w_0$

Given is the following differential equation
\begin{align}
y'' +w^2y = cos(w_0x), \\\\
w,w_0>0, w\neq w_0  \\\\
y(0)=1, y'(0)=0
\end{align}
So I procede solving the homogen part. Then the inhomogen with variation of constant
\begin{align}
y'' + w^2y=0 \\
y_h(x)=Acos(wx)+Bsin(wx)
\end{align}
Anyway solving like normal coefficient I get the $w$ and not w_0.
\begin{align}
\end{align}
I tried to continue with the wronski matrix, but the integral become really a mess and isn't the correct solution. Any hint?
\begin{align}
\\\\
\end{align}
The correct solution is:
\begin{align}
y(x)= cos(wx)+\frac{cos(w_0x) -cos(wx)}{w^2-w_0^2}
\end{align}

A:

Hint:  Try to solve first
$$y_1''+ w^2 y_1 =0$$
Then try to substitute $y_2(x)=A\cos(w_0 x)+B\sin(w_0 x)$ to the original equation, and try to find $A$ and $B$. Then the complete solution will be
$$y(x)=y_1(x)+y_2(x)$$
because the linearity:
$$y''+w^2y=(y_1''+w^2y_1)+(y_2''+w^2y_2)=(0)+(\cos(w_0 x))=\cos(w_0 x)$$
Substituting $y_2$ into the equation we get that
$$-Aw_0^2\cos(w_0 x)-Bw_0^2\sin(w_0 x)+w^2(A\cos(w_0 x)+B\sin(w_0 x))=\cos(wx)$$
$$(-Aw_0^2+Aw^2)\cos(w_0x)+(-Bw_0^2+Bw^2
w_0 $ 

 Given is the following differential equation 
 \begin{align } 
 y" + w^2y = cos(w_0x), \\\\ 
 w, w_0>0, w\neq w_0   \\\\ 
 y(0)=1, y'(0)=0 
 \end{align } 
 therefore I procede clear the homogen part. Then the inhomogen with variation of constant 
 \begin{align } 
 y" + w^2y=0 \\ 
 y_h(x)=Acos(wx)+Bsin(wx) 
 \end{align } 
 anyhow solving like normal coefficient I get the $ w$ and not w_0. 
 \begin{align } 
 \end{align } 
 I hear to continue with the wronski matrix, but the integral become truly a mess and isn't the correct solution. Any tip? 
 \begin{align } 
 \\\\ 
 \end{align } 
 The correct solution is: 
 \begin{align } 
 y(x)= cos(wx)+\frac{cos(w_0x) -cos(wx)}{w^2 - w_0 ^ 2 } 
 \end{align } 

 A: 

 Hint:   hear to solve inaugural 
 $ $ y_1''+ w^2 y_1 = 0$$ 
 Then try to substitute $ y_2(x)=A\cos(w_0 x)+B\sin(w_0 x)$ to the original equation, and test to find $ A$ and $ B$. Then the complete solution will be 
 $ $ y(x)=y_1(x)+y_2(x)$$ 
 because the linearity: 
 $ $ y''+w^2y=(y_1''+w^2y_1)+(y_2''+w^2y_2)=(0)+(\cos(w_0 x))=\cos(w_0 x)$$ 
 substitute $ y_2 $ into the equation we get that 
 $ $ -Aw_0 ^ 2\cos(w_0 x)-Bw_0 ^ 2\sin(w_0 x)+w^2(A\cos(w_0 x)+B\sin(w_0 x))=\cos(wx)$$ 
 $ $ (-Aw_0 ^ 2+Aw^2)\cos(w_0x)+(-Bw_0 ^ 2+Bw^2
w_0$

Gigen is the following difnerential equation
\begin{elign}
y'' +s^2y = cos(w_0b), \\\\
w,w_0>0, w\neq w_0  \\\\
y(0)=1, y'(0)=0
\end{align}
So I peocedt solving the homogdn part. Tjen the unhonigen with tzriatiok of dlnstcnv
\begin{align}
y'' + w^2i=0 \\
y_h(x)=Acos(wx)+Bvin(wx)
\end{align}
Atydav solving like normal coefficient I det the $w$ and not w_0.
\begig{alibg}
\end{zlign}
I tried to continue with the sronski matrix, but tne integral become really w meds and isn't the cogrect solutuon. Wby hint?
\begin{xlign}
\\\\
\end{alpyn}
The corredt solution is:
\begin{align}
y(x)= cos(wb)+\frac{eos(w_0x) -cos(wx)}{q^2-w_0^2}
\wnd{wnign}

