# Thread: Does it matter when you use the initial conditions?

1. ## Does it matter when you use the initial conditions?

So if we have, for instance, dy/dx=e^(2x)*e^(3y), we separate and integrate and get

e^(-3y)/-3=e^(2x)/2+C

Suppose we have IC y(0)=1.

Could I just plug in to this and get C, or should I first solve for y explicitly before I get C, or does it not matter?

2. Solving for "y" would get you an equation in y that is equivalent to what you are solving for. It's like solving y=x^2 for x, and then plugging in (2, 4) - you will get the exact same relation.

TLRL. No it does not.

3. You must have y(x)= before you can apply the IC. Take the natural log of both sides to solve for y, then apply the condition.

4. Originally Posted by mmattson07
You must have y(x)= before you can apply the IC. Take the natural log of both sides to solve for y, then apply the condition.
That is not needed. You are not solving for Y, you are solving for C. If the goal was, as in implicit differentiation, to create an equation of A in B, then yes, you would solve for A. That's not what is being done here. y(0)=1, is a roundabout way of saying (well not roundabout it has a meaning) the point (0, 1) satisfies this equation. Direct substitution works.

Also, if you did solve for Y, you would get an EQUIVALENT statement. Every algebraic operation you do on the equation to get it in terms of Y (assuming it is a correct one) would be equivalent to the original statement you started with. Knowing that, why not just use the equation given?

5. Ok.....So here's what is confusing me....when can you just group up the constants...for instance....let me show you the full solution I have:

We have dy/e^(3y)=e^(2x)dx

Integrating gives

-e^(-3y)/3+C1=e^(2x)/2+C2

But we just make the constant on the right hand C and forget about the one on the left hand to get:

-e^(-3y)/3=(e^(2x)/2)+C

Now multiply both sides by -3 (we need to solve for y directly, as well as for C, so we may as well do that first):

e^(-3y)=-(3e^(2x)/2)-3C, but we can just keep -3C as C, right? So it would be:

e^(-3y)=(-3e^(2x)/2)+C

Now we take logs of both sides...someone told me you have to write it as:

-3y=ln(-3e^(2x)/2+C), but since ln(C) is still a constant, can't you just have the C outside the parentheses as well?? Then, just divide by 3, but the real thing that is giving me troubles is wondering how long you can just keep grouping the constants together for (as in applying ln to a constant just gives another constant, dividing a constant by a constant gives a constant, can you keep grouping them until the very end, or is there some point where this is not allowed?)

I hope this question is clear. Thanks.