How Can I Solve This Initial Value Problem?

In summary, the conversation is about solving the equation ydx - (ytan(x/y) +x)dy = 0 with the initial condition y(1) = pi/4. The person asking for help is struggling to rearrange the equation to get all the x's on one side and the y's on the other. Another person suggests using the substitution a=x/y, which simplifies the equation to yda/dy = tan(a). The original poster thanks everyone for their help.
  • #1
phy
hi everyone. i need help solving this question

ydx - (ytan(x/y) +x)dy = 0, where y(1) = pi/4

i know how to do the question but my problem is just i don't know how to get all the x's on one side and the y's on the other. any help would be appreciated. thanks a lot.
 
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  • #2
phy said:
hi everyone. i need help solving this question

ydx - (ytan(x/y) +x)dy = 0, where y(1) = pi/4

i know how to do the question but my problem is just i don't know how to get all the x's on one side and the y's on the other. any help would be appreciated. thanks a lot.
[tex]ydx\ -\ (y\tan\left(\frac{x}{y}\right)\ +\ x)dy\ =\ 0[/tex]
[tex]ydx\ =\ (y\tan\left(\frac{x}{y}\right)\ +\ x)dy[/tex]
[tex]\frac{dx}{dy}\ =\ \tan\left(\frac{x}{y}\right)\ +\ \frac{x}{y}[/tex]
[tex]let\ a\ =\ \frac{x}{y}[/tex]
[tex]\frac{d(ya)}{dy}\ =\ \tan(a)\ +\ a[/tex]
[tex]a\ +\ y\frac{da}{dy}\ =\ \tan(a)\ +\ a[/tex]
[tex]y\frac{da}{dy}\ =\ \tan(a)[/tex]

From here you should know what to do. I probably gave away the real trick to the problem, which was the proper rearrangment of the equation. :frown:
 
  • #3
ok thanks a lot. i didn't know i was supposed to make the substitution a=x/y. but how do you know when to use it?
 
  • #4
phy said:
ok thanks a lot. i didn't know i was supposed to make the substitution a=x/y. but how do you know when to use it?
:wink: It's not a general rule or anything, it was just a helpful substitution. I think 40% of the problem was the getting the 2nd and 3rd lines, 50% of the problem was lines 4-7, and 10% was the rest. As for knowing when to make such substitutions, I didn't "know" that I was supposed to make it either. Sometimes you just see it. However, if you practice enough, the chances that you'll "just see it" tend to increase, for some reason. :wink:

Oh, and I suppose I should mention that this only holds for [tex]y\ \neq\ 0[/tex] because you divide by "y" at some point during the first three steps. However, I think it's simple enough to see from the original equation that y can never be zero anyways.
 
  • #5
Mostly when it is a homogeneous in x and y
 
  • #6
ok thanks guys :)
 

1. What is an initial value problem?

An initial value problem is a type of mathematical problem that involves finding a function or set of functions that satisfies a given set of conditions. These conditions typically include an equation or system of equations, along with one or more initial values for the function(s) being solved.

2. What is the purpose of solving an initial value problem?

The purpose of solving an initial value problem is to determine the behavior of a system or function over a given time period. By finding the solution to the initial value problem, we can understand how the system or function changes over time and make predictions about its future behavior.

3. How is an initial value problem solved?

An initial value problem is typically solved using various mathematical techniques, such as separation of variables, substitution, or integration. The specific method used will depend on the type of equation and initial conditions given in the problem.

4. What are some real-world applications of initial value problems?

Initial value problems have a wide range of applications in fields such as physics, engineering, and economics. They can be used to model and predict the behavior of physical systems, such as the motion of a falling object or the spread of a disease. They can also be used in financial modeling to predict stock prices or interest rates.

5. How do initial value problems differ from boundary value problems?

While initial value problems involve finding a function that satisfies a given set of conditions at a single point, boundary value problems involve finding a function that satisfies a set of conditions at multiple points. Additionally, initial value problems typically have a solution that is unique, while boundary value problems may have multiple solutions or no solution at all.

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