Second Order Nonhomogeneous Equation

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May 6th 2011, 05:03 AM
Nguyen

Second Order Nonhomogeneous Equation

Hey everyone, I need help understanding solving this equation as it deals with an input function that is discontinuous, which I don't know how to handle.

$g(x) =sin x$ for $0 \le x \le \frac{\pi}{2}$
$g(x)=0$ for $x > \frac{\pi}{2}$
Solve $y''+4y=g(x)$

The two intervals:
$y'' +4y = sin (x)$ for $0 \le x \le \frac{\pi}{2}$

and

$y'' +4y = 0$ for $x > \frac{\pi}{2}$

with $y(0)= 1$ and $y'(0)=2$

Can anyone help me?

Thanks
May 6th 2011, 05:51 AM
Prove It

Solve the homogeneous equation http://quicklatex.com/cache3/ql_ae3a...765eaaf_l3.png.

It has characteristic equation http://quicklatex.com/cache3/ql_1378...8f9c350_l3.png, which has solution http://quicklatex.com/cache3/ql_18f0...b29dce9_l3.png.

From this, we can determine that the solution to the homogeneous DE is http://quicklatex.com/cache3/ql_8dbf...aa5143a_l3.png. This is also the entire solution when http://quicklatex.com/cache3/ql_cf75...7b7f884_l3.png.

Now for the nonhomogeneous solution, i.e. when http://quicklatex.com/cache3/ql_07bb...bd960ba_l3.png, we try http://quicklatex.com/cache3/ql_0521...4106705_l3.png. This would mean http://quicklatex.com/cache3/ql_dbcc...cb5330d_l3.png and http://quicklatex.com/cache3/ql_5dc0...6348c2c_l3.png.

Substituting into the DE gives

http://quicklatex.com/cache3/ql_5997...721ec5d_l3.png.

http://quicklatex.com/cache3/ql_a0e5...cc26284_l3.png

So the particular solution is http://quicklatex.com/cache3/ql_ad3e...904f10c_l3.png.

Therefore when http://quicklatex.com/cache3/ql_07bb...bd960ba_l3.png, the solution of the DE is http://quicklatex.com/cache3/ql_b863...2575bd0_l3.png.

Now you will need to use your boundary conditions to evaluate A and B.
May 6th 2011, 06:15 AM
bugatti79

Quote:

Originally Posted by Prove It

Solve the homogeneous equation http://quicklatex.com/cache3/ql_ae3a...765eaaf_l3.png.

It has characteristic equation http://quicklatex.com/cache3/ql_1378...8f9c350_l3.png, which has solution http://quicklatex.com/cache3/ql_18f0...b29dce9_l3.png.

From this, we can determine that the solution to the homogeneous DE is http://quicklatex.com/cache3/ql_8dbf...aa5143a_l3.png. This is also the entire solution when http://quicklatex.com/cache3/ql_cf75...7b7f884_l3.png.

May I ask how do you know this is the entire solution when you dont know A and B at this stage?
May 6th 2011, 06:30 AM
Prove It

Because g(x) = 0 when x > pi/2, which means the DE is homogeneous in that domain.
May 6th 2011, 07:05 AM
TheEmptySet

Quote:

Originally Posted by Nguyen

Hey everyone, I need help understanding solving this equation as it deals with an input function that is discontinuous, which I don't know how to handle.

$g(x) =sin x$ for $0 \le x \le \frac{\pi}{2}$
$g(x)=0$ for $x > \frac{\pi}{2}$
Solve $y''+4y=g(x)$

The two intervals:
$y'' +4y = sin (x)$ for $0 \le x \le \frac{\pi}{2}$

and

$y'' +4y = 0$ for $x > \frac{\pi}{2}$

with $y(0)= 1$ and $y'(0)=2$

Can anyone help me?

Thanks

Since this is an IVP an alternative is to use the Laplace transform to solve note that

$\sin(x)=\cos\left( x-\frac{\pi}{2}\right)$

So the right hand side can be written as

$g(x)=\sin(x)-u\left(x-\frac{\pi}{2} \right)\cos\left( x-\frac{\pi}{2}\right)$

and this has Laplace transform

$G(s)=\frac{1}{s^2+1}-e^{-\frac{\pi}{2}s}\cdot \frac{s}{s^2+1}$
May 6th 2011, 05:00 PM
Nguyen

Thanks Prove It and TheEmptySet, I did the question the same way as Prove It but crossed it out becuase I didn't know how to deal with g(x) being discontinuous.

I got y= 5/6 sin(2x) +cos(2x)+1/3 sin(x)
May 7th 2011, 05:22 AM
topsquark

Quote:

Originally Posted by Nguyen

Thanks Prove It and TheEmptySet, I did the question the same way as Prove It but crossed it out becuase I didn't know how to deal with g(x) being discontinuous.

I got y= 5/6 sin(2x) +cos(2x)+1/3 sin(x)

Prove It's method was simply telling you to do the problem over two domains. Not that complicated! I'd recommend learning to do it that way. It's good practice.

-Dan