Question (1i)
z = 2 x + a + 2 y + b ⟶ p = ∂ z ∂ x = 2 2 2 x + a ≡ 1 2 x + a q = ∂ z ∂ y = 2 2 2 y + b ≡ 1 2 y + b z=\sqrt{2x+a}+\sqrt{2y+b}\longrightarrow\\[0.3cm]
p=\frac{\partial z}{\partial x}=\frac{2}{2\sqrt{2x+a}}\equiv\frac{1}{\sqrt{2x+a}}\\[0.3cm]
q=\frac{\partial z}{\partial y}=\frac{2}{2\sqrt{2y+b}}\equiv\frac{1}{\sqrt{2y+b}}\\[0.3cm] z = 2 x + a + 2 y + b ⟶ p = ∂ x ∂ z = 2 2 x + a 2 ≡ 2 x + a 1 q = ∂ y ∂ z = 2 2 y + b 2 ≡ 2 y + b 1
Then,
1 p + 1 q = 1 1 2 x + a + 1 1 2 y + b ⟶ 1 p + 1 q = 2 x + a + 2 y + b ≡ z ⟶ z = 2 x + a + 2 y + b ⟶ z = 1 p + 1 q \frac{1}{p}+\frac{1}{q}=\frac{1}{\frac{1}{\sqrt{2x+a}}}+\frac{1}{\frac{1}{\sqrt{2y+b}}}\longrightarrow\\[0.3cm]
\frac{1}{p}+\frac{1}{q}=\sqrt{2x+a}+\sqrt{2y+b}\equiv z\longrightarrow\\[0.3cm]
\boxed{z=\sqrt{2x+a}+\sqrt{2y+b}\longrightarrow z=\frac{1}{p}+\frac{1}{q}} p 1 + q 1 = 2 x + a 1 1 + 2 y + b 1 1 ⟶ p 1 + q 1 = 2 x + a + 2 y + b ≡ z ⟶ z = 2 x + a + 2 y + b ⟶ z = p 1 + q 1
Question (ii)
z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ 2 z ⋅ ∂ z ∂ x = 2 ( 1 + λ − 1 ) → p = 1 + λ − 1 z 2 z ⋅ ∂ z ∂ y = 2 λ ( 1 + λ − 1 ) → q = λ ( 1 + λ − 1 ) z z^2+\mu=2\left(1+\lambda^{-1}\right)\left(x+\lambda y\right)\longrightarrow\\[0.3cm]
2z\cdot\frac{\partial z}{\partial x}=2\left(1+\lambda^{-1}\right)\rightarrow p=\frac{1+\lambda^{-1}}{z}\\[0.3cm]
2z\cdot\frac{\partial z}{\partial y}=2\lambda\left(1+\lambda^{-1}\right)\rightarrow q=\frac{\lambda\left(1+\lambda^{-1}\right)}{z}\\[0.3cm] z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ 2 z ⋅ ∂ x ∂ z = 2 ( 1 + λ − 1 ) → p = z 1 + λ − 1 2 z ⋅ ∂ y ∂ z = 2 λ ( 1 + λ − 1 ) → q = z λ ( 1 + λ − 1 )
Then,
1 p + 1 q = 1 1 + λ − 1 z + 1 λ ( 1 + λ − 1 ) z ⟶ 1 p + 1 q = z 1 + λ − 1 + z λ ( 1 + λ − 1 ) ⟶ 1 p + 1 q = z 1 + 1 λ + z λ + 1 ⟶ 1 p + 1 q = z λ λ + 1 + z λ + 1 ⟶ 1 p + 1 q = z ( λ + 1 ) λ + 1 ≡ z z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ z = 1 p + 1 q \frac{1}{p}+\frac{1}{q}=\frac{1}{\frac{1+\lambda^{-1}}{z}}+\frac{1}{\frac{\lambda\left(1+\lambda^{-1}\right)}{z}}\longrightarrow\\[0.3cm]
\frac{1}{p}+\frac{1}{q}=\frac{z}{1+\lambda^{-1}}+\frac{z}{\lambda\left(1+\lambda^{-1}\right)}\longrightarrow\\[0.3cm]
\frac{1}{p}+\frac{1}{q}=\frac{z}{1+\frac{1}{\lambda}}+\frac{z}{\lambda+1}\longrightarrow\\[0.3cm]
\frac{1}{p}+\frac{1}{q}=\frac{z\lambda}{\lambda+1}+\frac{z}{\lambda+1}\longrightarrow\\[0.3cm]
