Solving a linear differential equation with constant coefficients

[Magnetons]

Mentor note: Moved from a technical math section, so missing the template.
TL;DR Summary: I'm trying to solve the differential equations (D^2
-4D + 3 )y = 2xexp(3x) + 3exp(x)Cos2x

On this page , I've solve particular integral by 2 ways, 1st is above the line in which ( see 7th equality from top ) firstly I've done the expansion and operate on x and then integrate ..

Below the line I do integration first and then done the expansion .
The answer is different by both ways.

Can someone tell me where I'm doing mistake?

 

[jedishrfu]

Have you checked that both solutions fit the DE?

If one doesn't then you know where your mistake lies.

 

[Magnetons]

is there is any rule that we can't integrate first ..

 

[Mark44]

is there is any rule that we can't integrate first ..

I taught differential equations a number of times, but out of the several textbooks we used, I never saw any technique like the one you used. In any case, it seems to me that you have an error in your work. In the first line you have a denominator of $D^2 - 4D + 3$. A couple lines later that changed to $D^2 + 2D$. Obviously those are different, and you can't get from the first expression to the latter one. Also, you seem to have ignored the fact that one of the terms in your nonhomogeneous equation is $3e^x\cos(2x)$.

The usual technique is to find solutions to the homogeneous equation, which would be $(D^2 - 4D + 3)y = 0$. This factors to $(D - 3)(D - 1)y = 0$, so a pair of solutions would be $y_1 = e^{3x}$ and $y_2 = e^x$.

Once you have found a set of solutions to the homogeneous problem, you can tackle the nonhomogeneous problem. One technique is the method of annihilators, which can be used to determine solutions to your nonhomogeneous problem. As an alternative, you could try Laplace Transforms.

 

[Magnetons]

As we can see the non homogeneous part is in the form of e^(ax).V( x) in our case v is a polynomial 2x .
In this case we change f(D) in the denominator to f(D+a) and then only operate it with polynomial and leave the e^ax.

 

[MidgetDwarf]

I taught differential equations a number of times, but out of the several textbooks we used, I never saw any technique like the one you used. In any case, it seems to me that you have an error in your work. In the first line you have a denominator of $D^2 - 4D + 3$. A couple lines later that changed to $D^2 + 2D$. Obviously those are different, and you can't get from the first expression to the latter one. Also, you seem to have ignored the fact that one of the terms in your nonhomogeneous equation is $3e^x\cos(2x)$.

The usual technique is to find solutions to the homogeneous equation, which would be $(D^2 - 4D + 3)y = 0$. This factors to $(D - 3)(D - 1)y = 0$, so a pair of solutions would be $y_1 = e^{3x}$ and $y_2 = e^x$.

Once you have found a set of solutions to the homogeneous problem, you can tackle the nonhomogeneous problem. One technique is the method of annihilators, which can be used to determine solutions to your nonhomogeneous problem. As an alternative, you could try Laplace Transforms.

Although Annihilator approach is nice.
I would tackle the particular solution using UC method [undetermined coefficients].

Although the Annihilator method is typically faster IRC.

OP. Have you taken linear algebra, and aware of linear independence? It cuts the works when using UC method.

 

[Mark44]

As we can see the non homogeneous part is in the form of e^(ax).V( x) in our case v is a polynomial 2x .

No, that isn't true. The right side is $2xe^{3x} + 3e^x \cos(2x)$. This is not something of the form $e^{ax}$ times a polynomial.

In this case we change f(D) in the denominator to f(D+a) and then only operate it with polynomial and leave the e^ax.

I have no idea what you're doing here. Note also that I pointed out at least two of your mistakes, neither of which you seem to have addressed.

 

[Magnetons]

No, that isn't true. The right side is $2xe^{3x} + 3e^x \cos(2x)$. This is not something of the form $e^{ax}$ times a polynomial.

I have no idea what you're doing here. Note also that I pointed out at least two of your mistakes, neither of which you seem to have addressed.

I'm applying this rule on my first partial integral..

Yes, I've addressed the mistakes that's why I'm trying to expalining what I'm doing in solution .