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

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

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