General Solution of DE

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April 11th 2010, 04:33 PM
cdlegendary

General Solution of DE

Find the general solution of http://homework.math.ucsb.edu/webwor...ea6562ac51.png, where y is a function of t.

Could someone help me out with this? I'm not sure how to go about solving for the general solution. All I've got is $dy/dt = 1-2y$ .

Thanks!
April 11th 2010, 04:38 PM
dwsmith

It is a linear non-homogeneous equation.

Solve for the integrating factor.
April 11th 2010, 04:45 PM
Math Major

Let $\mu = e^{ \int 2 dt} = e^{2t}$ . This will be our integrating factor.

Multiplying $\mu$ through to the whole equation we get that

$e^{2t} y' + 2e^{2t} y = e^{2t}$

You can work it from here.
April 11th 2010, 05:05 PM
Krizalid

Quote:

Originally Posted by cdlegendary

Could someone help me out with this? I'm not sure how to go about solving for the general solution. All I've got is $dy/dt = 1-2y$ .

good call, now, can you see your equation is separable?
April 11th 2010, 07:26 PM
cdlegendary

alright so this is what I ended up trying, but I don't think it's correct. any pointers on where I went wrong?

$e^{2t}y'+2e^{2t}y = e^{2t}$

$(e^{2t}y)' = e^{2t}$

$e^{2t}y = \int e^{2t} dt$

$e^{2t}y = (e^{2t})/2 +c$

$y= 1/2 +c$
April 11th 2010, 11:13 PM
deragon999

Dy/dt=1-2y
dy/(1-2y)=1 dt
integrate both sides
You should be able to get it from here
April 11th 2010, 11:31 PM
harish21

Quote:

Originally Posted by cdlegendary

alright so this is what I ended up trying, but I don't think it's correct. any pointers on where I went wrong?

$e^{2t}y'+2e^{2t}y = e^{2t}$

$(e^{2t}y)' = e^{2t}$

$e^{2t}y = \int e^{2t} dt$

$e^{2t}y = (e^{2t})/2 +c$

$y= 1/2 +c$

Look at what Krizalid said. Since your equation is separable, how about:

$\int \frac{dy}{1-2y} = \int dt$
April 12th 2010, 10:46 AM
cdlegendary

Quote:

Originally Posted by harish21

Look at what Krizalid said. Since your equation is separable, how about:

$\int \frac{dy}{1-2y} = \int dt$

Alright, so now I've got:

$\int dy/1-2y = \int dt$

$(-1/2)ln(1-2y) = t + c$

$ln(1-2y) = -2t - 2c$

$1-2y = e^{-2t-2c}$

$y=(e^{-2t-2c}-1)/-2$
April 12th 2010, 06:15 PM
harish21

Quote:

Originally Posted by cdlegendary

Alright, so now I've got:

$\int dy/1-2y = \int dt$

$(-1/2)ln(1-2y) = t + c$

$ln(1-2y) = -2t - 2c$

$1-2y = e^{-2t-2c}$

$y=(e^{-2t-2c}-1)/-2$

That looks good.I would put it this way:

$1-2y = e^{-2t-2c} = e^{-2t} e^{-2c}$

$y = \frac{1}{2}-\frac{e^{-2t} e^{-2c}}{2}$

$y = \frac{1}{2} + {e^{-2t}. C}$

Note : $C = \frac{-e^{-2c}}{2}$ is a constant
April 15th 2010, 12:43 AM
moiz

y' +2y =1

first find the characteristic roots

y'+2y=0
lambda=l

l^2+2=0
l=-2
l=+-sqrt(2)i
y=e^sqrt(2)i+e^-sqrt(2)i

expand with eulers formula get sin and cosine = yg, then
find particular solution
y=A
y'=0
plug into the question
2A=1
A=1/2

Now yp=1/2
y=yg+yp

Cheers;)