Differential Equation

**intomath** · December 8th 2008, 02:18 AM

Can someone help me with this question?

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

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

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

Thanks in advance

**mr fantastic** · December 8th 2008, 02:32 AM

Originally Posted by intomath

Can someone help me with this question?

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

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

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

Thanks in advance

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

You're given that the particular solution has the form $x = At^{2}e^{-t}$ (you should think about why that is) so you need to get the value of A. Start by calculating $\frac{dx}{dt}$ and $\frac{d^2x}{dt^2}$ . 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 $\frac{dx}{dt} = 0$ at t = 0 are used to get the value of the arbitrary constants.

**intomath** · December 8th 2008, 02:53 AM

This is how i worked it out so far:

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

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

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

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

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

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

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

x = Particular Integral + Complimentary Function

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

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

This is where i'm stuck, i don't know how to carry on from here.

**mr fantastic** · December 8th 2008, 03:10 AM

Originally Posted by intomath

This is how i worked it out so far:

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

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

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

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

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

Mr F says: None of the above is correct.

And there are various careless errors in your complementary solution as well:

should be $x = ({\color{red}a} + Bt) e^{-{\color{red}t}}$ . 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]

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

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

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

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

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

Now use the given boundary conditions x = 1 and

at t = 0 to solve for a and B.

**intomath** · December 8th 2008, 03:15 AM

Thanks a lot you saved my day

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