Mixing
Solution of complex differential equation involving $2$ variables
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
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
add a comment |
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
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
Didn’t you loose a negative sign in the equation for I?
– Makina
Nov 17 '18 at 15:52
add a comment |
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
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
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
differential-equations
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
asked Nov 17 '18 at 14:10
D Tiwari
5,3152630
5,3152630
Didn’t you loose a negative sign in the equation for I?
– Makina
Nov 17 '18 at 15:52
add a comment |
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
add a comment |
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
$$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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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
$$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
add a comment |
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
$$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
add a comment |
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
$$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
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
$$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
edited Nov 21 '18 at 9:29
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
answered Nov 20 '18 at 16:08
Dylan
12.3k31026
12.3k31026
add a comment |
add a comment |
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Didn’t you loose a negative sign in the equation for I?
– Makina
Nov 17 '18 at 15:52