- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

\begin{align*}\displaystyle

y'' +6y'&=-4xe^{-6x}\\

y_p&=Ax^2e^{-6x}+Bxe^{-6x}

\end{align*}

ok just wanted get this posted before I leave campus

so assume finding zeros is next

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,928

\begin{align*}\displaystyle

y'' +6y'&=-4xe^{-6x}\\

y_p&=Ax^2e^{-6x}+Bxe^{-6x}

\end{align*}

ok just wanted get this posted before I leave campus

so assume finding zeros is next

- Admin
- #2

- Thread starter
- #3

- Jan 31, 2012

- 2,928

$\tiny{17.2.05}$

\begin{align*}\displaystyle

y'' +6y'&=-4xe^{-6x}\\

\end{align*}

$\textrm{So the auxiliary equation is}$

\begin{align*}\displaystyle

r^2+6r&=0\\ r(r+6)&=0\\ r&=0\\ r&=-6

\end{align*}

$\textrm{then}$

\begin{align*}\displaystyle

y_h&=c_1e^{0x}+c_2e^{-6x}=c_1e+c_2e^{-6x}\\

y_p&=Ax^2e^{-6x}+Bxe^{-6x}

\end{align*}

is $y_p$ correct?

\begin{align*}\displaystyle

y'' +6y'&=-4xe^{-6x}\\

\end{align*}

$\textrm{So the auxiliary equation is}$

\begin{align*}\displaystyle

r^2+6r&=0\\ r(r+6)&=0\\ r&=0\\ r&=-6

\end{align*}

$\textrm{then}$

\begin{align*}\displaystyle

y_h&=c_1e^{0x}+c_2e^{-6x}=c_1e+c_2e^{-6x}\\

y_p&=Ax^2e^{-6x}+Bxe^{-6x}

\end{align*}

is $y_p$ correct?

Last edited:

- Admin
- #4

You correctly found the characteristic roots, and so that means the homogeneous solution is:$\tiny{17.2.05}$

\begin{align*}\displaystyle

y'' +6y'&=-4xe^{-6x}\\

\end{align*}

$\textrm{So the auxiliary equation is}$

\begin{align*}\displaystyle

r^2+6r&=0\\ r(r+6)&=0\\ r&=0\\ r&=-6

\end{align*}

$\textrm{then}$

\begin{align*}\displaystyle

y_h&=c_1e^{0x}+c_2e^{-6x}=c_1e+c_2e^{-6x}\\

y_p&=Ax^2e^{-6x}+Bxe^{-6x}

\end{align*}

is $y_p$ correct?

\(\displaystyle y_h(x)=c_1e^{0x}+c_2e^{-6x}=c_1+c_2e^{-6x}\)

The mistake you made was in using $e^0=e$ when we actually have $e^0=1$

Now, if we were not given the form for the particular solution, we would use:

\(\displaystyle y_p(x)=x^s\left(Ax+B\right)e^{-6x}\)

Since one of the terms of the homogeneous solution is $c_2e^{-6x}$, we need to let $s=1$ so that no term in the particular solution is in the homogeneous solution, and so we have:

\(\displaystyle y_p(x)=x\left(Ax+B\right)e^{-6x}\)

And this is equivalent to what you stated. Now you need to determine $A$ and $B$ using the method of undetermined coefficients.

- Thread starter
- #5

- Jan 31, 2012

- 2,928

y_p&=Ax^2e^{-6x}+Bxe^{-6x}\\

G(x)&=-4xe^{-6x}

\end{align*}

So

\begin{align*}\displaystyle

y_p(x)&=Ax^2+Bx+C\\

y_p^{'}& =2Ax+B\\

y_p^{''}& =2A

\end{align*}

substituting into the differential equation

$$(2A)+(2Ax+B)-2(Ax^2+Bx+C)=-4xe^{-6x}$$

got this from an example but ??

- Admin
- #6

\(\displaystyle y_p(x)=x\left(Ax+B\right)e^{-6x}\)

Using the general product rule for differentiation, we may state:

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

And hence:

\(\displaystyle y_p''(x)=\left(-12Ax+2(A-3B)\right)e^{-6x}-6\left(-6Ax^2+2(A-3B)x+B\right)e^{-6x}=2\left(18Ax^2-6(2A-3B)x+A-6B\right)e^{-6x}\)

Now, you want to substitute for $y_p''$ and $y_p'$ into the ODE, and then continue with the process of using the method of undetermined coefficients.

