stumped on finals studying... Please help :)

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coryshap

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been working away on practice problems for my final exam. I have a worksheet that has a bunch of differential equations with an initial value problem followed by some general solution ones. Ive been on a break and can't seem to shake off the proverbial "rust." I just need help on direction because I know the answer

I need to find the general solution of:
y''(x)+y'(x)-2y(x)=f(x) where f(x)=e^(-x)

I know the answer is supposed to be:
y(x) = c_1 e^(-2 x)+c_2 e^x-e^(-x)/2

do I start by rewriting the function as r^2+r-2=0?

Also, I already solved the initial value problem just before this one (same diff Eq with y(0)=0 and y'(0)=1 when f(x)=0). is any of the stuff I got in that one of use for this one or do I have to start from the beginning?
From the previous equation I have:
y=C_1_e^(x) + C_2_e^(-2x)
y"=C_1_e^(x) - 2C_2_e^(-2x)
as well as the final answer of the initial value problem being: y=(1/3)e^x - (1/3)e^(-2x)

Any help on direction is much appreciated. when it comes to math (math's in the UK where I am studying), figuring out a direction is what seems to get me. Thanks!
 
been working away on practice problems for my final exam. I have a worksheet that has a bunch of differential equations with an initial value problem followed by some general solution ones. Ive been on a break and can't seem to shake off the proverbial "rust." I just need help on direction because I know the answer

I need to find the general solution of:
y''(x)+y'(x)-2y(x)=f(x) where f(x)=e^(-x)

I know the answer is supposed to be:
y(x) = c_1 e^(-2 x)+c_2 e^x-e^(-x)/2

do I start by rewriting the function as r^2+r-2=0?
Click to expand...

Yes, you can use the the quadratic to find the auxiliary equation.

\(\displaystyle m^{2}+m-2=(m-1)(m+2)=0\)

\(\displaystyle m=1, \;\ m=-2\)

So, the complementary function is \(\displaystyle y_{c}=C_{1}e^{-2x}+C_{2}e^{x}\)

We can use undetermined coefficients to find the general solution.

\(\displaystyle y_{p}=Ae^{-x}\)

\(\displaystyle y'_{p}=-Ae^{-x}\)

\(\displaystyle y''_{p}=Ae^{-x}\)

Sub these into the DE and equate coefficients. You should then see that \(\displaystyle A=\frac{-1}{2}\)
 
Thanks, did it out and it all works. What would happen if the same DE was used when f(x)=e^x?

using the same characteristic equation Yc=C_1*e^(-2x) + C_2*e^(x)
and using:
y=Ae^x
y'=Ae^x
y"=Ae^x

you then plug in and get:

Ae^x + Ae^x - 2Ae^x = e^x
which simplifies to:
A + A -2A = 1
but the A's cancel out and its left with 0=1 :(
does this mean that there is no possible solution or
is there something I did wrong / I am missing some algebraic trickery?
 
The assumption \(\displaystyle Ae^{x}\) will fail in this case.

Try using \(\displaystyle y_{p}=Axe^{x}\)
 
Ok, and is there a way of figuring out what to use as y_p? I found another DE (with f(x)=e^(-2x) ) and using the assumption of Yp=Ae^(-2x) fails, as well as Yp=Axe^(-2x). I looked around online and am struggling on figuring out how to assume correctly for the arbitrary constant segment of solving the DE's.

if it helps, here is what I have on the case I still haven't solved:
DE: y"(x) + 4y'(x) + 4y(x) = e^(-2x)

I found the characteristic equation to be:
Yc= C_1*e^(-2x) + x*c_2*e^(-2x)

when I use Yp= e^(-2x) or Yp= Axe^(-2x) they both fail.
 
You can use Variation of Parameters instead of Undetermined Coefficients.

But, for the method we have been using, try \(\displaystyle y_{p}=Ax^{2}e^{-2x}\)

This will work.

Notice \(\displaystyle m^{2}+4m+4=(m+2)^{2}\)

So, the complementary is \(\displaystyle y_{c}=C_{1}e^{-2x}+C_{2}xe^{-2x}\)

Just as you have.
 
thanks. I thought that would be the end of my questions but I got two more....

1) in the equation: y"+4y = 0
how do you get the complimentary equation?
I turned the DE into:
r^2+4 = 0
r^2=-4
r=(-4)^.5
but that would mean that the complimentary equation has e to the power of an imaginary number instead of the y(x) = c_2 sin(2 x)+c_1 cos(2 x) that it is supposed to be.

2.) also, in another problem, I have the DE and complimentary equation as:
DE: y"-4y'=f(x) when f(x)=3
and Yc=(1/4)c_1*e^(4x) + c_2
the answer is supposed to be:
y(x) = 1/4 c_1 e^(4 x)+c_2-(3 x)/4

how did they solve to get the -(3x)/4 on the end of the equation?
If I am not mistaken, Variation of parameters as well as Undetermined coefficients won't work because f(x)=3 and doesnt have an x....
 
coryshap said:
thanks. I thought that would be the end of my questions but I got two more....

1) in the equation: y"+4y = 0
how do you get the complimentary equation?
I turned the DE into:
r^2+4 = 0
r^2=-4
r=(-4)^.5
but that would mean that the complimentary equation has e to the power of an imaginary number instead of the y(x) = c_2 sin(2 x)+c_1 cos(2 x) that it is supposed to be.
Click to expand...

Yes. Use \(\displaystyle y_{c}=C_{1}cos(2x)+C_{2}sin(2x)\). That is correct.

The solutions to \(\displaystyle m^{2}+4=0\) are \(\displaystyle -2i, \;\ 2i\)

2.) also, in another problem, I have the DE and complimentary equation as:
DE: y"-4y'=f(x) when f(x)=3
and Yc=(1/4)c_1*e^(4x) + c_2
the answer is supposed to be:
y(x) = 1/4 c_1 e^(4 x)+c_2-(3 x)/4

how did they solve to get the -(3x)/4 on the end of the equation?
If I am not mistaken, Variation of parameters as well as Undetermined coefficients won't work because f(x)=3 and doesnt have an x....
Click to expand...

Use \(\displaystyle y_{p}=Ax\)

\(\displaystyle y'_{p}=A\)

\(\displaystyle y''_{p}=0\)

Subbing into the DE gives:

\(\displaystyle 0-4A=3\)

\(\displaystyle A=\frac{-3}{4}\)

So, \(\displaystyle y_{p}=\frac{-3}{4}x\)
 
and last question. Or more like double checking,

The question gives a system of first order ODE's:
x'(t)=y(t)
y'(t)=4x(t)+3y(t)
then it asks to solve by reducing it to a second order ODE.

I substituted into the second equation the following:
y(t)=x'(t)
y'(t)=x"(t) (Is it correct to assume this?)
throwing everything onto one side, I get:
x"(t)-3x'(t)-4x(t)=0
which the complimentary equation/solution can be then found:
r^2-3r-4=0
(r+1)(-4)
r=-1 and r=4
solution: Yc=C_1*e^(-x) = C_2*e^(4x)

I also asks to "Plot qualitatively the phase portrait, identifying the stable and the unstable directions"

is that a vector plot?
 
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