# Differential Equation

• Dec 8th 2008, 02:18 AM
intomath
Differential Equation
Can someone help me with this question?

Using $\displaystyle x = At^{2}e^{-t}$ as a trial solution solve the differential equation ...

$\displaystyle \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = e^{-t}$ given that x = 1 and $\displaystyle \frac{dx}{dt} = 0$ at t = 0.

Use your solution to prove that for t>=0, x<=1

• Dec 8th 2008, 02:32 AM
mr fantastic
Quote:

Originally Posted by intomath
Can someone help me with this question?

Using $\displaystyle x = At^{2}e^{-t}$ as a trial solution solve the differential equation ...

$\displaystyle \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = e^{-t}$ given that x = 1 and $\displaystyle \frac{dx}{dt} = 0$ at t = 0.

Use your solution to prove that for t>=0, x<=1

First solve $\displaystyle \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = 0$ to get the homogenous solution. Can you do this?

You're given that the particular solution has the form $\displaystyle x = At^{2}e^{-t}$ (you should think about why that is) so you need to get the value of A. Start by calculating $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{d^2x}{dt^2}$. Can you do this?

Note: Once you have the value of A and hence the particular solution you can construct the complete solution by adding together the homogenous and particular solutions. The boundary conditions x = 1 and $\displaystyle \frac{dx}{dt} = 0$ at t = 0 are used to get the value of the arbitrary constants.
• Dec 8th 2008, 02:53 AM
intomath
This is how i worked it out so far:

Using trial solution $\displaystyle x = At^{2}e^{-t}$

$\displaystyle \frac{dx}{dt} = 0$
$\displaystyle \frac{d^{2}x}{dt^{2}} = 0$

therefore $\displaystyle At^{2}e^{-t} = e^{-t}$

comparing coefficients:
$\displaystyle At^{2} = 1$
$\displaystyle t = \sqrt{\frac{1}{A}}$ ....Particular Integral

Considering
$\displaystyle \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = 0$

Let $\displaystyle x = e^{ut}$ be a trial solution
therefore $\displaystyle u^{2} + 2u + 1 = 0$
u = -1(twice) therefore real equal roots

hence $\displaystyle x = (A+Bt)e^{-x}$ - Complimentary Function

x = Particular Integral + Complimentary Function

therefore $\displaystyle x = \sqrt{\frac{1}{A}} + (A+Bt)e^{-x}$ ....[A]

putting x = 1, t = 0 in [A]
$\displaystyle 1 = \sqrt{\frac{1}{A}} + (A+B(0))e^{-1}$

This is where i'm stuck, i don't know how to carry on from here.
• Dec 8th 2008, 03:10 AM
mr fantastic
Quote:

Originally Posted by intomath
This is how i worked it out so far:

Using trial solution $\displaystyle x = At^{2}e^{-t}$

$\displaystyle \frac{dx}{dt} = 0$
$\displaystyle \frac{d^{2}x}{dt^{2}} = 0$

Mr F says: I have no idea how you got the above results but they are wrong.

therefore $\displaystyle At^{2}e^{-t} = e^{-t}$

comparing coefficients:
$\displaystyle At^{2} = 1$
$\displaystyle t = \sqrt{\frac{1}{A}}$ ....Particular Integral

Mr F says: None of the above is correct.

And there are various careless errors in your complementary solution as well: http://www.mathhelpforum.com/math-he...2d59524e-1.gif should be $\displaystyle x = ({\color{red}a} + Bt) e^{-{\color{red}t}}$. The pronumeral A should not be used in the complementary solution because this symbol is already used in the question to mean something else.

[snip]

$\displaystyle x = At^2 e^{-t}$

$\displaystyle \Rightarrow \frac{dx}{dt} = Ae^{-t} (2t - t^2)$

$\displaystyle \Rightarrow \frac{d^2x}{dt^2} = Ae^{-t} (t^2 - 4t + 2)$.

Carefully substitute the above into the DE and simplify. You will be left with $\displaystyle 2A = 1 \Rightarrow A = \frac{1}{2}$.

So $\displaystyle x = (a + Bt) e^{-t} + \frac{t^2}{2} e^{-t}$.

Now use the given boundary conditions x = 1 and http://www.mathhelpforum.com/math-he...ffa3a8fe-1.gif at t = 0 to solve for a and B.
• Dec 8th 2008, 03:15 AM
intomath
Thanks a lot you saved my day