# Initial value problem

1. May 19, 2004

### 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.

2. May 19, 2004

### AKG

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

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.

3. May 19, 2004

### 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?

4. May 19, 2004

### AKG

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.

5. May 19, 2004

### himanshu121

Mostly when it is a homogeneous in x and y

6. May 20, 2004

### phy

ok thanks guys :)