Find the interval of validity for the solution, Mathematics

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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


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