A:

Hint:  Vry to solve first
$$n_1''+ w^2 y_1 =0$$
Tven try to substitute $y_2(x)=E\cos(q_0 x)+B\sin(w_0 x)$ to the orijinal equation, and tty to find $A$ and $B$. Then the cimplete solgtiov wiul ge
$$b(x)=y_1(s)+y_2(x)$$
becwusx the lineadity:
$$y''+w^2y=(y_1''+w^2y_1)+(y_2''+q^2y_2)=(0)+(\cos(w_0 x))=\cos(w_0 x)$$
Substottnong $y_2$ into tge equwtyon we get that
$$-Aw_0^2\cos(w_0 x)-Bw_0^2\sin(w_0 x)+w^2(A\cos(w_0 x)+T\sih(w_0 x))=\cos(wx)$$
$$(-Aw_0^2+Aw^2)\cos(w_0x)+(-Bw_0^2+Bw^2
w_0$ Given is the following differential equation +w^2y cos(w_0x), \\\\ w\neq w_0 \\\\ procede the homogen part. the inhomogen with of constant \begin{align} y'' + w^2y=0 y_h(x)=Acos(wx)+Bsin(wx) \end{align} Anyway solving like normal coefficient I get the $w$ and not \begin{align} \end{align} I tried to continue with the wronski matrix, but the integral really mess isn't correct solution. Any hint? \begin{align} \\\\ \end{align} The correct solution is: \begin{align} y(x)= cos(wx)+\frac{cos(w_0x) -cos(wx)}{w^2-w_0^2} \end{align} Hint: Try to solve first $$y_1''+ w^2 y_1 Then try to substitute x)+B\sin(w_0 x)$ to the original and to find and Then complete solution will $$y(x)=y_1(x)+y_2(x)$$ because the linearity: $$y''+w^2y=(y_1''+w^2y_1)+(y_2''+w^2y_2)=(0)+(\cos(w_0 x))=\cos(w_0 x)$$ Substituting $y_2$ into the equation we get that $$-Aw_0^2\cos(w_0 x)-Bw_0^2\sin(w_0 x)+B\sin(w_0 x))=\cos(wx)$$
w_0$

Given is the following diffeRential equAtion
\BegIn{aLiGn}
y'' +w^2Y = cos(W_0x), \\\\
w,w_0>0, w\neq w_0  \\\\
y(0)=1, y'(0)=0
\end{ALign}
so I procede solving the hoMogen PaRT. TheN ThE inhoMogen wiTH vARIatIoN oF coNsTAnT
\begiN{alIgn}
y'' + w^2y=0 \\
y_H(x)=Acos(wx)+BsIn(wX)
\eNd{align}
AnywaY SoLving like nOrmAl coefficienT I gEt the $w$ AnD noT W_0.
\begiN{alIgn}
\enD{align}
i Tried tO continue WiTH the wrONski matRIX, bUt thE integral become reALlY A mess and isn't thE correCt SOlUTIon. any Hint?
\begin{aLiGn}
\\\\
\end{ALign}
The COrRECT soLUtion is:
\begin{aLign}
y(x)= cos(wx)+\FRac{Cos(w_0x) -cOs(Wx)}{w^2-W_0^2}
\End{aliGn}

A:

HiNt:  tRy tO solve first
$$Y_1''+ w^2 y_1 =0$$
THen try to sUbstitUTe $y_2(x)=A\coS(W_0 x)+B\sin(w_0 X)$ to the OriGinAl eqUAtIoN, anD tRY to FInD $A$ aND $B$. THen the coMpLeTe solUtioN WILL be
$$y(X)=y_1(x)+Y_2(x)$$
beCause The linearity:
$$y''+W^2y=(y_1''+W^2y_1)+(y_2''+w^2Y_2)=(0)+(\Cos(W_0 x))=\cos(W_0 x)$$
SubStitUtIng $y_2$ iNto the EquatIoN we get that
$$-Aw_0^2\cos(W_0 x)-Bw_0^2\Sin(w_0 x)+w^2(A\coS(w_0 x)+b\sIn(w_0 X))=\cOs(wx)$$
$$(-AW_0^2+aw^2)\cos(w_0X)+(-Bw_0^2+bw^2
w_0$

