Question

How to answer "the non homo linear equation with constant coefficients" differential equation EASILY?

Answer 1

99% of people on this site are in 8th grade math lol. Check out chegg for diffy Q
What book are you using?

Answer 2

@silverxx

Answer 3

@perl
Do you know Diffy Q?

Answer 4

@lilshane

Answer 5

well what do you think

Answer 6

@jordan123321

Answer 7

yeahh what do you think

Answer 8

Post a question and i will look at it :)

Answer 9

yeah i kinda need a question

Answer 10

@perl can you think about making me a ambassor

Answer 11

r u still there

Answer 12

The first special case of first order differential equations that we will look is the linear first order differential equation. In this case, unlike most of the first order cases that we will look at, we can actually derive a formula for the general solution. The general solution is derived below. However, I would suggest that you do not memorize the formula itself. Instead of memorizing the formula you should memorize and understand the process that I'm going to use to derive the formula. Most problems are actually easier to work by using the process instead of using the formula.

Answer 13

does tht help you at all

Answer 14

@sierra!

Answer 15

@perl

Answer 16

perl

Answer 17

hi

Answer 18

@perl Have you taken diffy Q before?

Answer 19

@perl ???

Answer 20

@perl are you in calc 1?

Answer 21

(x^3D^3- 3x^2d^2 + 7xD - 8)y= 12x^2 ln^2 x

Answer 22

@perl =)

Answer 23

de by kells @Austin6i6

Answer 24

@perl

Answer 25

i think @perl is in calc 1

Answer 26

yes yes i think so

Answer 27

I only took up to calc 3 for computer science . Although i am debating on taking diffy q.
Is it worth it?

Answer 28

yes it is. diff eq would test your knowledge in integral and diff cal. Deq isnt hard just always remember the basic

Answer 29

@silverxx Whats your major?

Answer 30

\[\left(x^3D^3-3x^2D^2+7xD-8\right)y=12x^2\ln^2x\]
Is this really your ODE? Because those coefficients are not constant.

Answer 31

At any rate, we can still try to find a solution. This looks like an Euler-Cauchy equation to me, so we'll try to solve the associated homogeneous ODE using the substitution \(y=x^r\). I'll switch to prime notation for the derivatives:
\[\begin{align*}
0&=x^3y'''-3x^2y''+7xy'-8y\\\\
&=x^3\left(r(r-1)(r-2)x^{r-3}\right)-3x^2\left(r(r-1)x^{r-2}\right)+7x\left(rx^{r-1}\right)-8x^r\\\\
&=\left(r(r-1)(r-2)x^{r}\right)-3\left(r(r-1)x^{r}\right)+7\left(rx^{r}\right)-8x^r\\\\
&=r(r-1)(r-2)-3r(r-1)+7r-8\\\\
&=r^3-6r^2+12r-8\\\\
&=r^3-3(2)r^2+3(2^2)r-2^3\\\\
&=(r-2)^3
\end{align*}\]
With \(r=2\) having multiplicity \(3\), we have the characteristic solution
\[y_c=C_1x^2+C_2x^2\ln x+C_3x^2\ln^2x\]

Answer 32

The nonhomogeneous solution might take some work... On a whim, I would attempt to find that solution via reduction of order.

Answer 33

yes @SithsAndGiggles reduction of order would be better

Answer 34

i think you missed something.. ln^5 x /5 should be there after c3.. Answer's on the book

Answer 35

Right, I only found the homogeneous solution, but I'm still not sure how to obtain the nonhomogeneous one. I get the feeling we could use undetermined coefficients after a substitution, but that seems dicey.

Answer 36

Reduction of order: Let's try to find a solution of the form \(y(x)=z(x)\,x^2\ln^2x\).
\[\begin{cases}\begin{align*}
y'&=(x^2\ln^2x)z'+(2x\ln^2x+2x\ln x)z\\\\
y''&=(x^2\ln^2x)z''+(4x\ln^2x+4x\ln x)z'+(2\ln^2x+6\ln x+2)z\\\\
y'''&=(x^2\ln^2x)z'''+(6x\ln^2x+6x\ln x)z''\\
&\quad\quad+(6\ln^2x+18\ln x+6)z'+\left(4\frac{\ln x}{x}+\frac{6}{x}\right)z
\end{align*}\end{cases}\]
Substitute into the ODE:
\[\begin{align*}
12x^2\ln^2x&=x^3\bigg[(x^2\ln^2x)z'''+(6x\ln^2x+6x\ln x)z''\\
&\quad\quad\quad+(6\ln^2x+18\ln x+6)z'+\left(4\frac{\ln x}{x}+\frac{6}{x}\right)z\bigg]\\
&\quad-3x^2\bigg[(x^2\ln^2x)z''+(4x\ln^2x+4x\ln x)z'+(2\ln^2x+6\ln x+2)z\bigg]\\
&\quad+7x\bigg[(x^2\ln^2x)z'+(2x\ln^2x+2x\ln x)z\bigg]-8(x^2\ln^2x)z\\\\

12x^2\ln^2x&=(x^5\ln^2x)z'''+(3x^4\ln^2x+6x^4\ln x)z''+(x^3\ln^2x+6x^3\ln x+6x^3)z'
\end{align*}\]
Substituting \(t=z'\), we get the second order linear ODE,
\[12x^2\ln^2x=(x^5\ln^2x)t''+(3x^4\ln^2x+6x^4\ln x)t'+(x^3\ln^2x+6x^3\ln x+6x^3)t\]
I'm not sure where to go from here... I'm considering giving undetermined coefficients a shot.