\frac{1}{p}+\frac{1}{q}=\frac{z\left(\lambda+1\right)}{\lambda+1}\equiv z\\[0.3cm]
\boxed{z^2+\mu=2\left(1+\lambda^{-1}\right)\left(x+\lambda y\right)\longrightarrow z=\frac{1}{p}+\frac{1}{q}} p 1 + q 1 = z 1 + λ − 1 1 + z λ ( 1 + λ − 1 ) 1 ⟶ p 1 + q 1 = 1 + λ − 1 z + λ ( 1 + λ − 1 ) z ⟶ p 1 + q 1 = 1 + λ 1 z + λ + 1 z ⟶ p 1 + q 1 = λ + 1 z λ + λ + 1 z ⟶ p 1 + q 1 = λ + 1 z ( λ + 1 ) ≡ z z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ z = p 1 + q 1
Question 2
Let suppose that b = b ( a ; λ , μ ) b=b(a;\lambda,\mu) b = b ( a ; λ , μ )
F ( x , y , z , a , b ) = 2 x + a + 2 y + b − z = 0 ⟶ ∂ F ∂ a = 1 2 2 x + a + 1 2 2 y + b ⋅ d b d a = 0 ⟶ 1 2 2 x + a = − 1 2 2 y + b ⋅ d b d a ⟶ 2 2 y + b = − 2 2 x + a ⋅ d b d a ⟶ 2 y + b = − 2 x + a ⋅ d b d a F(x,y,z,a,b)=\sqrt{2x+a}+\sqrt{2y+b}-z=0\longrightarrow\\[0.3cm]
\frac{\partial F}{\partial a}=\frac{1}{2\sqrt{2x+a}}+\frac{1}{2\sqrt{2y+b}}\cdot\frac{db}{da}=0\longrightarrow\\[0.3cm]
\frac{1}{2\sqrt{2x+a}}=-\frac{1}{2\sqrt{2y+b}}\cdot\frac{db}{da}\longrightarrow\\[0.3cm]
2\sqrt{2y+b}=-2\sqrt{2x+a}\cdot\frac{db}{da}\longrightarrow\\[0.3cm]
\boxed{\sqrt{2y+b}=-\sqrt{2x+a}\cdot\frac{db}{da}} F ( x , y , z , a , b ) = 2 x + a + 2 y + b − z = 0 ⟶ ∂ a ∂ F = 2 2 x + a 1 + 2 2 y + b 1 ⋅ d a d b = 0 ⟶ 2 2 x + a 1 = − 2 2 y + b 1 ⋅ d a d b ⟶ 2 2 y + b = − 2 2 x + a ⋅ d a d b ⟶ 2 y + b = − 2 x + a ⋅ d a d b
Let
b ( a ) = − a λ + C ⟶ d b d a = − 1 λ b(a)=-\frac{a}{\lambda}+C\longrightarrow\frac{db}{da}=-\frac{1}{\lambda} b ( a ) = − λ a + C ⟶ d a d b = − λ 1
Then,
2 y + b = − 2 x + a ⋅ d b d a ⟶ 2 y + b = 1 λ 2 x + a ⟶ 2 y = 2 x + a λ 2 − b 2 x + a + 2 y + b ⏟ z = 2 x + a − 2 x + a ⋅ ( − 1 λ ) z = ( 1 + 1 λ ) 2 x + a ⟶ z 2 = ( 1 + λ λ ) 2 ( 2 x + a ) \sqrt{2y+b}=-\sqrt{2x+a}\cdot\frac{db}{da}\longrightarrow\\[0.3cm]
\sqrt{2y+b}=\frac{1}{\lambda}\sqrt{2x+a}\longrightarrow\\[0.3cm]
\boxed{2y=\frac{2x+a}{\lambda^2}-b}\\[0.3cm]
\underbrace{\sqrt{2x+a}+\sqrt{2y+b}}_{z}=\sqrt{2x+a}-\sqrt{2x+a}\cdot\left(-\frac{1}{\lambda}\right)\\[0.3cm]
z=\left(1+\frac{1}{\lambda}\right)\sqrt{2x+a}\longrightarrow\\[0.3cm]
\boxed{z^2=\left(\frac{1+\lambda}{\lambda}\right)^2(2x+a)} 2 y + b = − 2 x + a ⋅ d a d b ⟶ 2 y + b = λ 1 2 x + a ⟶ 2 y = λ 2 2 x + a − b z 2 x + a + 2 y + b = 2 x + a − 2 x + a ⋅ ( − λ 1 ) z = ( 1 + λ 1 ) 2 x + a ⟶ z 2 = ( λ 1 + λ ) 2 ( 2 x + a )
Consider the found function along with
