Question

hi
are there any people can solve this ques.
y''+9y=2cos3x+3sinx forthe method of undetermined coefficients

Answer 1

how you got your complementary solution ,
i.e. the solution for
y''+9y=0
?

Answer 2

we have a second order differential equation of this form
\[ay''+by'+cy=0\]
if we let
\[y=e^{mx}\]\[y'=me^{mx}\]\[y''=m^2e^{mx}\]
and substitute
\[am^2e^{mx}+bme^{mx}+ce^{mx}=0\]
\[(am^2+bm+c)e^{mx}=0\]
We have a product of two terms
If the product of any two terms equals zero , one of the terms must be zero,
e^mx is not zero

we arrive at
\[am^2+bm+c=0\]
the auxiliary equation

but you dont have to go through this rigmarole every time

just jump from
\[ay''+by'+cy=0\]\[\qquad\Downarrow\qquad\Downarrow\]to the auxiliary equation
\[\qquad\Downarrow\qquad\Downarrow\]\[am^2+bm+c=0\]

then you have to find the \(m\)'s