General solution y''+2y'+y=x(e^-x)+1

[ToastIQ]

Hi!

I'm having a hard time with general solutions of a certain type, maybe it has to do with the constant in the particular solution?

The equation is

y''+2y'+y=xe^-x+1

Is it right to assume that y_h=(C₁x+C₂)e^-x ?

For the particular solution, I've tried all kinds of guesses for the form:

y_p=
(C₁x+C₂)e^-x
(C₁x³+C₂x²)e^-x
C₁x³e^-x+C₂x²
C₁x³e^-x+C₂x²+C₃x+C₄

and more of the kind. I honestly have no clue what form's the right one. Help would be much appreciated.

Also, I've found some tables that show which form can be used for different kinds of equations. They're very handy, but usually just the kind of equation I'm looking for is missing. If some kind soul out there could help me understand when to use which form (like those tables), that would make my day. It was all good in the beginning but by now, I'm just confused about the different forms, when to use which one.

 

[MarkFL]

The characteristic equation association with the homogeneous solution has a repeated root, and so the homogeneous solution will be:

\(\displaystyle y_h()x)=c_1e^{-x}+c_2xe^{-x}\)

Because the term on the RHS $xe^{-x}$ is a homogeneous solution, our particular solution will take the form:

\(\displaystyle y_p(x)=x^2(Ax+B)e^{-x}+C\)

Do you see why the $x^2$ is required to ensure that no term in the particular solution is a solution to the corresponding homogeneous equation?

 

[ToastIQ]

The characteristic equation association with the homogeneous solution has a repeated root, and so the homogeneous solution will be:

\(\displaystyle y_h()x)=c_1e^{-x}+c_2xe^{-x}\)

Because the term on the RHS $xe^{-x}$ is a homogeneous solution, our particular solution will take the form:

\(\displaystyle y_p(x)=x^2(Ax+B)e^{-x}+C\)

Do you see why the $x^2$ is required to ensure that no term in the particular solution is a solution to the corresponding homogeneous equation?

Thanks for your reply.

So I believe my homogeneous solution was correct?

I know that there can't be any terms in the particular solution that are a solution to the homogeneous equation. I can't fully wrap my head around why though.

Can you see right when starting with the particular solution if it contains a solution to the homogeneous equation? Like, in this case, the particular solution has the form (before multiplying with x)
(Ax+B)e^-x+C
and I can see right away that both Axe^-x and Be^-x appear in the homogeneous equation.
But if I multiply with x and get the form
(Ax²+Bx)e^-x+C
I can't see how this also contains a solution to the homogeneous equation. So I don't understand why x¹ isn't enough, why it has to be x².

Do you have to go through all the steps for the particular solution, setting up y(x), y'(x) and y''(x) and substituting those into the equation to see if you need to multiply with one more x?
However. When I use the form y(x)=x²(Ax+B)e^-x+C I'm still making mistakes somewhere.

y(x)=x²(Ax+B)e^-x+C
y'(x)=(3Ax²-Ax³+2Bx-Bx²)e^-x
y''(x)=(6Ax-6Ax²+Ax³-4Bx+Bx²)e^-x

-> 6Axe^-x+C = xe^-x+1 (<- B canceled out?)

A=1/6​

C=1​

The general solution is supposed to be

y=e^-x(1/6x³-x-1)+1

 

[MarkFL]

Thanks for your reply.

So I believe my homogeneous solution was correct?

You stated:

\(\displaystyle y_h=\left(c_1+c_2\right)e^{-x}\)

But what we want is:

\(\displaystyle y_h(x)=c_1e^{-x}+c_2xe^{-x}\)

I know that there can't be any terms in the particular solution that are a solution to the homogeneous equation. I can't fully wrap my head around why though.

Can you see right when starting with the particular solution if it contains a solution to the homogeneous equation? Like, in this case, the particular solution has the form (before multiplying with x)
(Ax+B)e^-x+C
and I can see right away that both Axe^-x and Be^-x appear in the homogeneous equation.
But if I multiply with x and get the form
(Ax²+Bx)e^-x+C
I can't see how this also contains a solution to the homogeneous equation. So I don't understand why x¹ isn't enough, why it has to be x².

If we expand we get:

\(\displaystyle y_p(x)=Ax^2e^{-x}+Bxe^{-x}+C\)

The second term, \(\displaystyle Bxe^{-x}\) is a solution to the corresponding homogeneous equation. That's why we need:

\(\displaystyle y_p(x)=x^2(Ax+B)e^{-x}+C\)

Do you have to go through all the steps for the particular solution, setting up y(x), y'(x) and y''(x) and substituting those into the equation to see if you need to multiply with one more x?

No, you just need to ensure no terms of the particular solution are solutions of the homogeneous equation.

However. When I use the form y(x)=x²(Ax+B)e^-x+C I'm still making mistakes somewhere.

y(x)=x²(Ax+B)e^-x+C
y'(x)=(3Ax²-Ax³+2Bx-Bx²)e^-x
y''(x)=(6Ax-6Ax²+Ax³-4Bx+Bx²)e^-x

-> 6Axe^-x+C = xe^-x+1 (<- B canceled out?)

A=1/6​

C=1​

The general solution is supposed to be

y=e^-x(1/6x³-x-1)+1

Okay, with the particular solution, we have:

\(\displaystyle y_p(x)=x^2(Ax+B)e^{-x}+C\)

\(\displaystyle y_p'(x)=-x\left(Ax^2-3Ax+Bx-2B\right)e^{-x}\)

\(\displaystyle y_p''(x)=\left(Ax^3-6Ax^2+6Ax+Bx^2-4Bx+2B\right)e^{-x}\)

It looks like we have different results for \(\displaystyle y_p''(x)\).

