Answer to Question #232366 in Differential Equations for Maloh

Question

How do we solve y''+y=cos x by variation of parameters method?

Answer 1

The homogeneous differential equation

$$y''+y=0$$

Characteristic (auxiliary) equation

$$r^2+1=0$$

$$r=\pm i$$

The general solution of the homogeneous differential equation is

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

$$W(y_1, y_2)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=\begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix}$$

$$=\cos^2x+\sin^2x=1$$

$$W_1=\begin{vmatrix} 0 & y_2 \\ \cos x& y_2' \end{vmatrix}=\begin{vmatrix} 0 & \sin x \\ \cos x & \cos x \end{vmatrix}=-\sin x\cos x$$

$$W_2=\begin{vmatrix} y_1 & 0 \\ y_1' & \cos x \end{vmatrix}=\begin{vmatrix} \cos x & 0 \\ -\sin x & \cos x \end{vmatrix}=\cos^2 x$$

$$C_1'=\dfrac{W_1}{W(y_1, y_2)}=\dfrac{-\sin x\cos x}{1}=-\sin x\cos x$$

$$C_1=\int(-\sin x\cos x)dx=-\dfrac{1}{2}\sin^2x+C_3$$

$$C_2'=\dfrac{W_2}{W(y_1, y_2)}=\dfrac{\cos^2 x}{1}=\cos^2x$$

$$C_2=\int(\cos^2 x)dx=\dfrac{1}{2}\int(1+\cos(2 x))dx$$

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

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

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

The general solution of the given differential equation is

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