# Thread: 2nd ODE problem

1. ## 2nd ODE problem

find the general solution:

y" + 4y = 3sin(2x)

I'm able to solve for the homogeneous by:

y" + 4y = 0

y(x) = Asin(2x) + Bcos(2x)

however I'm unsure about the particular solution for 3sin(2x).

any hints to point me in the right direction would be much appreciated. Thank you

2. consider the homogenous equation

$\displaystyle \frac{d^{2}y}{dx^{2}}+4y=0$

this has roots repeated at -2 since

$\displaystyle r^{2}+4=(r+2)^{2}$

the solution to the homogenous equation will then be of the form

$\displaystyle y=(c_{1}+c_{2}x)e^{-2x}$

now we consider the particular solution, since we are differentiating twice if you try the following solution

$\displaystyle y_{particular}=(c_{1}+c_{2}x)e^{-2x}+c_{3}\sin (2x)$

you should find the correct answer.

Plug it into your DE and solve for the constant $\displaystyle c_{3}$

3. r^2 +4 is not (r+2)^2

Tasukete you are correct

y(x) = Asin(2x) + Bcos(2x)

is the correct homogeneous solution

For the particular solution if you use undetermined cofficients
use yp = Axsin(2x) + Bxcos(2x)

multiply by x since sin(2x) and cos(2x) are homogeneous solutions.

Personally I would use variation of parameters if you have learned that method if not see Laplace Transforms for the notes on var of parameters

4. Originally Posted by Mauritzvdworm
consider the homogenous equation

$\displaystyle \frac{d^{2}y}{dx^{2}}+4y=0$

this has roots repeated at -2 since

$\displaystyle r^{2}+4=(r+2)^{2}$

the solution to the homogenous equation will then be of the form

$\displaystyle y=(c_{1}+c_{2}x)e^{-2x}$

now we consider the particular solution, since we are differentiating twice if you try the following solution

$\displaystyle y_{particular}=(c_{1}+c_{2}x)e^{-2x}+c_{3}\sin (2x)$

you should find the correct answer.

Plug it into your DE and solve for the constant $\displaystyle c_{3}$
Sorry but this is completely wrong.

1. The roots of $\displaystyle r^{2}+4 = 0$ are $\displaystyle r = \pm 2i$. Therefore the homogenous solution is $\displaystyle y_h = A \cos (2x) + B \sin (2x)$.

@OP: The difficulty in getting a particular solution is that the term on the right hand side of the differential equation is part of the homogenous solution .... Therefore try this one:

$\displaystyle y_p = (ax + b) \sin (2x) + (cx + d) \cos (2x)$

Your job is to determine the value of the coefficients a, b, c and d.

5. Mr fantastic you don't need the b and d in

to find the particular solution.

though in the general solution you will have yc(x) = Asin(2x) + Bcos(2x)

as part of the general solution

6. Originally Posted by Calculus26
Mr fantastic you don't need the b and d in

to find the particular solution.

though in the general solution you will have yc(x) = Asin(2x) + Bcos(2x)

as part of the general solution
Yes.

When f(x) is part of the homogeneous solution, the rule of thumb is to multiply the 'usual form' of the particular solution by x^m, where m is the smallest positive integer needed to ensure that the 'new' particular solution is not part of the homogeneous solution.

7. sorry, that was a bit careless of me. we will have imaginary roots

### math parameters y=asin2x bcos2x solved.

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