How to Separate Variables in an 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 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.

 

[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 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.

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

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.

 

[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?

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.

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 :)