initial value problem

[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 dont 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.

[AKG]

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 dont 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.
ydx\ -\ (y\tan\left(\frac{x}{y}\right)\ +\ x)dy\ =\ 0
ydx\ =\ (y\tan\left(\frac{x}{y}\right)\ +\ x)dy
\frac{dx}{dy}\ =\ \tan\left(\frac{x}{y}\right)\ +\ \frac{x}{y}
let\ a\ =\ \frac{x}{y}
\frac{d(ya)}{dy}\ =\ \tan(a)\ +\ a
a\ +\ y\frac{da}{dy}\ =\ \tan(a)\ +\ a
y\frac{da}{dy}\ =\ \tan(a)

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:

[phy]

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?

[AKG]

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 y\ \neq\ 0 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.

[himanshu121]

Mostly when it is a homogeneous in x and y

[phy]

ok thanks guys :)