Answer to Question #295410 in Differential Equations for Moni

Question

Find a particular solution of y''+y=sinx, using the method of undetermined coefficients

Answer 1

From $\displaystyle y^{\prime\prime}+y=\sin x$, the auxiliary equation is;

$m^2+1=0\\ \Rightarrow m=\pm i$

Thus, the complementary function is;

$\displaystyle y_c=c_1\cos x+c_2 \sin x$ where $c_1$ and $c_2$ are arbitrary constant.

Next, to obtain the particular integral using method of undetermined coefficients.

Let $\displaystyle y_p=ax\cos x+bx\sin x$, then

$\displaystyle y_p^{\prime\prime}=2b\cos x-2a\sin x-ax\cos x-bx\cos x$

Substituting $y_p$ and $y_p^{\prime\prime}$ into the given DE yields;

$\displaystyle (2b\cos x-2a\sin x-ax\cos x-bx\cos x)+(ax\cos x+bx\sin x)=\sin x\\ \Rightarrow 2b\cos x-2a\sin x=\sin x$

comparing yields;

$\displaystyle b=0,$ and $\displaystyle a=-\frac{1}{2}$

Thus, $\displaystyle y_p=ax\cos x+bx\sin x=-\frac{x\cos x}{2}+0\times x\sin x=-\frac{x\cos x}{2}$

Hence, the general solution is;

$\displaystyle y=y_c+y_p=c_1\cos x+c_2 \sin x-\frac{x\cos x}{2}$