Differential Equations

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January 13th 2009, 01:56 PM
Lonehwolf

Differential Equations

i Find the value of the constant k such that $y=kx^2e^{-2x}$ is a particular integral of the differential equation: $\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=2e^{-2x}$

ii Find the solution of this differential equation for which y=1 and $\frac{dy}{dx}=0$ when x=0.

iii Use the differential equation to determine the value of $\frac{d^2y}{dx^2}$ when x=0. Hence prove that 0<=1 for x>0
January 13th 2009, 02:14 PM
Mush

Quote:

Originally Posted by Lonehwolf

i Find the value of the constant k such that $y=kx^2e^{-2x}$ is a particular integral of the differential equation: $\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=2e^{-2x}$

ii Find the solution of this differential equation for which y=1 and $\frac{dy}{dx}=0$ when x=0.

iii Use the differential equation to determine the value of $\frac{d^2y}{dx^2}$ when x=0. Hence prove that 0<=1 for x>0

[
If $y=kx^2e^{-2x}$ , then:

$\frac{dy}{dx} = k(2xe^{-2x} -2 x^2e^{-2x}) = 2kxe^{-2x}(1-x)$

$\frac{d^2y}{dx^2} = k(2e^{-2x} + -4xe^{-2x} -4xe^{-2x}+4x^2e^{-2x}) = 2ke^{-2x}(1-4x-3x+4x^2)$

Now plug it into the equation and compare the RHS and LHS coefficients! Try this, and then if you succeed you can try the other 2 parts on your own. If you have trouble with them again, just post!
January 14th 2009, 02:50 AM
Lonehwolf

Quote:

Originally Posted by Mush

[
If $y=kx^2e^{-2x}$ , then:

$\frac{dy}{dx} = k(2xe^{-2x} -2 x^2e^{-2x}) = 2kxe^{-2x}(1-x)$

$\frac{d^2y}{dx^2} = k(2e^{-2x} + -4xe^{-2x} -4xe^{-2x}+4x^2e^{-2x}) = 2ke^{-2x}(1-4x-3x+4x^2)$

Now plug it into the equation and compare the RHS and LHS coefficients! Try this, and then if you succeed you can try the other 2 parts on your own. If you have trouble with them again, just post!

I got this far:

$2ke^{-2x}(4x2-7x+1)+8kxe^{-2x}(1-x)+4(kx^2e^{-2x})=2e^{-2x}$

Stuck at this part now.

Trying to sum LHS together and I have no idea how to do that, while trying to expand the RHS to the LHS or get both sides to meet somewhere in the middle is still pretty much impossible for me :(
January 14th 2009, 05:04 AM
Mush

Quote:

Originally Posted by Lonehwolf

I got this far:

$2ke^{-2x}(4x2-7x+1)+8kxe^{-2x}(1-x)+4(kx^2e^{-2x})=2e^{-2x}$

Stuck at this part now.

Trying to sum LHS together and I have no idea how to do that, while trying to expand the RHS to the LHS or get both sides to meet somewhere in the middle is still pretty much impossible for me :(

Okay, let's write the expressions for $y, \frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in a different form, and see if that helps!

$y = kx^2e^{-2x}$

$\frac{dy}{dx} = ke^{-2x}(2x-2x^2)$

$\frac{d^2y}{dx^2} = ke^{-2x}(4x^2-8x+2)$

Put them all into the original ODE:

$\frac{d^2y}{dx^2}+4\frac{dy}{dx} +4y = 2e^{-2x}$

$\therefore ke^{-2x}(4x^2-8x+2)+4ke^{-2x}(2x-2x^2) +4kx^2e^{-2x} =2e^{-2x}$

Divide both sides by $e^{-2x}$ :

$\therefore k(4x^2-8x+2)+4k(2x-2x^2) +4kx^2 =2$

Now take out a factor of k:

$\therefore k(4x^2-8x+2 +8x-8x^2 +4x^2) =2$

$\therefore k(2) =2$

$\therefore k=1$

For part two you must find the general solution of the equation (in part 1 you only found a particular solution!), and then implement the given conditions to find another particular solution. You do this using normal methods.

First of all, use the auxiliary equation:

$\lambda^2+4\lambda+4 = 0$

Solve this for $\lambda$ and you will get 1 solution! (Since the discriminant is 0!). That means that your complimentary function is

$y_c(x) = Ae^{\lambda x } + Bxe^{\lambda x} = e^{\lambda x}(A+Bx)$ , where B and A are the arbitrary constants! You then need to consider the particular integral to account for the RHS of the ODE, to find $y_p(x)$ . You should find, however, that the RHS is linearly dependent on your complimentary function, so you have to make adjustments! Do you know what basis function you use for when the RHS is $e^{-2x}$ , considering the linear dependency on the complimentary function?

Once you have found $y_p(x)$ and $y_c(x)$ , then you find the total function $y(x)$ simply by adding these two together! Because your function is linear, then the solution can be superposed of lots of little solutions. Hence $y(x) = y_c(x) + y_p(x)$