# Thread: differential equation using the method of undetermined coefficient coefficients

1. ## differential equation using the method of undetermined coefficient coefficients

Solve the following differential equation using the method of undetermined coefficient coefficients

d^2y/dx^2 + 9y = x^2 cos3x

This was the most difficult assignments i had encounted as this was not taught in our module yet we are required to solve it . what i had learn before are the right hand side x^2 + cos3x but not multiply .

r^2 + 9 = 0
r(r+ 9) = 0
r=+/- 3j
Yc =C1cos3x + C2sin3x

Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?

2. Originally Posted by higoodhi
Solve the following differential equation using the method of undetermined coefficient coefficients

d^2y/dx^2 + 9y = x^2 cos3x

This was the most difficult assignments i had encounted as this was not taught in our module yet we are required to solve it . what i had learn before are the right hand side x^2 + cos3x but not multiply .

r^2 + 9 = 0
r(r+ 9) = 0
r=+/- 3j
Yc =C1cos3x + C2sin3x

Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?
Almost. Since cos 3x and sin 3x are already solutions to the homogeneous equation, multiply that polynomial by x:
$Y_p= (Ax^3+ Bx^2+ Cx)(D cos(3x)+ E sin(3x))$.

3. why is it Yp= (Ax^3 + Bx^2 + Cx) (Dcos3x + Esin3x) n not Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?

to solve the equation we need to find yp,yp1,yp2
yp1 differential yp by dx , and yp2 is differential by yp1.

how to solve it,since it will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......... (Rem this is only yp ) still need to solve yp1 and yp2 there will be a lot of unknown .....

is there any simple method ?

aft substutite Yp2 and yp into d^2y/dx^2 + 9y = x^2 cos3x

where we need to find the value of A,B,C,D,E

4. Originally Posted by higoodhi
why is it Yp= (Ax^3 + Bx^2 + Cx) (Dcos3x + Esin3x) n not Yp= (Ax^2 + Bx + C) (Dcos3x + Esin3x) ?
you can't have your particular solution of the same form as your homogeneous solution. you will just get zero when you plug everything in, which gets you nowhere. to avoid this, multiplying through by x is the standard trick.

to solve the equation we need to find yp,yp1,yp2
yp1 differential yp by dx , and yp2 is differential by yp1.
no, you need to find yh (the homogeneous solution) and yp. that's it

how to solve it,since it will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......... (Rem this is only yp ) still need to solve yp1 and yp2 there will be a lot of unknown .....

is there any simple method ?

aft substutite Yp2 and yp into d^2y/dx^2 + 9y = x^2 cos3x

where we need to find the value of A,B,C,D,E
what you typed is confusing. anyway, the method of undetermined coefficients says to find $y_p$, $y_p'$ and $y_p''$ and plug them into your original differential equation. equate the coefficients to solve for A, B, C, D and E

5. Originally Posted by Jhevon
anyway, the method of undetermined coefficients says to find $y_p$, $y_p'$ and $y_p''$ and plug them into your original differential equation. equate the coefficients to solve for A, B, C, D and E
how to solve the product unkown since the product unkown will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......

6. Originally Posted by higoodhi
how to solve the product unkown since the product unkown will become very complex as Ax^3.Dcos3x =ADx^3cos 3x(2 unkown is inside AD), Ax^3.Esin3x =AEx^sin 3x(2 unkown is inside AE)......
leave off D and E. just forget about those. solve for A, B and C

7. Originally Posted by Jhevon
leave off D and E. just forget about those. solve for A, B and C
so do u mean
Yp= (Ax^3 + Bx^2 + Cx) (cos3x + sin3x) ?

8. Originally Posted by higoodhi
so do u mean
Yp= (Ax^3 + Bx^2 + Cx) (cos3x + sin3x) ?
yes

9. thank