Solution of complex differential equation involving $2$ variables












1















Solve the differential equation



$displaystyle frac{dy}{dx} = sqrt{frac{1}{2}+int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}},$



where $theta = x+y$ and $displaystyle x+yin bigg(frac{pi}{4};,frac{3pi}{4}bigg)$




Try: First we will sove $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots cdots (1)$$



Replace $phi = cos^4 theta-sin^4 theta-phi = cos (2theta)-phi.$



So $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(cos 2 theta -phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots (2)$$



So $$2I = int^{-sin^4 theta}_{cos^4 theta}dphi = phibigg|^{-sin^4 theta}_{cos^4 theta}$$



$$I = frac{1}{2}bigg[sin^4 theta +cos ^4 thetabigg] = frac{1}{2}-sin^2 theta cos^2 theta$$



So $$frac{dy}{dx} = sqrt{1-sin^2 theta cos^2 theta}$$



could some help me how to solve further, i don,t get any clue



thanks



answer given as $displaystyle frac{2}{sqrt{3}}tan^{-1}bigg(frac{2tan(x+y)+1}{sqrt{3}}bigg)=x+c.$






differential-equations







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  • Didn’t you loose a negative sign in the equation for I?
    – Makina
    Nov 17 '18 at 15:52
















1















Solve the differential equation



$displaystyle frac{dy}{dx} = sqrt{frac{1}{2}+int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}},$



where $theta = x+y$ and $displaystyle x+yin bigg(frac{pi}{4};,frac{3pi}{4}bigg)$




Try: First we will sove $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots cdots (1)$$



Replace $phi = cos^4 theta-sin^4 theta-phi = cos (2theta)-phi.$



So $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(cos 2 theta -phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots (2)$$



So $$2I = int^{-sin^4 theta}_{cos^4 theta}dphi = phibigg|^{-sin^4 theta}_{cos^4 theta}$$



$$I = frac{1}{2}bigg[sin^4 theta +cos ^4 thetabigg] = frac{1}{2}-sin^2 theta cos^2 theta$$



So $$frac{dy}{dx} = sqrt{1-sin^2 theta cos^2 theta}$$



could some help me how to solve further, i don,t get any clue



thanks



answer given as $displaystyle frac{2}{sqrt{3}}tan^{-1}bigg(frac{2tan(x+y)+1}{sqrt{3}}bigg)=x+c.$






differential-equations







share|cite|improve this question






















  • Didn’t you loose a negative sign in the equation for I?
    – Makina
    Nov 17 '18 at 15:52














1












1








1








Solve the differential equation



$displaystyle frac{dy}{dx} = sqrt{frac{1}{2}+int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}},$



where $theta = x+y$ and $displaystyle x+yin bigg(frac{pi}{4};,frac{3pi}{4}bigg)$




Try: First we will sove $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots cdots (1)$$



Replace $phi = cos^4 theta-sin^4 theta-phi = cos (2theta)-phi.$



So $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(cos 2 theta -phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots (2)$$



So $$2I = int^{-sin^4 theta}_{cos^4 theta}dphi = phibigg|^{-sin^4 theta}_{cos^4 theta}$$



$$I = frac{1}{2}bigg[sin^4 theta +cos ^4 thetabigg] = frac{1}{2}-sin^2 theta cos^2 theta$$



So $$frac{dy}{dx} = sqrt{1-sin^2 theta cos^2 theta}$$



could some help me how to solve further, i don,t get any clue



thanks



answer given as $displaystyle frac{2}{sqrt{3}}tan^{-1}bigg(frac{2tan(x+y)+1}{sqrt{3}}bigg)=x+c.$






differential-equations







share|cite|improve this question














Solve the differential equation



$displaystyle frac{dy}{dx} = sqrt{frac{1}{2}+int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}},$



where $theta = x+y$ and $displaystyle x+yin bigg(frac{pi}{4};,frac{3pi}{4}bigg)$




Try: First we will sove $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots cdots (1)$$



Replace $phi = cos^4 theta-sin^4 theta-phi = cos (2theta)-phi.$



So $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(cos 2 theta -phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots (2)$$



So $$2I = int^{-sin^4 theta}_{cos^4 theta}dphi = phibigg|^{-sin^4 theta}_{cos^4 theta}$$



$$I = frac{1}{2}bigg[sin^4 theta +cos ^4 thetabigg] = frac{1}{2}-sin^2 theta cos^2 theta$$



So $$frac{dy}{dx} = sqrt{1-sin^2 theta cos^2 theta}$$



could some help me how to solve further, i don,t get any clue



thanks



answer given as $displaystyle frac{2}{sqrt{3}}tan^{-1}bigg(frac{2tan(x+y)+1}{sqrt{3}}bigg)=x+c.$






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asked Nov 17 '18 at 14:10









D Tiwari

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  • Didn’t you loose a negative sign in the equation for I?
    – Makina
    Nov 17 '18 at 15:52


















  • Didn’t you loose a negative sign in the equation for I?
    – Makina
    Nov 17 '18 at 15:52
















Didn’t you loose a negative sign in the equation for I?
– Makina
Nov 17 '18 at 15:52




Didn’t you loose a negative sign in the equation for I?
– Makina
Nov 17 '18 at 15:52










1 Answer
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I'll assume that you meant to have $f(phi)$ and not $f(theta)$ in the denominator of the integral



