Solving Linear First Order D.E. y' - y*tan[x] = 2sin[x]

In summary, the equation y’ – y*tan[x] = 2sin[x] has two valid solutions: y = 2ln(sec[x]) / sec[x] - c/sec[x] and y = (c - cos[x]) * sec[x]. The latter is obtained by taking into account the natural logarithm and using the method of variation of parameters or guessing a solution.
  • #1
splitendz
32
0
Solve y’ – y*tan[x] = 2sin[x].

I keep arriving at the answer: y = 2ln(sec[x]) / sec[x] - c/sec[x]
for this question.

According to my textbook the correct answer is: y = (c - cos[x]) * sec[x]. Can anyone explain how to obtain this answer?

Thanks :)
 
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  • #2
Show us your work.Because of that natural logarithm,the two answers are not equivalent (equal for some value of the integration constant),therefore at least one of them is wrong.

Daniel.
 
  • #3
The text answer is correct. I suspect you overlooked the fact that integrating will give you a ln(y) as well as "ln(sec(x))".

Since this is a linear equation, you can do the homogeneous and non-homogeneous parts separately. The associated homogeneous equation is just y'- y tan x= 0 or
y'= y tan x so dy/y= tan x dx= (sin x dx)/cos x.
To find a solution to the non-homogeneous part you could use "variation of parameters": try a solutution of the form y(x)= u(x) sec(x) (I'm giving away part of the answer there!), differentiate and plug into the equation to find a simple equation for u(x).
Or just guess a simple solution!
 
  • #4
Ah right. Thanks a lot for your help HallsofIvy! I've got it now.
 

FAQ: Solving Linear First Order D.E. y' - y*tan[x] = 2sin[x]

1. What is a linear first order differential equation?

A linear first order differential equation is a type of mathematical equation that involves a function and its first derivative. It can be written in the form y' + p(x)y = q(x), where p(x) and q(x) are functions of x. The term "linear" refers to the fact that the equation is linear in terms of y and its derivative.

2. What is the general solution to a linear first order differential equation?

The general solution to a linear first order differential equation is y = Ce-∫p(x)dx + e-∫p(x)dx∫q(x)e∫p(x)dxdx, where C is a constant. This solution can be found by using the method of integrating factors.

3. How do you solve a linear first order differential equation?

To solve a linear first order differential equation, you can use the method of integrating factors. This involves multiplying both sides of the equation by a specific function, called the integrating factor, in order to simplify the equation and make it easier to solve. Once the equation is simplified, you can find the general solution by integrating both sides and solving for y.

4. What is an initial value problem for a linear first order differential equation?

An initial value problem for a linear first order differential equation involves finding a specific solution to the equation that satisfies a given initial condition. This condition usually takes the form of y(x0) = y0, where x0 is a specific value and y0 is the value of the function at that point.

5. What are some real-world applications of linear first order differential equations?

Linear first order differential equations can be used to model a variety of real-world phenomena, such as population growth, chemical reactions, and electrical circuits. They are also commonly used in engineering, physics, and economics to analyze and predict the behavior of systems.

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