# Math Help - General Solution of DE

1. ## General Solution of DE

Find the general solution of , 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!

2. It is a linear non-homogeneous equation.

Solve for the integrating factor.

3. 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.

4. 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?

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

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

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

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

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

$y= 1/2 +c$

6. Dy/dt=1-2y
dy/(1-2y)=1 dt
integrate both sides
You should be able to get it from here

7. 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:

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

8. 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:

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

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

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

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

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

9. 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

10. 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