- Admin
- #7

Yes.this ?

$y'' +6y'=-4xe^{-6x}$

$(-6Ax^2+2(A-3B)x+B)e^{-6x}+6[2(18Ax^2-6(2A-3B)x+A-6B)e^{-6x}]=-4xe^{-6x}$

- Thread starter
- #8

- Jan 31, 2012

- 2,928

$$(-6Ax^2+2(A-3B)x+B)e^{-6x}+6[2(18Ax^2-6(2A-3B)x+A-6B)e^{-6x}]=-4xe^{-6x}\\

\textrm{expanded}$$

$$210Ae^{-6x}x^2-142Ae^{-6x}x+12Ae^{-6x}+210Be^{-6x}x-71Be^{-6x}

=-4e^{-6x}x$$

$$210Ax^2-142Ax+12A+210Bx-71B=-4x$$

\textrm{expanded}$$

$$210Ae^{-6x}x^2-142Ae^{-6x}x+12Ae^{-6x}+210Be^{-6x}x-71Be^{-6x}

=-4e^{-6x}x$$

$$210Ax^2-142Ax+12A+210Bx-71B=-4x$$

Last edited:

- Admin
- #9

\(\displaystyle 2\left(18Ax^2-6(2A-3B)x+A-6B\right)e^{-6x}+6\left(\left(-6Ax^2+2(A-3B)x+B\right)e^{-6x}\right)=-4xe^{-6x}\)

Divided through by $4^{-6x}$ since it cannot be zero:

\(\displaystyle 2\left(18Ax^2-6(2A-3B)x+A-6B\right)+6\left(\left(-6Ax^2+2(A-3B)x+B\right)\right)=-4x\)

Arrange both sides in standard form:

\(\displaystyle -12Ax+2A-6B=-4x+0\)

Okay, now what do you get when you equate coefficients?

- Admin
- #10

What you want to do is equate corresponding coefficients, like so:$\displaystyle-12Ax+2A-6B=-4x+0$\\

$\displaystyle x=0$\\

$\displaystyle 2A-6B=0$\\

$\displaystyle x=1$\\

$\displaystyle -12A+2A-6B=0$\\

$\displaystyle -10A-6B=4$

this doesn't look to good!

\(\displaystyle -12A=-4\)

\(\displaystyle 2A-6B=0\)

Now you can uniquely determine $A$ and $B$...

- Jan 29, 2012

- 1,151

Subtracting the second equation from the first, 12A= -4 so A= -1/3. Then 2A- 6B= -2/3- 6B= 0 so 6B= -2/3 and B= -1/9[FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]10[/FONT][FONT=MathJax_Math]

- Thread starter
- #12

- Jan 31, 2012

- 2,928

updated

$A=\frac{1}{3}$

$B=\frac{1}{9}$

so

$$\displaystyle

y_p=\frac{1}{3}x^2e^{-6x}+\frac{1}{9}x e^{-6x}$$

$A=\frac{1}{3}$

$B=\frac{1}{9}$

so

$$\displaystyle

y_p=\frac{1}{3}x^2e^{-6x}+\frac{1}{9}x e^{-6x}$$

Last edited:

- Admin
- #13

I got:I got

$A=-\frac{1}{3}$

$B=-\frac{1}{9}$

i was expecting integers

\(\displaystyle (A,B)=\left(\frac{1}{3},\frac{1}{9}\right)\)

And so, this would give me:

\(\displaystyle y_p(x)=x\left(\frac{1}{3}x+\frac{1}{9}\right)e^{-6x}=\frac{x}{9}(3x+1)e^{-6x}\)

Now, suppose I wish to check my answer, I would then compute:

\(\displaystyle y_p'=\frac{1}{9}(1-18x^2)e^{-6x}\)

\(\displaystyle y_p''=\frac{2}{3}(18x^2-6x-1)e^{-6x}\)

Now plug into the LHS of the original ODE and simplify to make sure we get the RHS of the original ODE:

\(\displaystyle \frac{2}{3}(18x^2-6x-1)e^{-6x}+6\left(\frac{1}{9}(1-18x^2)e^{-6x}\right)=\frac{2}{3}e^{-6x}\left(18x^2-6x-1+1-18x^2\right)=\frac{2}{3}e^{-6x}\left(-6x\right)=-4xe^{-6x}\quad\checkmark\)

So, we find this particular solution works. Check your algebra to find where you went wrong with the signs.

