kindly receive the complete solution ; if any query , please ask me.RegardsPallvee

Question 2:

1
y(t )  0 ; y(0)  10
2
y (t ) 1

y (t ) 6

3 y (t ) 

Integrating both sides, with respect to 't ' , we have

ln  y (t )  

t
 C...............(1)
6

y (t )  e t / 6C
When t  0 ; y(0)  10

10  e C ; C  ln(10)....................(2)
From equation (1) and (2), we have

ln( y (t )) 

t
 ln(10)
6

y (t )  10e t / 6

Question 4:

y (t )  0.5 y (t )  0
y (t )  0.5 y (t )

It is homogenous second order differential equation.
Characteristic equation is:
2  0.5  0
1 1

,
2 2
Thus, general solution is:
y(t )  Ae



t
2

t

 Be

2

 Ae



t
2

t

 Be

2

Given y(0)  1
y(0)  A  B  1...............(2)
Given y' (0)  1

y (0)  

A
2



B
2

 1................(3)

.............................(1)

From equation (2) and (3), we have
B

1 2
and A 
2

2 1
2

Thus, solution is:

y (t ) 





2  1  2 (1  2 ) 2
e 
e
2
2
t

t

Question 6:
y (t )  0.5 y ' (t )  0
y (t )  0.5 y ' (t )

It is homogenous second order differential equation.
Characteristic equation is:
2  0.5  0
1
  0,
2
Thus, general solution is:
t
2

t
2

y(t )  Ae  Be  Ae  Be .............................(1)
0

0

Given y(0)  1
y(0)  A  B  1...............(2)
Given y' (0)  1

y (0) 

B
 1................(3)
2

From equation (2) and (3), we have
B  2...