-17.2.05 by undetermined coefficients.

[karush]

$\textrm{Solve the given equation by the method of undetermined coefficients.}$
\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

 

[MarkFL]

Yes, in order to get the general solution to the given ODE, you will need the homogeneous solution, and this involves finding the roots (zeroes) of the characteristic equation. :D

 

[karush]

$\tiny{17.2.05}$
\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?

 

[MarkFL]

$\tiny{17.2.05}$
\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?

You correctly found the characteristic roots, and so that means the homogeneous solution is:

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

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

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

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

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:

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

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

 

[karush]

\begin{align*}\displaystyle
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 ??

 

[MarkFL]

We have:

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

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

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

And hence:

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

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. :D

 

[MarkFL]

this ?

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

Yes. :D

 

[karush]

$$(-6Ax^2+2(A-3B)x+B)e^{-6x}+6[2(18Ax^2-6(2A-3B)x+A-6B)e^{-6x}]=-4xe^{-6x}\\
\textrm{expanded}$$
$$210Ae^{-6x}x^2-142Ae^{-6x}x+12Ae^{-6x}+210Be^{-6x}x-71Be^{-6x}
=-4e^{-6x}x$$
$$210Ax^2-142Ax+12A+210Bx-71B=-4x$$

 

[MarkFL]

I mistakenly told you "Yes" when I should have pointed out that you substituted incorrectly (sorry about that )...you should have:

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

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

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

Arrange both sides in standard form:

$$-12Ax+2A-6B=-4x+0$$

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

 

[MarkFL]

$\displaystyle-12Ax+2A-6B=-4x+0$\\
$\displaystyle x=0$\\
$\displaystyle 2A-6B=0$\\
$\displaystyle x=1$\\
$\displaystyle -12A+2A-6B=0$\\
$\displaystyle -10A-6B=4$
this doesn't look to good!

What you want to do is equate corresponding coefficients, like so:

$$-12A=-4$$

$$2A-6B=0$$

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

 

[HallsofIvy]

When Karush set x= 0, he got the "constant terms", 2A- 6B= 0. When he set x= 1 he got the sum of the constant term and the coefficient of x, -10A- 6B= 4, effectively the same thing. I don't know why he says "this doesn't look too good".

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=MathJax_Main]10[FONT=MathJax_Math]A[FONT=MathJax_Main]−[FONT=MathJax_Main]6[FONT=MathJax_Math]B[FONT=MathJax_Main]=[FONT=MathJax_Main]4[FONT=MathJax_Main]−[FONT=MathJax_Main]10[FONT=MathJax_Math]A[FONT=MathJax_Main]−[FONT=MathJax_Main]6[FONT=MathJax_Math]B[FONT=MathJax_Main]=[FONT=MathJax_Main]4

 

[karush]

updated
$A=\frac{1}{3}$
$B=\frac{1}{9}$
so
$$\displaystyle
y_p=\frac{1}{3}x^2e^{-6x}+\frac{1}{9}x e^{-6x}$$

 

[MarkFL]

I got

$A=-\frac{1}{3}$
$B=-\frac{1}{9}$

i was expecting integers

I got:

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

And so, this would give me:

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

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

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

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

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

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

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

 

[karush]

much mahalo
these are ? max problems

note
I had to repeatedly log in?

 

[MarkFL]

Here's an alternative approach to working the problem...we are given:

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

Let:

$$u=y'\implies u'=y''$$

And we have:

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

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

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

We obtain:

$$e^{6x}u'+6e^{6x}u=-4x$$

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

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

Integrate w.r.t $x$:

$$e^{6x}u=-2x^2+c_1$$

And so we have:

$$u=-2x^2e^{-6x}+c_1e^{-6x}$$

Back-substitute for $u$:

$$y'=-2x^2e^{-6x}+c_1e^{-6x}$$

Integrate w.r.t $x$:

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

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

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

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

$$u=x^2\implies du=2x\,dx$$

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

And so we have:

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

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

$$u=x\implies du=dx$$

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

And so we have:

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

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

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

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

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

 

[karush]

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

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

did you mean $e^{-6x}$

 

[MarkFL]

did you mean $e^{-6x}$

Yes, 'twas a typo. :D