# Differential Equations General Solution

• May 5th 2009, 07:19 AM
jokke22
Differential Equations General Solution
Hello!

My differential equation is given as following:

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

I am then to find the general solution.

Which is:

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

Is this correct? (Wondering)

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

Any help would be greatly appreciated!
• May 5th 2009, 07:26 AM
Jester
Quote:

Originally Posted by jokke22
Hello!

My differential equation is given as following:

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

I am then to find the general solution.

Which is:

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

Is this correct? (Wondering)

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

Any help would be greatly appreciated!

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

$\displaystyle \frac{dy}{y} = \lambda \,dx$ so $\displaystyle \ln|y| = \lambda \,x + \ln c$ or $\displaystyle y = ce^{\lambda x}$.
• May 5th 2009, 07:59 AM
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:
$\displaystyle \frac{dy}{dt}= y^2 cos t$

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

(Wondering)
• May 5th 2009, 09:50 AM
Jester
Quote:

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:
$\displaystyle \frac{dy}{dt}= y^2 cos t$

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

(Wondering)

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

thus $\displaystyle y = \frac{1}{c - \sin t}$. Since $\displaystyle \sin t$ is $\displaystyle 2 \pi$ - periodic then so is $\displaystyle y$.
• May 5th 2009, 04:53 PM
HallsofIvy
Quote:

Originally Posted by jokke22
Hello!

My differential equation is given as following:

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

I am then to find the general solution.

Which is:

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

Is this correct? (Wondering)

You already know this is not correct. The reason you can't do it is because on the right you have $\displaystyle \int\lambda y dt$ 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$\displaystyle \lambda$ is positive, negative or 0 what would happen in these three cases if you let $\displaystyle t \rightarrow \infty$

Any help would be greatly appreciated![/QUOTE]
• May 6th 2009, 01:05 AM
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!

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

which satisfy $\displaystyle y(1) = e$

(Wondering)
• May 6th 2009, 06:07 AM
Jester
Quote:

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!

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

which satisfy $\displaystyle y(1) = e$

(Wondering)

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

$\displaystyle x y^{/} + y = xe^{x^2}$ or $\displaystyle \frac{d}{dx} (x y) = xe^{x^2}$. If you let u = xy then $\displaystyle \frac{d}{dx} u = xe^{x^2}$ which is separable. I think you can take it from here.
• May 7th 2009, 09:56 AM
jokke22
I can't seem to get it right...

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

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

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

Go detailed since i just plain suck at this, if I could bother you yet again! (Happy)
• May 22nd 2009, 01:04 AM
jokke22
Okey, in stead of replacing... Could I go this way:

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

How would this end up?

Help is greatly appreciated!

Maybe:

$\displaystyle 2xe^{x}-e^x+C$ (Wondering)
• May 22nd 2009, 05:30 AM
Jester
Quote:

Originally Posted by jokke22
Okey, in stead of replacing... Could I go this way:

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

How would this end up?

Help is greatly appreciated!

Maybe:

$\displaystyle 2xe^{x}-e^x+C$ (Wondering)

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

$\displaystyle \int x e^{x^2}dx = \frac{1}{2} e^{x^2} + c$