You made a sign error



$$ I = frac12 left(-sin^4 theta - cos^4 theta right) = -frac12 +sin^2thetacos^2theta $$



Then
$$ frac{dy}{dx} = sqrt{frac12 + I} = bigvertsinthetacosthetavert $$



You also have



$$ frac{dtheta}{dx} = 1 + frac{dy}{dx} = 1 + bigvertsinthetacosthetabigvert
= begin{cases}
1 + sinthetacostheta, & theta in left(frac{pi}{4}, frac{pi}{2} right) \
1 - sinthetacostheta, & theta in left( frac{pi}{2}, frac{3pi}{4} right)
end{cases} $$



Separating leads to the integral



$$ int frac{1}{1+sinthetacostheta}dtheta = int frac{sec^2theta}{sec^2theta+tantheta}dtheta = intfrac{1}{u^2+u+1}du $$



where $u = tantheta$. Completing the square on the denominator gives



$$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$






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    1 Answer
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    1 Answer
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    active

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    1














    I'll assume that you meant to have $f(phi)$ and not $f(theta)$ in the denominator of the integral



    You made a sign error



    $$ I = frac12 left(-sin^4 theta - cos^4 theta right) = -frac12 +sin^2thetacos^2theta $$



    Then
    $$ frac{dy}{dx} = sqrt{frac12 + I} = bigvertsinthetacosthetavert $$



    You also have



    $$ frac{dtheta}{dx} = 1 + frac{dy}{dx} = 1 + bigvertsinthetacosthetabigvert
    = begin{cases}
    1 + sinthetacostheta, & theta in left(frac{pi}{4}, frac{pi}{2} right) \
    1 - sinthetacostheta, & theta in left( frac{pi}{2}, frac{3pi}{4} right)
    end{cases} $$



    Separating leads to the integral



    $$ int frac{1}{1+sinthetacostheta}dtheta = int frac{sec^2theta}{sec^2theta+tantheta}dtheta = intfrac{1}{u^2+u+1}du $$



    where $u = tantheta$. Completing the square on the denominator gives



    $$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$






    share|cite|improve this answer




























      1














      I'll assume that you meant to have $f(phi)$ and not $f(theta)$ in the denominator of the integral



      You made a sign error



      $$ I = frac12 left(-sin^4 theta - cos^4 theta right) = -frac12 +sin^2thetacos^2theta $$



      Then
      $$ frac{dy}{dx} = sqrt{frac12 + I} = bigvertsinthetacosthetavert $$



      You also have



      $$ frac{dtheta}{dx} = 1 + frac{dy}{dx} = 1 + bigvertsinthetacosthetabigvert
      = begin{cases}
      1 + sinthetacostheta, & theta in left(frac{pi}{4}, frac{pi}{2} right) \
      1 - sinthetacostheta, & theta in left( frac{pi}{2}, frac{3pi}{4} right)
      end{cases} $$



      Separating leads to the integral



      $$ int frac{1}{1+sinthetacostheta}dtheta = int frac{sec^2theta}{sec^2theta+tantheta}dtheta = intfrac{1}{u^2+u+1}du $$



      where $u = tantheta$. Completing the square on the denominator gives



      $$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$






      share|cite|improve this answer


























        1












        1








        1






        I'll assume that you meant to have $f(phi)$ and not $f(theta)$ in the denominator of the integral



        You made a sign error



        $$ I = frac12 left(-sin^4 theta - cos^4 theta right) = -frac12 +sin^2thetacos^2theta $$



        Then
        $$ frac{dy}{dx} = sqrt{frac12 + I} = bigvertsinthetacosthetavert $$



        You also have



        $$ frac{dtheta}{dx} = 1 + frac{dy}{dx} = 1 + bigvertsinthetacosthetabigvert
        = begin{cases}
        1 + sinthetacostheta, & theta in left(frac{pi}{4}, frac{pi}{2} right) \
        1 - sinthetacostheta, & theta in left( frac{pi}{2}, frac{3pi}{4} right)
        end{cases} $$



        Separating leads to the integral



        $$ int frac{1}{1+sinthetacostheta}dtheta = int frac{sec^2theta}{sec^2theta+tantheta}dtheta = intfrac{1}{u^2+u+1}du $$



        where $u = tantheta$. Completing the square on the denominator gives



        $$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$






        share|cite|improve this answer














        I'll assume that you meant to have $f(phi)$ and not $f(theta)$ in the denominator of the integral



        You made a sign error



        $$ I = frac12 left(-sin^4 theta - cos^4 theta right) = -frac12 +sin^2thetacos^2theta $$



        Then
        $$ frac{dy}{dx} = sqrt{frac12 + I} = bigvertsinthetacosthetavert $$



        You also have



        $$ frac{dtheta}{dx} = 1 + frac{dy}{dx} = 1 + bigvertsinthetacosthetabigvert
        = begin{cases}
        1 + sinthetacostheta, & theta in left(frac{pi}{4}, frac{pi}{2} right) \
        1 - sinthetacostheta, & theta in left( frac{pi}{2}, frac{3pi}{4} right)
        end{cases} $$



        Separating leads to the integral



        $$ int frac{1}{1+sinthetacostheta}dtheta = int frac{sec^2theta}{sec^2theta+tantheta}dtheta = intfrac{1}{u^2+u+1}du $$



        where $u = tantheta$. Completing the square on the denominator gives



        $$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 '18 at 9:29

























        answered Nov 20 '18 at 16:08









        Dylan

        12.3k31026




        12.3k31026






























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