how to solve this

**Sambit** · November 1st 2010, 07:01 AM

find the continuous solution of $\frac{dy}{dx}+y=g(x), 0<x<\infty, y(0)=2$ where $g(x)=3, 0<x<\frac{\pi}{2}$
$=\cos x, x\geq\frac{\pi}{2}$

for the first part, i got $-ln(3-y)=x$ . but how can i solve it for the second part?

**Ackbeet** · November 1st 2010, 07:13 AM

Solving for $y$ in the first part of your solution yields

$y=3-e^{-x}.$

We get $y(\pi/2)=3-e^{-\pi/2}.$ This is the initial condition for the next region's DE.

So now you're solving $y'(x)+y(x)=\cos(x),$ subject to $y(\pi/2)=3-e^{-\pi/2}.$ Ideas?

**Sambit** · November 1st 2010, 07:19 AM

unfortunately, no

. i have done solving algebraic equation subject to some constraint, but not differential equation. How to do this?

**Ackbeet** · November 1st 2010, 07:23 AM

Hmm. I suggest you take a look at Chris's DE tutorial. You can find the technique you need there.

**chisigma** · November 1st 2010, 07:36 AM

Originally Posted by Ackbeet

Solving for $y$ in the first part of your solution yields

$y=3-e^{-x}.$

We get $y(\pi/2)=3-e^{-\pi/2}.$ This is the initial condition for the next region's DE.

So now you're solving $y'(x)+y(x)=\cos(x),$ subject to $y(\pi/2)=3-e^{-\pi/2}.$ Ideas?

The solution You proposed is not properly adequate because the 'queue' of the y(x) due to the 'head' of g(x) doesn't vanish for x>3. The best approach to a problem of this type is the Laplace Transform...

Kind regards

$\chi$ $\sigma$

**Sambit** · November 1st 2010, 07:41 AM

solving by integrating factor, i got $ye^x=\frac{e^x(\sin x + \cos x)}{2} + C$
from the condition given in question, i am getting C=4.
now what about "continuous solution"?

**Sambit** · November 1st 2010, 07:42 AM

solving by integrating factor, i got $ye^x=\frac{e^x(\sin x + \cos x)}{2} + C$
from the condition given in question, i am getting C=4.
now what about "continuous solution"?

**Ackbeet** · November 1st 2010, 07:43 AM

Reply to chisigma:

I'm not following you. What do you mean by "queue" and "head"? I would agree that the LT is a very convenient method of solution for this problem (non-homogeneous IVP's, especially with respect to piece-wise defined functions). However, as Sambit appears not to be familiar with DE's in general, I think the problem is more basic than that.

I would argue that the way I've set the problem up, you will theoretically get a continuous solution on the desired interval. I mean, the initial condition for the region x >= 3 is as I've given, correct?

**Ackbeet** · November 1st 2010, 07:45 AM

Reply to Sambit:

Good so far. What next?

**Sambit** · November 1st 2010, 07:50 AM

as you said, $y(\pi/2)$ should be equal to $3-e^{-\pi/2}$ but it does not hold here.

**Ackbeet** · November 1st 2010, 07:53 AM

...but it does not hold here...

Why not? Are there no values of C which can make that happen? Incidentally, in post # 6, I was confused by your assertion that C = 4. How did you arrive at that? Is there a part of the problem statement that you haven't posted?

**Sambit** · November 1st 2010, 07:57 AM

in my question, i have $y(0)=2$ . What does this mean? i think it means at x=0, y is 2. or am i wrong?

**Ackbeet** · November 1st 2010, 08:00 AM

Right, but you've already found the solution for the region 0 < x < pi/2. You're now finding a solution for pi/2 < x < infty. You'll have a piece-wise defined solution, just like you have a piece-wise defined RHS of the DE. Therefore, the condition y(0) = 2 does not apply to the piece you're trying to find now.

**Sambit** · November 1st 2010, 08:03 AM

so am i now to find a value of C such that $y(\pi/2)=3-e^{-\pi/2}.$ holds?

**Sambit** · November 1st 2010, 08:08 AM

is $C=6-2e^{-\pi/2}$ ?

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