Respuesta :
Answer:
(1)
Therefore the complete solution is
[tex]y= cos2x+\frac{11}{12}sin2x+\frac13 sinx[/tex]
(2)
Therefore the complete solution is
[tex]y= cos2x+2sin2x[/tex]
Step-by-step explanation:
Given differential equation is
y''+4y = g(x) is associated with y''+4y=0.
The auxiliary equation is
m²+4=0
⇒m²= -4
[tex]\Rightarrow m= \pm 2i[/tex]
The complimentary function is
[tex]y_c=Acos 2x+Bsin 2x[/tex]
Case 1:
[tex]0\le x \le \frac\pi2[/tex]
g(x)= sinx
Then y''+4 y= sin x
F(D)≡ (D²+1) [Here [tex]D=\frac{d}{dx}[/tex] ]
Therefore
[tex]y_p=\frac{1}{D^2+4} (sin\ x)[/tex]
[tex]=\frac{1}{-1^2+4} sin x[/tex] [ replace D² by -1² and F(D)≠0]
[tex]=\frac{1}{3}sin x[/tex]
The complete solution is
[tex]y=A cos 2x+Bsin2 x+\frac13sinx[/tex]
Differentiating with respect to x
[tex]y'=-2Asin2x+2Bcos2 x+\frac13cosx[/tex]
The initial condition are y(0) =1 and y'(0)=4
[tex]1=y(0)= A cos(2. 0) +B sin(2. 0)+\frac 13sin0[/tex]
[tex]\Rightarrow 1=A[/tex]
and
[tex]4=y'(0)=-2Asin(2.0)+2Bcos(2. 0)+\frac13cos0[/tex]
[tex]\Rightarrow 2B+\frac13= 4[/tex]
[tex]\Rightarrow 2B= 4-\frac13[/tex]
[tex]\Rightarrow B= \frac{11}{12}[/tex]
Therefore the complete solution is
[tex]y= cos2x+\frac{11}{12}sin2x+\frac13 sinx[/tex]
Case 2:
[tex]x>\frac\pi2[/tex]
g(x)=0
Then [tex]y_p=0[/tex]
The complete solution is
[tex]y=A cos 2x+Bsin2 x[/tex]
Differentiating with respect to x
[tex]y'=-2Asin2x+2Bcos2 x[/tex]
The initial condition are y(0) =1 and y'(0)=4
[tex]1=y(0)= A cos(2. 0) +B sin(2. 0)[/tex]
[tex]\Rightarrow 1=A[/tex]
and
[tex]4=y'(0)=-2Asin(2.0)+2Bcos(2. 0)[/tex]
[tex]\Rightarrow 2B= 4[/tex]
⇒B=2
Therefore the complete solution is
[tex]y= cos2x+2sin2x[/tex]
