Answer to Question #195121 in Differential Equations for fhumu

Question

d^2y/dx^2 + 2dy/dx +5y=34sinxcosx

Answer 1

First we solve the related homogeneous equation 

$y''+2y'+5=0$

The characteristic equation is

$r^2+2r+5=0$

$r_1=-1-2i, r_2=-1+2i$

Hence, the general solution of the homogeneous equation is given by

$y_0(x)=C_1e^{-x}\cos 2x+C_2e^{-x}\sin 2x$

where $C_1, C_2$  are constant numbers.

Let’s go back to the nonhomogeneous equation

$y''+2y'+5y=34\sin x \cos x$

$y''+2y'+5=17\sin 2x$

Find a particular solution of the nonhomogeneous differential equation.

$y_1=(Ax+B)\cos 2x+(Cx+D)\sin 2x$

The derivatives are given by

$y_1'=A\cos 2x-2(Ax+B)\sin 2x$

$+C\sin 2x+2(Cx+D)\cos 2x$

$y_1''=-2A\sin 2x-2A\sin 2x-4(Ax+B)\cos 2x$

$+2C\cos 2x+2C\cos 2x-4(Cx+D)\sin 2x$

Substituting the function $y_1$ and its derivatives in the differential equation yields:

$-2A\sin 2x-2A\sin 2x-4(Ax+B)\cos 2x$

$+2C\cos 2x+2C\cos 2x-4(Cx+D)\sin 2x$

$+2A\cos 2x-4(Ax+B)\sin 2x$

$+2C\sin 2x+4(Cx+D)\cos 2x$

$+5(Ax+B)\cos 2x+5(Cx+D)\sin 2x=17\sin 2x$

$-4A-4Cx-4D-4Ax-4B+2C=17$

$-4Ax-4B+4C+2A+4Cx+4D=0$

$-4A-4C=0$

$-4A-4B+2C-4D=17$

$-4A+4C=0$

$2A-4B+4C+4D=0$

$A=C=0$

$B=D$

$-8B=17$

Therefore we will look for a particular solution of the form

$y_1=-\dfrac{17}{8}\cos 2x-\dfrac{17}{8}\sin 2x$

Now we can write the full solution of the nonhomogeneous equation:

$y=C_1e^{-x}\cos 2x+C_2e^{-x}\sin 2x$

$-\dfrac{17}{8}\cos 2x-\dfrac{17}{8}\sin 2x$