Answer to Question #243501 in Differential Equations for Micau

Question

Find the general solution of the differential equations using method of unndetermined coefficients (MUC) and method of variation of parameters (MVP) .

y" +y = 2x sin x

Answer 1

1. Homogeneous equation

$$y''+y=0$$

Corresponding (auxiliary) equation

$$r^2+r=0$$

$$r=\pm i$$

The general solution of the homogeneous differential equation is

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

Find the particular solution of the non homogeneous differential equation

$$y_p=(Ax^2+Bx+C)\cos x+(Dx^2+Ex+F)\sin x$$

$$y_p'=-(Ax^2+Bx+C)\sin x+(2Ax+B)\cos x$$

$$+(Dx^2+Ex+F)\cos x+(2Dx+E)\sin x$$

$$y_p''=-(Ax^2+Bx+C)\cos x-2(2Ax+B)\sin x$$

$$+2A\cos x-(Dx^2+Ex+F)\sin x$$

$$+2(2Dx+E)\cos x+2D\sin x$$

Substitute

$$-(Ax^2+Bx+C)\cos x-2(2Ax+B)\sin x$$ $$+2A\cos x-(Dx^2+Ex+F)\sin x$$ $$+2(2Dx+E)\cos x+2D\sin x$$ $$+(Ax^2+Bx+C)\cos x+(Dx^2+Ex+F)\sin x$$ $$=2x\sin x$$

$\sin x:$

$$-4Ax-2B-Dx^2-Ex-F+2D$$

$$+Dx^2+Ex+F=2x$$

$\cos x:$

$$-Ax^2-Bx-C+2A+4Dx+2E$$

$$+Ax^2+Bx+C=0$$

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

$$y_p=-\dfrac{1}{2}x^2\cos x+\dfrac{1}{2}x\sin x$$

The general solution of the non homogeneous differential equation is

$$y=y_h+y_p$$

$$y=c_1\cos x+c_2\sin x-\dfrac{1}{2}x^2\cos x+\dfrac{1}{2}x\sin x$$

2. Homogeneous equation

$$y''+y=0$$

Corresponding (auxiliary) equation

$$r^2+r=0$$

$$r=\pm i$$

Find the general solution of the non homogeneous differential equation in form

$$y=C_1\cos x+C_2\sin x$$

$$y'=-C_1\sin x+C_2\cos x+C_1'\cos x+C_2'\sin x$$

Let

$$C_1'\cos x+C_2'\sin x=0$$

Then

$$y'=-C_1\sin x+C_2\cos x$$

$$y''=-C_1\cos x-C_2\sin x-C_1' \sin x+C_2'\cos x$$

Substitute

$$-C_1\cos x-C_2\sin x-C_1' \sin x+C_2'\cos x$$

$$+C_1\cos x+C_2\sin x=2x\sin x$$

$$C_1'=-C_2'\tan x$$

$$C_2' \tan x\sin x+C_2'\cos x=2x\sin x$$

$$C_2' \sin^2 x+C_2'\cos^2 x=2x\sin x\cos x$$

$$C_2'=2x\sin x\cos x$$

$$C_2=\int2x\sin x\cos xdx$$

$$\int2x\sin x\cos xdx$$

$$\int udv=uv-\int vdu$$

$$u=x, du=dx$$

$$dv=2\sin x\cos xdx, v=\int2\sin x\cos xdx$$

$$=\int\sin (2x)dx=-\dfrac{1}{2}\cos(2x)$$

$$\int2x\sin x\cos xdx=-\dfrac{1}{2}x\cos(2x)+\int\dfrac{1}{2}\cos(2x)dx$$

$$=-\dfrac{1}{2}x\cos(2x)+\dfrac{1}{4}\sin(2x)+C_3$$

$$C_2=-\dfrac{1}{2}x\cos(2x)+\dfrac{1}{4}\sin(2x)+C_3$$

$$C_1'=-2x\sin x\cos x\tan x=-2x\sin^2x$$

$$=-x(1-\cos (2x)=x\cos (2x)-x$$

$$C_1=\int(x\cos (2x)-x)dx$$

$$\int x\cos (2x)dx$$

$$\int udv=uv-\int vdu$$

$$u=x, du=dx$$

$$dv=\cos (2x)dx, v=\int\cos(2x)dx=\dfrac{1}{2}\sin(2x)$$

$$\int x\cos (2x)dx=\dfrac{1}{2}x\sin(2x)-\int\dfrac{1}{2}\sin(2x)dx$$

$$=\dfrac{1}{2}x\sin(2x)+\dfrac{1}{4}\cos(2x)+C_4$$

$$C_1=\dfrac{1}{2}x\sin(2x)+\dfrac{1}{4}\cos(2x)-\dfrac{1}{2}x^2+C_4$$

$$y=(\dfrac{1}{2}x\sin(2x)+\dfrac{1}{4}\cos(2x)-\dfrac{1}{2}x^2+C_4)\cos x$$

$$+(-\dfrac{1}{2}x\cos(2x)+\dfrac{1}{4}\sin(2x)+C_3)\sin x$$

$$y=C_4\cos x+C_3\sin x-\dfrac{1}{2}x^2\cos x$$

$$+\dfrac{1}{2}x(\sin(2x)\cos x-\cos(2x)\sin x)$$

$$+\dfrac{1}{4}(\cos(2x)\cos x+\sin(2x)\sin x)$$ $$y=C_4\cos x+C_3\sin x-\dfrac{1}{2}x^2\cos x+\dfrac{1}{2}x+\dfrac{1}{4}\cos x$$

The general solution of the non homogeneous differential equation is

$$y=C_5\cos x+C_3\sin x-\dfrac{1}{2}x^2\cos x+\dfrac{1}{2}x$$