# Need help with this differential equations please

• May 8th 2013, 05:46 AM
Sunex
Need help with this differential equations please
Obtain the general solution of the differential equations by using the method of undetermined coefficients
y''' - 2y'' + 3y' - 5y = 5sin2x + 10x^2 - 3x - 7
y''' + y' - 5y = 4sinx + 2x^2
• May 8th 2013, 06:02 AM
HallsofIvy
Re: Need help with this differential equations please
Quote:

Originally Posted by Sunex
Obtain the general solution of the differential equations by using the method of undetermined coefficients
y''' - 2y'' + 3y' - 5y = 5sin2x + 10x^2 - 3x - 7

The characteristic equation is [tex]r^3- 2r^2+ 3r- 5= 0[/quote]. What are its roots?
Those are not easy to find but it is clear that neither 0 nor 2i is a root so you need to look for a "particular solution" of the form
y= Acos(2x)+ Bsin(2x)+ Cx^2+ Dx+ E

Quote:

y''' + y' - 5y = 4sinx + 2x^2
Here the characteristic equation is $r^3+ r'- r= 0$. Again, that has only irrational roots which will be hard to find but we know the "particular solution" must be of the form $Asin(x)+ Bcos(x)+ Cx^2+ Dx+ E$

Are you sure you have copied the problems correctly? It looks to me like the general solution to either of those will be very hard to find.
• May 8th 2013, 06:50 AM
Sunex
Re: Need help with this differential equations please
that is just the way the question is
• May 8th 2013, 12:49 PM
HallsofIvy
Re: Need help with this differential equations please
Well, this reduces to solving the cubic equations $r^3- 2r^2+ 3r- 5= 0$ and $r^3+ r- 5= 0$. Graphing shows that each has a single real root between 1 and 2 and two complex roots.
• May 8th 2013, 03:28 PM
Sunex
Re: Need help with this differential equations please
the question is a assignment which has been submitted, would be happy if i could get the complete solution to the first one
• May 8th 2013, 04:20 PM
Sunex
Re: Need help with this differential equations please
Y"'+y'=4sinx +2x^2 and Y'''+y"+3y'-5y=5sin2x+10x^2-3x+7. Is the correct question I just got...am so sorry sir
• May 8th 2013, 05:00 PM
Prove It
Re: Need help with this differential equations please
Have you at least solved for the homogeneous equations yet?
• May 8th 2013, 11:06 PM
Sunex
Re: Need help with this differential equations please
No sir, am requesting for a step by step solution + explanation.
• May 8th 2013, 11:42 PM
MarkFL
Re: Need help with this differential equations please
Let's look at the first one:

(1) $y'''+y'=4\sin(x)+2x^2$

The differential operator $D^2+1$ annihilates $\4\sin(x)$ and $D^3$ annihilates $2x^2$ hence the operator:

$A\equiv D^3(D^2+1)$

annihilates $4\sin(x)+2x^2$.

Thus, applying $A$ to both sides of (1) gives us:

(2) $D^4(D^2+1)^2[y]=0$

The characteristic roots are then:

$r=0,\,\pm i$

where $r=0$ is of multiplicity 4 and $r=\pm i$ are of multiplicity 2, and so the general solution to (2) is given by:

(3) $y(x)=c_1+c_2x+c_3x^2+c_4x^3+(c_5+c_6x)\cos(x)+(c_7 +c_8x)\sin(x)$

Now, recall that a general solution to (1) is of the form $y(x)=y_h(x)+y_p(x)$. Since every solution to (1) is also a solution to (2), then $y(x)$ must have the form displayed on the right-hand side of (3). But we recognize that:

$y_h(x)=c_1+c_2\cos(x)+c_3\sin(x)$

and so there must exist a particular solution of the form:

$y_p(x)=c_2x+c_3x^2+c_4x^3+c_6x\cos(x)+c_8x\sin(x)$

Now it is a matter of using the method of undetermined coefficients to determine the particular solution that satisfies (1).

So, what you need to do is compute $y_p'(x)$ and $y_p'''(x)$, substitute these into (1), and solve the resulting linear system that arises when you equate coefficients. Can you proceed?
• May 9th 2013, 01:04 AM
Prove It
Re: Need help with this differential equations please
Quote:

Originally Posted by Sunex
No sir, am requesting for a step by step solution + explanation.

You are at the wrong site. Here we provide hints and guidance once students have shown they have made some effort to attempt the questions themselves, shown what they have tried and where they are stuck, and shown that they are willing to use the guidance given to master their questions themselves.
• May 9th 2013, 03:34 AM
topsquark
Re: Need help with this differential equations please
Quote:

Originally Posted by MarkFL
Let's look at the first one:

(1) $y'''+y'=4\sin(x)+2x^2$

The differential operator $D^2+1$ annihilates $\4\sin(x)$ and $D^3$ annihilates $2x^2$ hence the operator:

$A\equiv D^3(D^2+1)$

annihilates $4\sin(x)+2x^2$.

Thus, applying $A$ to both sides of (1) gives us:

(2) $D^4(D^2+1)^2[y]=0$

The characteristic roots are then:

$r=0,\,\pm i$

where $r=0$ is of multiplicity 4 and $r=\pm i$ are of multiplicity 2, and so the general solution to (2) is given by:

(3) $y(x)=c_1+c_2x+c_3x^2+c_4x^3+(c_5+c_6x)\cos(x)+(c_7 +c_8x)\sin(x)$

Now, recall that a general solution to (1) is of the form $y(x)=y_h(x)+y_p(x)$. Since every solution to (1) is also a solution to (2), then $y(x)$ must have the form displayed on the right-hand side of (3). But we recognize that:

$y_h(x)=c_1+c_2\cos(x)+c_3\sin(x)$

and so there must exist a particular solution of the form:

$y_p(x)=c_2x+c_3x^2+c_4x^3+c_6x\cos(x)+c_8x\sin(x)$

Now it is a matter of using the method of undetermined coefficients to determine the particular solution that satisfies (1).

So, what you need to do is compute $y_p'(x)$ and $y_p'''(x)$, substitute these into (1), and solve the resulting linear system that arises when you equate coefficients. Can you proceed?

Wow. I haven't seen that method in a loooong time. Nice!

-Dan