?
  3. How does the method of undetermined coefficients differ from variation of parameters?
  4. What happens if the non-homogeneous term is not of a standard form like secx\sec x?
  5. How do you verify the correctness of the solution to a differential equation?

Tip: When solving integrals like secxexdx\int \sec x e^x dx, use integration by parts and remember that secxdx=lnsecx+tanx\int \sec x dx = \ln | \sec x + \tan x |.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
First-Order Linear Differential Equations
Variation of Parameters
Homogeneous Equations
Particular Solutions

Formulas

y' + y = sec x
y' + y = 0
y = C_1 e^{-x}
v(x) = e^x ln |sec x + tan x| + C_2
y_p = v(x) e^{-x}
y = C e^{-x} + ln |sec x + tan x|

Theorems

Method of Variation of Parameters
First-Order Linear Differential Equation Solution
Integration by Parts

Suitable Grade Level

Undergraduate (Mathematics/Engineering)