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Ans 1)
f ͼ C(s)
S = [ξ-a, ξ+a] x R
S satisfies Lipscitz Condition with respect to y in
S : |f(x,y) – f(x,z)| <= L|y-z|
Let y’ = f(t,x), for ξ<=x<= ξ+a, y(ξ)= η
Let C(J), J = [ξ-a, ξ+a]
||y|| = maxxͼJ |y(x) e-α|x-ξ||
According to Lipscitz Condition;
|f(t,y1) – f(t,y2)| <= L|y1 – y2|
Whenever (t, y1), (t,y2) are in D and D ͼ R2
Lipscitz condition is sufficient for uniqueness
||y|| = max|y(x) e- α|x-ξ| |
If e-α|x- ξ| = 0 then ||y|| can be maximized.
Now this is a modulus function, so x- ξ is always positive.
If x= ξ then e- α|x-ξ| = maximum
So, ||y|| = y(x)
|y’| = y’(x) and y(ξ) = η from equation that we got from S;
So we can say that in interval of J there exist an unique solution to this IVP.
Ans 2. Dy/dx=y 2
y 2 dy=dx
y 3 /3= x + c
If
y(0)=1

1/3=c
y 3 /3= x + 1/3
y 3 = 3x + 1
x=(y 3 -1)/3 &lt;= a
y-1 = cube root(3x+1) -1 &lt;= b
Solution exits on interval (y 3 -1)/3 to cube root(3x+1) -1.
Ans3)
f(x,y) satisfies a local Lipscitz condition; wrt y
D ϲ= R;
If A ϲ D is compact and f bounded on A then f satisfies a Lipscitz condition
w.r.t. y in A.
➔ f(x,y) is bounded on A; - (1)
➔ A c D -(2)
➔ And f(x,y) satisfies Lipscitz condition w.r.t. y in D where D c= R ----(3)
From analysing these three equations in order 3>1>2 then we can say that
actual domain of f(x,y) lies in A while satisfying the Lipscitz equation.
Ans 4)
Lipscitz condition is :
|f(x,y) – f(x,z)| <= L(y-z) w.r.t. y;
v ϵ ([a,b]) and w ϵ ([a,b]) and x ϵ [a,b];
If we want to establish the relation w.r.t. x then x <= [a,b]
V <= C[a,b] and w<= [a,b]
So Lipscitz condition is

|f(x, v(x)) – f(x, w(x))| <= L|v(x) – w(x)|
Ans 5. y = xy + 2/6 ( 1+y 2 ) 3/2 + c
y(0) = 1
1 = 0 + 2/6 (1) + c
3 = c
y = xy + 2/6 ( 1+y 2 ) 3/2 + 3 Lower Solution
y = xy + 2/6 (( 1+y 2 ) 3/2 )/y + c
y(0) = 1
y= 0 + 2/6 +c
y = xy + 2/6 (( 1+y 2 ) 3/2 )/y + 3 Upper Solution