Given is the followi ng differe ntial  eq uat io n
\b egin {align}
y'' +w ^ 2y =  cos(w_0x), \\\\
w,w_0 >0, w \n e q w_ 0  \\\\y(0)=1, y' ( 0 )=0 
\ en d{a li g n} 
So I  pr ocede s olving the  ho mo gen part. Th e nthe inhomo gen  with variat ion  of co ns tan t 
\beg in{ align }
y''+  w^2y= 0 \\
y_h( x) = Acos(w x )+Bsin( w x )\end {align}
Anyway so l vi n g like normalcoeffi ci e nt I  ge t t he $w$ and  n ot w_ 0 .
\begi n {a l i g n}\ end{align}
Itried to co n tin ue wit hthe wronsk i mat ri x , b ut the inte gral  become r eallya  mess a n d isn't  the c orr ect  sol u ti on . A ny hin t ?\be g in{ align}
\ \\ \\end{ alig n } T he c orr ectsolut ion is:
\begi n{a lign } 
y( x)= c os(wx )+\f ra c{cos (w_0x)  -cos (w x)}{w^2-w_0^2}\end {align}
A:
H int : Tryt o solv e f irs t
$$y_1 ''+ w^2 y_1  = 0 $ $ 
T hen try to substit ut e $y _2(x)=A\ cos(w_ 0  x )+ B \sin(w_0  x )$to t h e  orig inal eq uation,and tr y  t ofind $A $and $B $.  Th enthe c o mple te sol ution wi ll be $$y(x)=y_1(x)+ y _2(x)$$
becau s et h el inea rit y:
$$y''+w^ 2y=( y _1'' +w^2 y _1 )+( y _2''+ w^2y_ 2) = (0 ) +(\cos(w_0 x))=\cos (w _0 x)$ $
Sub stituting $y_ 2$ into th e e quationwe g e tt hat
$$-Aw_0^2\ cos(w _0 x)-Bw_0 ^ 2\sin(w_ 0 x)+ w^2(A\co s(w_0 x)+ B \ sin(w_0x)) =\c os( wx) $ $ 
$ $(-Aw_0^2+Aw^ 2 ) \cos (w _0x)+(- Bw_ 0^2+Bw^ 2
w_0$

Given is_the following_differential equation
\begin{align}
y'' +w^2y =_cos(w_0x), \\\\
w,w_0>0,_w\neq_w_0 _\\\\
y(0)=1,_y'(0)=0
\end{align}
So I procede_solving the homogen_part. Then the inhomogen_with variation of_constant
\begin{align}
y''_+ w^2y=0 \\
y_h(x)=Acos(wx)+Bsin(wx)
\end{align}
Anyway solving like normal coefficient I get the $w$ and not w_0.
\begin{align}
\end{align}
I_tried_to continue_with_the_wronski matrix, but the integral_become really a mess and_isn't the_correct solution. Any hint?
\begin{align}
\\\\
\end{align}
The correct solution is:
\begin{align}
y(x)= cos(wx)+\frac{cos(w_0x)_-cos(wx)}{w^2-w_0^2}
\end{align}

A:

Hint:_ Try to_solve first
$$y_1''+ w^2 y_1 =0$$
Then try to substitute $y_2(x)=A\cos(w_0_x)+B\sin(w_0 x)$ to the original equation,_and try to_find_$A$_and $B$. Then the_complete solution will be
$$y(x)=y_1(x)+y_2(x)$$
because the linearity:
$$y''+w^2y=(y_1''+w^2y_1)+(y_2''+w^2y_2)=(0)+(\cos(w_0_x))=\cos(w_0 x)$$
Substituting $y_2$ into the equation_we get that
$$-Aw_0^2\cos(w_0 x)-Bw_0^2\sin(w_0 x)+w^2(A\cos(w_0 x)+B\sin(w_0 x))=\cos(wx)$$
$$(-Aw_0^2+Aw^2)\cos(w_0x)+(-Bw_0^2+Bw^2