How to get a particular solution.

In summary, the conversation discusses how to solve the initial value problem of y'' + y = 2cosx - 3sinx. The person is able to solve everything except for determining the particular solution (yp). They suggest using Lagrange's method of variation of constants and solving the homogeneous equation. They also mention using the differential operator (D^2+1) and substituting in the particular solution to equate coefficients.
  • #1
Cafka
6
0
I have a "Solve the initial value problem" It is:

y'' + y = 2cosx - 3sinx

I know how to do everything except for get yp. I know it has to be something so when i substitute yp into the left hand side of the equation, i get the right hand side, 2cosx - 3sinx.

By this definition i would think yp = Acosx - Bsinx, where A = 2 and B = 3. But that doesn't seem right.

Can anyone help me out?

Thank you.
 
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  • #2
Yeah,use Lagrange's method of variation of constants...Solve the homogenous ODE and then make the 2 constants (associated with the 2 independent solution spanning the solutions space) become functions of "x"...


Daniel.
 
  • #3
For the equation:

[tex]y^{''}+y=2Cos[x]-3Sin[x][/tex]

Whenever the right-hand side is a particular solution of a linear homogeneous equation, then you can apply that differential operator to both sides of the above equation to collapse it into a homogeneous one:

Since RHS is a particular solution to homogeneous equation (in operator notation):

[tex](D^2+1)y=0[/tex]

Applying this opearator to both sides of the original equation leads to:

[tex](D^2+1)(D^2+1)R=0[/tex]

Solution of this one is:

[tex]y(x)=c_1Sin[x]+c_2Cos[x]+AxSin[x]+BxCos[x][/tex]

but:

Well, read the book to fill in the details, but then just substitute:

[tex]y_p(x)=AxSin[x]+BxCos[x][/tex]

into the original equation, equate coefficients. You're done.
 

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