Find the interval of validity for the solution, Mathematics

Assignment Help:

Solve the subsequent IVP and find the interval of validity for the solution

xyy' + 4x2 + y2 = 0,       y(2) = -7,          x > 0

Solution:

Let's first divide on both sides by x2 to rewrite the differential equation as given,

(y/x)y' = -4 - (y2/x2)= - 4 - (y/x)2

Here, it is not in the officially exact form as we have listed above, though we can see that everywhere the variables are listed they put in an appearance as the ratio, y/x and thus this is truly as far as we require to go. Therefore, let's plug the substitution in this form of the differential equation to find,

n (n+ x n') = - 4 - n2

Subsequently, rewrite the differential equation to determine everything separated out.

n x n' = - 4 - 2n2

x n' = - (4 + 2n2)/ n

n/(4 + 2n2) dv = - (1/x) dx

Integrating on both sides we find,

¼ In (4 2n2) = - In (x) + c

We require doing a little rewriting using fundamental logarithm properties in order to be capable to easily solve this for n.

In (4 2n2)¼ = In (x)-1 + c

Then exponentiates on both sides and do a little rewriting,

(4 + 2n2)¼ 

= eIn(x)-1 + c

446_Find the interval of validity for the solution.png

= c/x

Remember that as c is an unknown constant so next is ec and so we may also just call this c as we did above.

At last, let's solve for v and after that plug the substitution back in and we'll play a little fast and loose along with constants again.

4 + 2n2 = c4/x4 = c/x4

n2 = ½ ((c/x4)- 4)

y2/x2 = ½ ((c - x4)/x4)

y2 = ½ x2 ((c - x4)/x4)

y2 = (c - x4)/2x2

At this point this would probably be best to go in front and apply the initial condition. Doing this gives as,

49 = (c- 4(16))/(2(4))

⇒ c = 456

Remember that we could have also transformed the original initial condition in one in terms of v and after that applied it upon solving the separable differential equation. Under this case though, it was probably a little easier to do this in terms of y provided all the logarithms in the solution to the separable differential equation.

At last, plug in c and solve for y to find:

y2 = (228 - 2x4) /x2

⇒ Y(x) = + √((228 - 2x4) /x2)

Here the initial condition tells us that the "-" should be the correct sign and thus the actual solution is as,

y(x) = - √((228 - 2x4) /x2)

For the interval of validity we can notice that we need to ignore x = 0 and since we can't allow negative numbers in the square root we also want to need,

228 - 2x4 > 0

x4 < 114 ⇒ - 3.2676 < x< 3.2676

Therefore, we have two possible intervals of validity as:

- 3.2676 < x < 0,                   x < 0< 3.2676

And the initial condition tells us that this should be 0 < x ≤ 3.2676

The graph of the solution is as:

104_Find the interval of validity for the solution1.png


Related Discussions:- Find the interval of validity for the solution

Statisctics, I would like to know what a symbol in my homework means?

I would like to know what a symbol in my homework means?

Types of series - telescoping series, Telescoping Series  It's now tim...

Telescoping Series  It's now time to look at the telescoping series.  In this section we are going to look at a series that is termed a telescoping series.  The name in this c

Factor Fiction, Ok this is true or false wit a definition. The GCF of a pai...

Ok this is true or false wit a definition. The GCF of a pair of numbers can never be equal to one of the numbers.

Extreme value theorem, Extreme Value Theorem : Assume that f ( x ) is cont...

Extreme Value Theorem : Assume that f ( x ) is continuous on the interval [a,b] then there are two numbers a ≤ c, d ≤ b so that f (c ) is an absolute maximum for the function and

Error analysis: describle and correct the error in plotting, to plot (5,-4)...

to plot (5,-4), start at (0,0) and move 5 units left and 4 units down

Vectors, A plane is flying at 200 mph with a heading of 45degrees and encou...

A plane is flying at 200 mph with a heading of 45degrees and encounters a wind mph from the west. What is the velocity and heading?

Testing the hypothesis equality of two variances, Testing the hypothesis eq...

Testing the hypothesis equality of two variances The test for equality of two population variances is based upon the variances in two independently chosen random samples drawn

Triangle treat, what letters to fill in the boxes

what letters to fill in the boxes

Square root., i dont get these questions they are hard for me

i dont get these questions they are hard for me

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd