Answer to Question #244970 in Differential Equations for sam

Question

5. Find the general solution to the differential equation y'' + y= sin² x

Answer 1

Solution.

$$y''+y=\sin^2(x)$$

1) Solve linear equation with constant coefficients

$y''+y=0.$

Сompose a characteristic equation $\lambda^2+1=0.$

From here $\lambda=i,$ or $\lambda=-i.$

So, we have

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

2) Find particular solution by the method of undefined coefficients. Such as $\sin^2(x)=\frac{1-\cos{2x}}{2},$ particular solution for the right side equal to the sum of particular solutions for $\frac{1}{2}$ and $-\frac{\cos(2x)}{2}.$

Solution for $\frac{1}{2}$ is:

$y=A.$ Then $y''=0.$ Substitute in original equation $A=\frac{1}{2}.$ So,

$$y=\frac{1}{2}.$$

Solution for $-\frac{\cos(2x)}{2}$ is:

$y=B\sin(2x)+D\cos(2x).$ Then $y''=-4B\sin(2x)-4D\cos(2x).$ Substitute in original equation $-6B\sin(2x)-6D\cos(2x)=-\cos(2x).$ From here $B=0, D=\frac{1}{6}.$ So,

$$y=\frac{\cos(2x)}{6}.$$

3)

$$y=C_1\sin x+C_2\cos x+\frac{1}{2}+\frac{\cos(2x)}{6}.$$

Answer. $y=C_1\sin x+C_2\cos x+\frac{1}{2}+\frac{\cos(2x)}{6}.$