Differential Equations : Solution Curves

mehtamonica
Messages
26
Reaction score
0
I have to solve the differential equation (y')^2= 4y to verify the general solution curves and singular solution curves.
Determine the points (a,b) in the plane for which the initial value problem (y')^2= 4y, y(a)= b has
(a) no solution ,
(b) infinitely many solutions (that are defined for all values of x )
(c) on some neighborhood of the point x=a , only finitely many solutions.

general solution that i am getting is y (x) = (x-c)^2 and singular solution is y(x)=0.

I am able to get part (a), as if b < 0, the problem has no solution.

Please help me figure out (b) and (c) .
 
Physics news on Phys.org
Think about a function g(x) defined piecewise with g(x) = 0 for x < c and g(x) = (x-c)2 if x ≥ c.
 
For (c) consider the situation when b= 0.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top