 

[ToastIQ]

You stated:

\(\displaystyle y_h=\left(c_1+c_2\right)e^{-x}\)

But what we want is:

\(\displaystyle y_h(x)=c_1e^{-x}+c_2xe^{-x}\)

I don't know if I'm missing out on something here but I think it said
y_h=(C₁x+C₂)e^-x
which after expanding the brackets should become
C₁xe^-x+C₂e^-x
which is basically the same as C₁e^-x+C₂xe^-x or am I wrong? Just that the x is with C₁ instead of C₂.
Does it make a difference if the x is with C₁ or C₂ as long as you stick to the same setup all the way to the end? I'm wondering because in my class we learned to put the x with the first coefficient, in this case C₁.

If we expand we get:

\(\displaystyle y_p(x)=Ax^2e^{-x}+Bxe^{-x}+C\)

The second term, \(\displaystyle Bxe^{-x}\) is a solution to the corresponding homogeneous equation. That's why we need:

\(\displaystyle y_p(x)=x^2(Ax+B)e^{-x}+C\)

Ah, I see. Thanks.

Okay, with the particular solution, we have:

\(\displaystyle y_p(x)=x^2(Ax+B)e^{-x}+C\)

\(\displaystyle y_p'(x)=-x\left(Ax^2-3Ax+Bx-2B\right)e^{-x}\)

\(\displaystyle y_p''(x)=\left(Ax^3-6Ax^2+6Ax+Bx^2-4Bx+2B\right)e^{-x}\)

It looks like we have different results for \(\displaystyle y_p''(x)\).

True, the 2B term was actually there but I totally lost it in this mess of letters and numbers. So I get
(6Ax+2B)e^-x+C=xe^-x+1

Which gives me

A=1/6​

B=0 (?)​

C=1​

If that is correct, then y_p=1/6x³e^-x+1

But when I put y_h and y_p together, it doesn't look like the right solution (y=e^-x(1/6x³-x-1)+1). I don't understand how to get the -x and -1 aswell. I've done so many similar problems, I don't know why this one is so hard for me.

 

[MarkFL]

I don't know if I'm missing out on something here but I think it said
y_h=(C₁x+C₂)e^-x

Okay, so it did. I missed that before, even when I double checked. I would advise you to use $\LaTeX$ which is easier to read in general, and easier to use rather than all those superscript and subscript BBCodes. :)

...True, the 2B term was actually there but I totally lost it in this mess of letters and numbers. So I get
(6Ax+2B)e^-x+C=xe^-x+1

Which gives me

A=1/6​

B=0 (?)​

C=1​

If that is correct, then y_p=1/6x³e^-x+1

But when I put y_h and y_p together, it doesn't look like the right solution (y=e^-x(1/6x³-x-1)+1). I don't understand how to get the -x and -1 aswell. I've done so many similar problems, I don't know why this one is so hard for me.

Okay, if your particular solution is:

\(\displaystyle y_p(x)=\frac{1}{6}x^3e^{-x}+1\)

Then the general solution is:

\(\displaystyle y(x)=c_1e^{-1}+c_2xe^{-x+}+\frac{1}{6}x^3e^{-x}+1\)

Did you give initial conditions, because I didn't see any?

 

[ToastIQ]

Okay, so it did. I missed that before, even when I double checked. I would advise you to use $\LaTeX$ which is easier to read in general, and easier to use rather than all those superscript and subscript BBCodes. :)

Yeah sorry, you're right, I really need to learn how to use Latex! :D

Okay, if your particular solution is:

\(\displaystyle y_p(x)=\frac{1}{6}x^3e^{-x}+1\)

Then the general solution is:

\(\displaystyle y(x)=c_1e^{-1}+c_2xe^{-x+}+\frac{1}{6}x^3e^{-x}+1\)

Did you give initial conditions, because I didn't see any?

Jesus. I moved on from this problem long ago, always thinking "There have to be initial conditions" but I couldn't see any. Went back to look again for the 3rd time when you asked and THAT'S when I found out I had been looking at a different problem (very similar, but without conditions). Been so fed up with this problem and busy trying to figure out the general solution I couldn't even get the most basic things taken care of.

The initial conditions were:
\(\displaystyle y(0)=y'(0)=0
\\y=(c_1+c_2x+\frac{1}{6}x^3)e^{-x}+1
\\y'=(-c_1-c_2x+c_2-\frac{1}{6}x^3+\frac{1}{2}x^2)e^{-x}
\\ ~\\
y(0)=0:
\\(c_1+c_2\cdot0+\frac{1}{6}\cdot0+\frac{1}{2}\cdot0)e^{0}+1=0
\\c_1+1=0
\\c_1=-1
\\~\\
y'(0)=0:
\\(-c_1-c_2\cdot0+c_2-\frac{1}{6}\cdot0+\frac{1}{2}\cdot0)e^{0}=0
\\-c_1+c_2=0
\\c_2=c_1=-1

\\~\\
y=(-1-1\cdot x+\frac{1}{6}x^3)e^{-x}+1
\\y=(\frac{1}{6}x^3-x-1)e^{-x}+1
\)

(First try on Latex, hope it will show up okay)
Thanks so much for bearing with my confused mind haha :D

 

[MarkFL]

I'm glad you got it all sorted out, and excellent job on your use of $\LaTeX$. I appreciate it, as it is easier on the eyes for math, and you appear to have picked it up very quickly.

 

[ToastIQ]

I'm glad you got it all sorted out, and excellent job on your use of $\LaTeX$. I appreciate it, as it is easier on the eyes for math, and you appear to have picked it up very quickly.

Thanks! What can I say, Google rarely ever disappoints me as a friend :D I totally agree with you. It just seemed so complicated at first, but it's so worth it.