- Thread starter
- #14

- Jan 31, 2012

- 2,928

much mahalo

these are ??? max problems

note

I had to repeatedly log in???

these are ??? max problems

note

I had to repeatedly log in???

- Admin
- #15

\(\displaystyle y''+6y'=-4xe^{-6x}\)

Let:

\(\displaystyle u=y'\implies u'=y''\)

And we have:

\(\displaystyle u'+6u=-4xe^{-6x}\)

Now, if we multiply through by an integrating factor of:

\(\displaystyle \mu(x)=e^{6x}\)

We obtain:

\(\displaystyle e^{6x}u'+6e^{6x}u=-4x\)

The LHS may now be rewritten as the differentiation of a product:

\(\displaystyle \frac{d}{dx}\left(e^{6x}u\right)=-4x\)

Integrate w.r.t $x$:

\(\displaystyle e^{6x}u=-2x^2+c_1\)

And so we have:

\(\displaystyle u=-2x^2e^{-6x}+c_1e^{-6x}\)

Back-substitute for $u$:

\(\displaystyle y'=-2x^2e^{-6x}+c_1e^{-6x}\)

Integrate w.r.t $x$:

\(\displaystyle y(x)=\int -2x^2e^{-6x}+c_1e^{-6x}\,dx=-2\int x^2e^{-6x}\,dx-\frac{1}{6}c_1e^{-6x}\)

Now, if we note that the factor in the second term of \(\displaystyle -\frac{1}{6}c_1\) is just an arbitrary constant, we may write:

\(\displaystyle y(x)=-2\int x^2e^{-6x}\,dx+c_2e^{-6x}\)

Now, on the remaining integral, let's use IBP, where:

\(\displaystyle u=x^2\implies du=2x\,dx\)

\(\displaystyle dv=e^{-6x}\,dx\implies v=-\frac{1}{6}e^{-6x}\)

And so we have:

\(\displaystyle y(x)=-2\left(-\frac{1}{6}x^2e^{-6x}+\frac{1}{3}\int xe^{-6x}\,dx\right)+c_2e^{-6x}=\frac{1}{3}x^2e^{-6x}-\frac{2}{3}\int xe^{-6x}\,dx+c_2e^{-6x}\)

For the remaining integral, let's use IBP again, where:

\(\displaystyle u=x\implies du=dx\)

\(\displaystyle dv=e^{-6x}\,dx\implies v=-\frac{1}{6}e^{-6x}\)

And so we have:

\(\displaystyle y(x)=\frac{1}{3}x^2e^{-6x}-\frac{2}{3}\left(-\frac{1}{6}xe^{-6x}+\frac{1}{6}\int e^{-6x}\,dx\right)+c_2e^{-6x}=\frac{1}{3}x^2e^{-6x}+\frac{1}{9}xe^{-6x}+\frac{1}{18^2}e^{-6x}+c_1+c_2e^{-6x}\)

\(\displaystyle y(x)=c_1+\left(c_2+\frac{1}{18^2}\right)e^{-6x}+\frac{x}{9}\left(3x+1\right)e^{-6x}\)

Noting that \(\displaystyle \left(c_2+\frac{1}{18^2}\right)\) is still just an arbitrary constant, we may write:

\(\displaystyle y(x)=c_1+c_2e^{-6x}+\frac{x}{9}\left(3x+1\right)e^{-6x}\)

And this is equivalent to the solution obtain through the method of undetermined coefficients (but with a lot more work).

- Thread starter
- #16

- Jan 31, 2012

- 2,928

did you mean $e^{-6x}$Divided through by $4^{-6x}$ since it cannot be zero:

\(\displaystyle 2\left(18Ax^2-6(2A-3B)x+A-6B\right)+6\left(\left(-6Ax^2+2(A-3B)x+B\right)\right)=-4x\)

- Admin
- #17

Yes, 'twas a typo.did you mean $e^{-6x}$