What can you say about the graph of a solution of the equation y' = #(xy^3)/36# when x is close to 0?

Question

(a) What can you say about the graph of a solution of the equation y' = #(xy^3)/36# when x is close to 0? What if x is large?

If x is close to 0, then y' = #(xy^3)/36# is (choices are: large, close to 0, and close to 36) , and hence y' is (choices are: large, close to 0, and close to 36) Thus, the graph of y must have a tangent line that is nearly (choices are: horizontal, vertical). If x is large, then #(xy^3)/36# is (choices are: large, close to 0, and close to 36), and the graph of y must have a tangent line that is nearly (choices are: horizontal, vertical). (In both cases, we assume reasonable values for y.)

(b) Verify that all members of the family y = 6#(c - x^2)^(-1/2)# are solutions of the differential equation y' = #(xy^3)/36#.

y = 6#(c - x^2)^(-1/2)# => y' = (blank) #(c - x^2)^(-3/2)#.

RHS = #(xy^3)/36# = #(x[6(c - x^2)^(-1/2))]/36# = (blank) #(c - x^2)^(-3/2)# = y' = LHS.

(c) Graph several members of the family of solutions on a common screen.

or

Do the graphs confirm what you predicted in part (a)?

(A) |y'| gets close to 0 if either x gets close to 0 or x gets larger.

(B) As x gets close to 0, |y'| gets larger. As x gets larger, y' gets close to 0.

(C) y' gets close to 0 if either x gets close to 0 or x gets larger.

(D) When x is close to 0, y' is also close to 0. As x gets larger, so does |y'|.

(d) Find a solution of the initial-value problem.

y' = #(xy^3)/36#, y(0) = 12

y = ???

How can I do parts a, b, and c? I think I got the graphing part of c just fine, but I'm not sure. How can I do the rest?

Emily Ammerman · Accepted Answer

(a) Close to 0 / Close to 0 / Horizontal / Large / Vertical

(b) Put 6x for both boxes.

(c)

(d) y = #12(1 - 4x^2)^(-1/2)#

Emilia Aguilar · Answer

undefined When $ x $ is close to 0, the behavior of the graph of the solution of the equation $ y' = \frac{xy^3}{36} $ depends on the initial condition of the differential equation.

If the initial condition is such that $ y(0) = 0 $, then the solution will remain at $ y = 0 $ as $ x $ approaches 0.

If the initial condition is such that $ y(0) \neq 0 $, then the graph of the solution will approach a vertical asymptote as $ x $ approaches 0. This is because the denominator of the equation $ y' = \frac{xy^3}{36} $ becomes very small as $ x $ approaches 0, causing the derivative $ y' $ to become large. As a result, the graph of the solution will steeply increase or decrease, depending on the sign of $ y(0) $, near $ x = 0 $.

In summary, when $ x $ is close to 0:
- If $ y(0) = 0 $, the solution remains at $ y = 0 $.
- If $ y(0) \neq 0 $, the solution approaches a vertical asymptote.