Finding particular integrals

[OllyRutts]

Homework Statement

(d^2y/dx^2) + (dy/dx) = cos x


So you have the trial solution
y= p*cos(x)+q*sin(x)
(dy/dx) = -p*sin(x)+q*cos(x)
(d^2y/dx^2)=-p*cos(x)-q*sin(x)

The issue I am having is equating the coefficients when after I have subbed them into the initial equation:
-p*cos(x)-q*sin(x)-p*sin(x)+q*cos(x)=cos(x)

Equating coefficients of sin x gives -q-p=0
Equating coefficients ofcos x gives -p+q=1
My question is why is
-q*sin(x)-p*sin(x)=cos(x)
= q-p=0


and same for equating for cos x?

Thanks
Olly

 

[Mark44]

Homework Statement

(d^2y/dx^2) + (dy/dx) = cos x


So you have the trial solution
y= p*cos(x)+q*sin(x)
(dy/dx) = -p*sin(x)+q*cos(x)
(d^2y/dx^2)=-p*cos(x)-q*sin(x)

It's much simpler to write the 2nd and 3rd equations as
y' = -p * sin(x) + q * cos(x)
y'' = -p * cos(x) - q * sin(x)

The issue I am having is equating the coefficients when after I have subbed them into the initial equation:
-p*cos(x)-q*sin(x)-p*sin(x)+q*cos(x)=cos(x)



Equating coefficients of sin x gives -q-p=0
Equating coefficients ofcos x gives -p+q=1
My question is why is
-q*sin(x)-p*sin(x)=cos(x)
= q-p=0

Because the equation $(-p + q)\cos(x) + (-q - p)\sin(x) = \cos(x)$ has to be identically true; i.e., true independent of the values of x.
So for the equation to be true for all values of x, it must be the case that the coefficient of cos(x) on the left side has to be 1, and the coefficient of sin(x) on the left side has to be 0, since sin(x) doesn't appear on the right side.
It's also true (and germane to my reasoning here) that the sin and cos functions are linearly independent. That is, neither one is a constant multiple of the other.

and same for equating for cos x?

Thanks
Olly

 

[Charles Link]

Homework Statement

(d^2y/dx^2) + (dy/dx) = cos x


So you have the trial solution
y= p*cos(x)+q*sin(x)

.
Don't forget the possibility of solutions to the homogeneous equation, where instead of $\cos x$, the right side of the differential equation is equal to zero. The complete solution consists of the particular solution, (which you found), plus any homogeneous solution.