Question

Please Annswer Q:5
Solve above Differential Equation :
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Answer 1

The answer is given in the attachment.

Thank you for your question.

Answer 2

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♦ Ordinary Differential Equations ♦
◙ Bernoulli ODE ◙

→ First Order Bernoulli ODE has the form : y' + p( x )y = q( x )yⁿ 
    • p( x ) and q( x ) are continuous functions and n ∈ R

→ Solving Bernoulli involves subtituting v = y¹⁻ⁿ , so we use that in our                   solution.
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→ THE ATTACHMENT TO BE REFERRED : 

    • General Solution for the First Order Linear ODE is given by : 

$y(x) = \frac{ \int\limits {e^{ \int\limits {p(x)} \, dx }} \ q(x) \, dx \ + \ C }{e^{ \int\limits {p(x)} \, dx }}$  
    
   → where p( x ), q( x ) have their usual representation when the ODE is                written in the simplest form.

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$\ \textgreater \ y(x)= \sqrt{x^2 \ ( 2 \ sin ( x^2 ) + C ) } \\ \\ At \ x = \sqrt{\frac{ \pi }{2} } : \\ \\ \ \textgreater \ y( x) = \sqrt{ \frac{ \pi }{2}( 2 \ sin \frac{ \pi }{2} + C ) } \\ \\ \ \textgreater \ \sqrt{ \pi } = \sqrt{\frac{ \pi }{2}} \sqrt{( 2 + C )}$

→ Just work up with the Algebra and you get : C = 0

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♦Result :      

$y( x) = \sqrt{ x^2 ( 2 \ sin (x^2) ) }$

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→ For further details on Bernoulli ODE's :   ♦ http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx  

  • The website above mentioned 0_0 gives an insight into all sorts of ODE's and their generalized Problem Solving Methods    

 ♦ http://mathworld.wolfram.com/BernoulliDifferentialEquation.html 
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^_^ Hope it helps !