z 2 = 2 ( 1 + λ − 1 ) ( x + y λ ) − μ ⟶ z 2 = 2 ( 1 + λ λ ) ( x + y λ ) − μ z^2=2(1+\lambda^{-1})(x+y\lambda)-\mu\longrightarrow\\[0.3cm]
\boxed{z^2=2\left(\frac{1+\lambda}{\lambda}\right)(x+y\lambda)-\mu} z 2 = 2 ( 1 + λ − 1 ) ( x + y λ ) − μ ⟶ z 2 = 2 ( λ 1 + λ ) ( x + y λ ) − μ
Then,
z 2 = ( 1 + λ λ ) 2 ( 2 x + a ) z 2 = 2 ( 1 + λ λ ) ( x + y λ ) − μ ( 1 + λ λ ) 2 ( 2 x + a ) = 2 ( 1 + λ λ ) ( x + y λ ) − μ ( 1 + λ λ ) ( 2 x + a ) = 2 ( x + y λ ) − λ 1 + λ ⋅ μ ( 1 + λ ) ( 2 x + a ) = 2 λ ( x + y λ ) − λ 2 1 + λ ⋅ μ 2 x + a + 2 x λ + a λ = 2 x λ + 2 y λ 2 − λ 2 1 + λ ⋅ μ a ( 1 + λ ) = 2 y λ 2 − 2 x − λ 2 1 + λ ⋅ μ a ( 1 + λ ) = ( 2 x + a λ 2 − b ) λ 2 − 2 x − λ 2 1 + λ ⋅ μ a ( 1 + λ ) = 2 x + a − b λ 2 − 2 x − λ 2 1 + λ ⋅ μ b λ 2 = − a λ − λ 2 1 + λ ⋅ μ b = − a λ − μ 1 + λ z^2=\left(\frac{1+\lambda}{\lambda}\right)^2(2x+a)\\[0.3cm]
z^2=2\left(\frac{1+\lambda}{\lambda}\right)(x+y\lambda)-\mu\\[0.3cm]
\left(\frac{1+\lambda}{\lambda}\right)^2(2x+a)=2\left(\frac{1+\lambda}{\lambda}\right)(x+y\lambda)-\mu\\[0.3cm]
\left(\frac{1+\lambda}{\lambda}\right)(2x+a)=2(x+y\lambda)-\frac{\lambda}{1+\lambda}\cdot\mu\\[0.3cm]
(1+\lambda)(2x+a)=2\lambda(x+y\lambda)-\frac{\lambda^2}{1+\lambda}\cdot\mu\\[0.3cm]
2x+a+2x\lambda+a\lambda=2x\lambda+2y\lambda^2-\frac{\lambda^2}{1+\lambda}\cdot\mu\\[0.3cm]
a(1+\lambda)=2y\lambda^2-2x-\frac{\lambda^2}{1+\lambda}\cdot\mu\\[0.3cm]
a(1+\lambda)=\left(\frac{2x+a}{\lambda^2}-b\right)\lambda^2-2x-\frac{\lambda^2}{1+\lambda}\cdot\mu\\[0.3cm]
a(1+\lambda)=2x+a-b\lambda^2-2x-\frac{\lambda^2}{1+\lambda}\cdot\mu\\[0.3cm]
b\lambda^2=-a\lambda-\frac{\lambda^2}{1+\lambda}\cdot\mu\\[0.3cm]
\boxed{b=-\frac{a}{\lambda}-\frac{\mu}{1+\lambda}} z 2 = ( λ 1 + λ ) 2 ( 2 x + a ) z 2 = 2 ( λ 1 + λ ) ( x + y λ ) − μ ( λ 1 + λ ) 2 ( 2 x + a ) = 2 ( λ 1 + λ ) ( x + y λ ) − μ ( λ 1 + λ ) ( 2 x + a ) = 2 ( x + y λ ) − 1 + λ λ ⋅ μ ( 1 + λ ) ( 2 x + a ) = 2 λ ( x + y λ ) − 1 + λ λ 2 ⋅ μ 2 x + a + 2 x λ + aλ = 2 x λ + 2 y λ 2 − 1 + λ λ 2 ⋅ μ a ( 1 + λ ) = 2 y λ 2 − 2 x − 1 + λ λ 2 ⋅ μ a ( 1 + λ ) = ( λ 2 2 x + a − b ) λ 2 − 2 x − 1 + λ λ 2 ⋅ μ a ( 1 + λ ) = 2 x + a − b λ 2 − 2 x − 1 + λ λ 2 ⋅ μ b λ 2 = − aλ − 1 + λ λ 2 ⋅ μ b = − λ a − 1 + λ μ
Conclusion,
The complete integral (ii) is the envelope of one parameter sub-system obtained by taking
b = − a λ − μ 1 + λ b=-\frac{a}{\lambda}-\frac{\mu}{1+\lambda} b = − λ a − 1 + λ μ
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