Question

hi can you help me in higher order linear differential equations problems?

Answer 1

9y''+3y'+6=xcosx

Answer 2

Have you worked on it at all? Do you have the homogeneous solution? Anything?

Answer 3

i have done others in which both acts separately like 9y''+3y'+6=cox etc etc .

Answer 4

If the nonhomogeneous part is \(x\cos x\), you can't find the nonhomogeneous solution by splitting up \(x\) and \(\cos x\).

First thing to do is solve the homogeneous part, i.e.
\[9y''+3y'+6=0\]
which has the characteristic equation
\[9r^2+3r+6=0\]
Find your roots \(r_1\) and \(r_2\), then the general homogeneous solution takes on the form \(\large C_1e^{r_1x}+C_2e^{r_2x}\).

Answer 5

For the nonhomogeneous part, i.e. \(9y''+3y'+6y=x\cos x\), I would try setting \(y=Ax\sin x+Bx\cos x\).

Answer 6

i can solve left side but right side does not many any sense to me.

Answer 7

Have you learned about the method of undetermined coefficients?

Answer 8

like?

Answer 9

i mean any example of that?

Answer 10

I'll give you a link with plenty of examples: http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 11

yes i can solve that given problem by myself.

Answer 12

Okay. If \(y=Ax\sin x+Bx\cos x\) doesn't work out, try \(y=(Ax+B)\sin x+(Cx+D)\cos x\).

Answer 13

i need only Yp , i hope you know Yp=F(x)/f(D) where f(D) is the left side.

Answer 14

I'm not familiar with the notation you're using. Here's how I would proceed:
\[\begin{cases}
y=(Ax+B)\sin x+(Cx+D)\cos x\\
y'=(A-Cx-D)\sin x+(Ax+B+C)\cos x\\
y''=(-Ax-B-2C)\sin x+(2A-Cx-D)\cos x
\end{cases}\]
Substitute this into the original ODE:
\[\begin{align*}
9((-Ax-B-2C)\sin x+(2A-Cx-D)\cos x)&\\
+3((A-Cx-D)\sin x+(Ax+B+C)\cos x)&\\
+6((Ax+B)\sin x+(Cx+D)\cos x)&=x\cos x\\\\

(-9Ax-9B-18C)\sin x+(18A-9Cx-9D)\cos x&\\
+(3A-3Cx-3D)\sin x+(3Ax+3B+3C)\cos x&\\
+(6Ax+6B)\sin x+(6Cx+6D)\cos x&=x\cos x\\\\

(-6Ax-3B-18C)\sin x+(18A-9Cx-9D)\cos x&\\
+(3A+3Cx+3D)\sin x+(3Ax+3B+3C)\cos x&=x\cos x
\end{align*}\]
You have the system
\[\begin{cases}
-6A+3C=0&(x\sin x)\\
3A-3B-18C+3D=0&(\sin x)\\
3A-9C=1&(x\cos x)\\
18A+3B+3C-9D=0&(\cos x)
\end{cases}\]

Answer 15

There are actually a few minor errors in the algebra above, but I'm sure you can iron them out for yourself.

Answer 16

after getting values of A ,B,C,D whats next step?

Answer 17

Once you find \(A,B,C,D\), you need only add the homogeneous solution to the nonhomogeneous solution, and you are done.

Answer 18

some shortcuts that will come in handy whn finding particular solutions of higher order DE with constant coefficients.
\[a_ny^{(n)}+a_{n-1}y^{(n-1)}+....+a_1y^{(1)}+a_0=Q(x)\]
\[y=\frac{ Q(x) }{ F(D) }\]
(1)If
\[Q(x)=e^{ax}\]
If
\[F(a)\ne0\], then
\[y_p=\frac{ 1 }{ F(a) }e^{ax}\].
If\[F(a)=0\], then
(D-a) is a factor of F(D).
Suppose \((D-a)^n\) is a factor of F(D). Then
\[F(D)=(D-a)^nG(D),G(a)\ne0.\]
Now,
\[\frac{ 1 }{ (D-a)^n }e^{ax}=\frac{ x^n }{ n! }e^{ax}\]

Answer 19

(2)If
\[Q(x)=C, \]
where C is a constant.
Then,
\[Q(x)=Ce^{0x}\]
If \(F(0)\ne0\), then \(y_p=\frac{1}{F(D)}Ce^{0x}=\frac{C}{F(0)}\)
If \(F(0)=0\), then D is a factor of F(D). Suppose \(D^n\) is a factor of F(D).
Then, \(F(D)=D^nG(D), G(0)\ne0.\)
Now,
\(\frac{1}{D^n}Ce^{0x}=C\frac{x^n}{n!}\)

Answer 20

(3)If \(Q(x)=xV(x)\) where
\(V(x)=\cos ax\) or \(V(x)=\sin ax\) or \(V(x)=x^m\)
\[y_p=\frac{ 1 }{ F(D) }xV(x)-x \frac{ 1 }{ F(D)}V(x)-\frac{ F'(D) }{ [F(D)]^2 } V(x)\]