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Differential Equations Compute y’ and y’’ and then combine these derivatives with y as a linear second-order differential equation that is free of the symbols c1 and c2 and has the form F(y, y’,...

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Differential Equations Compute y’ and y’’ and then combine these derivatives with y as a linear second-order differential equation that is free of the symbols c1 and c2 and has the form F(y, y’, y’’)=0. The symbols c1 and c2 represent constants. Y= c1ex + c2xex Verify that the indicated function is a particular solution of the given differential equation. Give an interval of definition I for each solution.
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Differential Equations Compute y’ and y’’ and then combine these derivatives with y as a linear second-order differential equation that is free of the symbols c1 and c2 and has the form F(y, y’, y’’)=0. The symbols c1 and c2 represent constants. Y= c1ex + c2xex Verify that the indicated function is a particular solution of the given differential equation. Give an interval of definition I for each solution. y’’ + y = sexx; y=xsinx + (cosx)ln(cosx) Solve the differential equation (2x+y+1)y’ =1 Solve the given initial-value problem and give the largest interval I on which the solution is defined. dy/dt + 2(t+1)y2 = 0, y(0) = -1/8 According to Stefan’s law of radiation the absolute temperature T of a body cooling in a medium at constant absolute temperature Tm is given by dT/dt = k(T4 – T4m), solve the differential equation. Using the information provided in problem #5, show that when T-Tm is small in comparison to Tm then Newton’s law of cooling approximates Stefan’s law. Find the general solution of the differential equation. 2y’’ + 2y’ +3y =0 Solve the differential equation subject to the indicated conditions. Y’’ + 2y’ +y=0, y(-1) = 0, y’(0) = 0 Amass weighing 32 pounds stretches a spring 6 inches. The mass moves thought a medium offering a damping force that is numerically equal to ? times the instantaneous velocity. Determine the values of ? >0 for which the spring? Mass system will exhibit oscillatory motion. The vertical motion of a mass attached to a spring is describe by the IVP 1/4x’’ + x’ + x = 0, x(0)=4, x’(0)=2. Determine the maximum vertical displacement of the mass.

Answered Same Day Dec 29, 2021

Solution

Robert answered on Dec 29 2021
111 Votes
Sol: (1)
1 2The symbol and reptresent constants,C C
1 2 ..................... (1),
x xY C e C xe 
On taking a differentiation with respect to x ,

 1 2
1 2 2 ........................ (2),
x x x
x x x
Y C e C xe e
Y C e xe C C e
   
   

From equation (1) and (2), we get
2 ......................... (3),
xY Y C e  
On taking a differentiation with respect to x ,
2 .........................(4),
xY Y C e  
From equation (3), 2 =
xC e Y Y 
From equation (4),
2
2 0
Y Y Y Y
Y Y Y
Y Y Y
    
  
   
Sol: (2) The given differential equation is
secY Y x  
The particular solution is
   
   
   
2
1
. sec
1
1
sec
1 1 1
sec
2
1 1 1 1
. sec sec ............... (1),
2 2
P I x
D
x
D i D i
x
i D i D i
P I x x
i D i i D i



 
 
  
  
 
 

Now,
 
 
 
1
sec sec
cos sin
cos
1 tan
ln cos
ix ix
ix
ix
ix
x e e xdx
D i
x i x
e dx
x
e i x dx
e x i x




 
 




Changing to , we havei i

 
 
1
sec ln cosixx e x i x
D i
 


Putting these values in equation (1), we get
   
1 1
. ln cos ln cos
2 2
ln cos ln cos
2 2 2 2
ln cos
2 2
. sin cos ln cos
ix ix
ix ix
ix ix
ix ix ix ix
P I e x i x e x i x
i i
x e x x e x
e e
i i
e e e e
x x
i
P I x x x x



 
         
   
    
    
   
 

Verified.
 Interval: , 
Sol: (3) The given differential equation is
 2 1 1x y y   …………….. (1),
Let
   
   
   
2
2
2
v x y x x
dv x dy x
dx dx
dy x dv x
dx dx
 
 
 

From equation (1), we get
  
 
1 2 1
dv x
v x
dx
 
   
 

Solve fo
 dv x
dx
:
   
 
 
  
 
2 3
1
1
2 3
dv x v x
dx v x
dv x
v x
dx dx
v x







Integrate both sides with respect tov :
      1
1
log 2 3 2 3
4
v x v x x c     
Solve forv :
     141 3
2
x c
v x W e
 
   
Substitute back for     2y x v x x 
     141 4 3
2
x c
y x x W e
 
    
Sol: (4) The differential equation is
   2
1
2 1 0, 0 ,
8
dy
t y y
dt
    
 
 
2
2
2 1
2 1
dy
t y
dt
dy
t dt
y
  
 ...
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