Answer to Question #218091 in Differential Equations for Syed Hassan abdull

Question

(D^2+1)y=x^3+e^xcosx

Answer 1

$y''+y=x^3+e^x\cos x$

Homogeneous equation

$y''+y=0$

Characteristic equation

$r^2+1=0$

$r=\pm i$

The general solution of the homogeneous equation is

$y_h=c_1\cos x+c_2\sin x$

Find the partial solution of the nonhomogeneous equation in the form

$y_p=Ax^3+Bx^2+Cx+D+Ee^x\cos x+Fe^x\sin x$

$y_p'=3Ax^2+2Bx+C+Ee^x\cos x-Ee^x\sin x$

$+Fe^x\sin x+Fe^x\cos x$

$y_p''=6Ax+2B+Ee^x\cos x-Ee^x\sin x$

$-Ee^x\sin x-Ee^x\cos x+Fe^x\sin x+Fe^x\cos x$

$+Fe^x\cos x-Fe^x\sin x$

Substitute

$6Ax+2B-2Ee^x\sin x+2Fe^x\cos x$

$+Ax^3+Bx^2+Cx+D+Ee^x\cos x+Fe^x\sin x$

$=x^3+e^x\cos x$

$x^3: A=1$

$x^2: B=0$

$x^1: 6A+C=0$

$x^0: 2B+D=0$

$e^x\cos x:E+2F=1$

$e^x\sin x:-2E+F=0$

$A=1, B=0, C=-6, D=0, E=\dfrac{1}{5}, F=\dfrac{2}{5}$

$y_p=x^3-6xe^x+\dfrac{1}{5}\cos x+\dfrac{2}{5}e^x\sin x$

The general solution of the given nonhomogeneous equation is

$y=c_1\cos x+c_2\sin x$

$+x^3-6xe^x+\dfrac{1}{5}\cos x+\dfrac{2}{5}e^x\sin x$