# Thread: Need help with this differential equations please

1. ## 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

2. ## Re: Need help with this differential equations please

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

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.

3. ## Re: Need help with this differential equations please

that is just the way the question is

4. ## 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.

5. ## 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

6. ## 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

7. ## Re: Need help with this differential equations please

Have you at least solved for the homogeneous equations yet?

8. ## Re: Need help with this differential equations please

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

9. ## 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?

10. ## Re: Need help with this differential equations please

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.

11. ## Re: Need help with this differential equations please

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