# Thread: y remains finite for large values of x

1. ## y remains finite for large values of x

Hi again, just a quick question about the wording of a question.

I have the General Solution to a differential equation:

$y = Ae^{3x} + Be^{-2x} + \frac{1}{6} (\cos{3x} - \sin{3x})$

We are given the fact that when $y=1$, $x=0$.

From this fact I've got the equation $A+B = \frac{5}{6}$

The only other information we are given is that y remains finite for large values of x.

Here's how I interpreted this, can someone please just verify that this is right.

When $x$ is large, then $Be^{-2x}$ tends towards 0, we can then discount this from out equation.

When $x$ gets larger, $Ae^{3x}$ eventually goes towards infinity, but the question states that y remains finite, therefore $A = 0$.

The final solution I have got then is:

$y = \frac{5}{6}e^{-2x} + \frac{1}{6} (\cos{3x} - \sin{3x})$.

Is this correct?

Thanks

2. Originally Posted by craig
Hi again, just a quick question about the wording of a question.

I have the General Solution to a differential equation:

$y = Ae^{3x} + Be^{-2x} + \frac{1}{6} (\cos{3x} - \sin{3x})$

We are given the fact that when $y=1$, $x=0$.

From this fact I've got the equation $A+B = \frac{5}{6}$

The only other information we are given is that y remains finite for large values of x.

Here's how I interpreted this, can someone please just verify that this is right.

When $x$ is large, then $Be^{-2x}$ tends towards 0, we can then discount this from out equation.

When $x$ gets larger, $Ae^{3x}$ eventually goes towards infinity, but the question states that y remains finite, therefore $A = 0$.

The final solution I have got then is:

$y = \frac{5}{6}e^{-2x} + \frac{1}{6} (\cos{3x} - \sin{3x})$.

Is this correct?

Thanks
looks like you covered all the bases to me.