Question

giải phương trình y′′ − 7y′ + 6y = sin x.

Answer 1

Start with the underlying homogeneous equation:

$y''-7y'+6y=0$

which has characteristic equation

$r^2-7r+6=(r-6)(r-1)=0$

with roots at r = 6 and r = 1. So the characteristic solution is

$y_c=C_1e^{6x}+C_2e^x$

Now for the particular solution, we can use the method of undetermined coefficients, with the following ansatz (the "guess" solution) and its derivatives,

$y_p=a\cos x+b\sin x$

${y_p}'=-a\sin x+b\cos x$

${y_p}''=-a\cos x-b\sin x$

Substituting these into the original equation gives

$(-a\cos x-b\sin x)-7(-a\sin x+b\cos x)+6(a\cos x+b\sin x)=\sin x$

$(5a-7b)\cos x+(7a+5b)\sin x=\sin x$

$\implies\begin{cases}5a-7b=0\\7a+5b=1\end{cases}\implies a=\dfrac7{74},b=\dfrac5{74}$

So the particular solution is

$y_p=\dfrac7{74}\cos x+\dfrac5{74}\sin x$

and hence the general solution is

$y=y_c+y+p=\boxed{C_1e^{6x}+C_2e^x+\dfrac7{74}\cos x+\dfrac5{74}\sin x}$