Solve. y^(′′) − 4y′ + 4y = sin2x


Question and Answers Forum

All Questions Topic List
Differential Equation Questions
Previous in All Question Next in All Question
Previous in Differential Equation Next in Differential Equation
Question Number 6457 by sanusihammed last updated on 27/Jun/16

$S o l v e .$ $y^{^{″}} - 4 y^{'} + 4 y = s i n 2 x$
	Answered by nburiburu last updated on 27/Jun/16
	$homogeneus solution: r^2 −4r+4=0 ⇒r_1 =r_2 =2 ⇒B_h ={e^(2x) , xe^(2x) } y_h =c_1 e^(2x) +c_2 xe^(2x) particular: g(x)=sin2x ⇒B_p ={cos2x,sin2x} y_p =Acos2x +Bsin2x y_p ′=−2Asin2x+2Bcos2x y_p ′′=−4Acos2x−4Bsin2x seeing cos2x and sin2x separately −4A −4(2B)+4(A)=0 −4B −4(−2A)+4(B)=1 B=0 ; A=(1/8) y_p =(1/8)cos 2x y= y_h +y_p =c_1 e^(2x) +c_2 xe^(2x) +(1/8)cos 2x$
	$h o m o g e n e u s s o l u t i o n :$ $r^{2} - 4 r + 4 = 0 \Rightarrow r_{1} = r_{2} = 2 \Rightarrow B_{h} = {e^{2 x}, {x e}^{2 x}}$ $y_{h} = c_{1} e^{2 x} + c_{2} {x e}^{2 x}$ $p a r t i c u l a r :$ $g (x) = s i n 2 x \Rightarrow B_{p} = {c o s 2 x, s i n 2 x}$ $y_{p} = A c o s 2 x + B s i n 2 x$ $y_{p}^{'} = - 2 A s i n 2 x + 2 B c o s 2 x$ $y_{p}^{″} = - 4 A c o s 2 x - 4 B s i n 2 x$ $s e e i n g c o s 2 x a n d s i n 2 x s e p a r a t e l y$ $- 4 A - 4 (2 B) + 4 (A) = 0$ $- 4 B - 4 (- 2 A) + 4 (B) = 1$ $B = 0; A = \frac{1}{8}$ $y_{p} = \frac{1}{8} c o s 2 x$ $y = y_{h} + y_{p} = c_{1} e^{2 x} + c_{2} {x e}^{2 x} + \frac{1}{8} c o s 2 x$
	Commented by sanusihammed last updated on 27/Jun/16

	$W o w t h a n k s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum