Differential Equations General Solution

**jokke22** · May 5th 2009, 07:19 AM

Hello!

My differential equation is given as following:

$\frac{dy}{dt}=\lambda y$

I am then to find the general solution.

Which is:

$\int\frac{dy}{dt}\,dt=\int \lambda y\,dt \implies y+C_1 = \lambda^2+C_2 \implies \color{red}\boxed{y=\lambda^2+C}$

Is this correct?

Addition, if the $\lambda$ is positive, negative or 0 what would happen in these three cases if you let $t \rightarrow \infty$

Any help would be greatly appreciated!

**Danny** · May 5th 2009, 07:26 AM

Originally Posted by jokke22

Hello!

My differential equation is given as following:

$\frac{dy}{dt}=\lambda y$

I am then to find the general solution.

Which is:

$\int\frac{dy}{dt}\,dt=\int \lambda y\,dt \implies y+C_1 = \lambda^2+C_2 \implies \color{red}\boxed{y=\lambda^2+C}$

Is this correct?

Addition, if the $\lambda$ is positive, negative or 0 what would happen in these three cases if you let $t \rightarrow \infty$

Any help would be greatly appreciated!

Sorry. $\lambda$ is just a parameter. Your equation is separable

$\frac{dy}{y} = \lambda \,dx$ so $\ln|y| = \lambda \,x + \ln c$ or $y = ce^{\lambda x}$ .

**jokke22** · May 5th 2009, 07:59 AM

I just plain suck at these equation.

Thanks yet again Danny!

If I could trouble you yet again..

If you were to solve this one then:
$\frac{dy}{dt}= y^2 cos t$

I may be able to find a solution, but how do you show that its periodic with $2\pi$

**Danny** · May 5th 2009, 09:50 AM

Originally Posted by jokke22

I just plain suck at these equation.

Thanks yet again Danny!

If I could trouble you yet again..

If you were to solve this one then:
$\frac{dy}{dt}= y^2 cos t$

I may be able to find a solution, but how do you show that its periodic with $2\pi$

$\frac{dy}{y^2} = \cos t\,dt$ so $- \frac{1}{y} = \sin t - c$

thus $y = \frac{1}{c - \sin t}$ . Since $\sin t$ is $2 \pi$ - periodic then so is $y$ .

**HallsofIvy** · May 5th 2009, 04:53 PM

Originally Posted by jokke22

Hello!

My differential equation is given as following:

$\frac{dy}{dt}=\lambda y$

I am then to find the general solution.

Which is:

$\int\frac{dy}{dt}\,dt=\int \lambda y\,dt \implies y+C_1 = \lambda^2+C_2 \implies \color{red}\boxed{y=\lambda^2+C}$

Is this correct?

You already know this is not correct. The reason you can't do it is because on the right you have $\int\lambda y dt$ and that has to be integrated with respect to t. But y is an unknown function of t. Since you don't know what y is in terms of t, you can't integrate with respect to t.

Notice that danny arigo's method gives a function of y time "dy" equal to a function of t times "dt". That means you integrate the function of y with respect to y and the function of t with respect to t.

Addition, if the $\lambda$ is positive, negative or 0 what would happen in these three cases if you let $t \rightarrow \infty$

Any help would be greatly appreciated![/QUOTE]

**jokke22** · May 6th 2009, 01:05 AM

Grateful for all the help so far!

I do have one more though I can't solve... Appreciate all the help I can get!

$y^{/}+\frac{1}{x}y=e^{x^2}$
where
$x>0$

which satisfy $y(1) = e$

**Danny** · May 6th 2009, 06:07 AM

Originally Posted by jokke22

Grateful for all the help so far!

I do have one more though I can't solve... Appreciate all the help I can get!

$y^{/}+\frac{1}{x}y=e^{x^2}$
where
$x>0$

which satisfy $y(1) = e$

This is a linear ODE in standard form. The integrating factor is $\mu = exp \int \frac{dx}{x} = x$ so our equation comes

$x y^{/} + y = xe^{x^2}$ or $\frac{d}{dx} (x y) = xe^{x^2}$ . If you let u = xy then $\frac{d}{dx} u = xe^{x^2}$ which is separable. I think you can take it from here.

**jokke22** · May 7th 2009, 09:56 AM

I can't seem to get it right...

$\int\frac{d}{dx} u = \int xe^{x^2}$

which
$\int xe^{x^2} dx$

Sorry mate, just getting confused... Apply conditions for $x$ , $xe^{x^2}$ and $dx$ ?

Go detailed since i just plain suck at this, if I could bother you yet again!

**jokke22** · May 22nd 2009, 01:04 AM

Okey, in stead of replacing... Could I go this way:

$\frac{d}{dx}(xy)=xe^{x^2} \rightarrow xy = \int xe^{x^2}dx+C$

How would this end up?

Help is greatly appreciated!

Maybe:

$2xe^{x}-e^x+C$

**Danny** · May 22nd 2009, 05:30 AM

Originally Posted by jokke22

Okey, in stead of replacing... Could I go this way:

$\frac{d}{dx}(xy)=xe^{x^2} \rightarrow xy = \int xe^{x^2}dx+C$

How would this end up?

Help is greatly appreciated!

Maybe:

$2xe^{x}-e^x+C$

The left hand step is good but there's a mistake on the right.

$<br /> \int x e^{x^2}dx = \frac{1}{2} e^{x^2} + c<br />$

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