Question #295410

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

1
Expert's answer
2022-02-10T15:22:15-0500


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

m2+1=0m=±im^2+1=0\\ \Rightarrow m=\pm i

Thus, the complementary function is;

yc=c1cosx+c2sinx\displaystyle y_c=c_1\cos x+c_2 \sin x where c1c_1 and c2c_2 are arbitrary constant.

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

Let yp=axcosx+bxsinx\displaystyle y_p=ax\cos x+bx\sin x, then

yp=2bcosx2asinxaxcosxbxcosx\displaystyle y_p^{\prime\prime}=2b\cos x-2a\sin x-ax\cos x-bx\cos x

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

(2bcosx2asinxaxcosxbxcosx)+(axcosx+bxsinx)=sinx2bcosx2asinx=sinx\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;

b=0,\displaystyle b=0, and a=12\displaystyle a=-\frac{1}{2}

Thus, yp=axcosx+bxsinx=xcosx2+0×xsinx=xcosx2\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;

y=yc+yp=c1cosx+c2sinxxcosx2\displaystyle y=y_c+y_p=c_1\cos x+c_2 \sin x-\frac{x\cos x}{2}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS