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

Precalculus, describe the end behavior of the following function using Limi...

describe the end behavior of the following function using Limit notation f(x)= 2x-1/x-1

Discrete-time signals as energy or power signals, Classify the following di...

Classify the following discrete-time signals as energy or power signals. If the signal is of energy type, find its energy. Otherwise, find the average power of the signal. X 1

Monomial, express the area of a square with sides of length 5ab as monomial...

express the area of a square with sides of length 5ab as monomial

How much money does she have left, Mary has $2 in her pocket. She does yard...

Mary has $2 in her pocket. She does yard work for four various neighbors and earns $3 per yard. She then spends $2 on a soda. How much money does she have left? This translates

integration: if f(x)+f(x+1/2) =1 find limit 0 to 2, f(x)+f(x+1/2) =1 f(x...

f(x)+f(x+1/2) =1 f(x)=1-f(x+1/2) 0∫2f(x)dx=0∫21-f(x+1/2)dx 0∫2f(x)dx=2-0∫2f(x+1/2)dx take (x+1/2)=v dx=dv 0∫2f(v)dv=2-0∫2f(v)dv 2(0∫2f(v)dv)=2 0∫2f(v)dv=1 0∫2f(x)dx=1

Boundary value problem, solve the in-homogenous problem where A and b are c...

solve the in-homogenous problem where A and b are constants on 0 ut=uxx+A exp(-bx) u(x,0)=A/b^2(1-exp(-bx)) u(0,t)=0 u(1,t)=-A/b^2 exp(-b)

Estimate how much should every friend pay, A group of ?ve friends gone out ...

A group of ?ve friends gone out to lunch. The total bill for the lunch was $53.75. Their meals all cost about the similar, so they needed to split the bill evenly. Without consider

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