Answer on Question #52706 – Math – Differential Calculus | Equations
Solve the following by Bernoulli equation?
d y d x + x y 1 − x 2 = x ⋅ y 2 1 − x 2 \frac{dy}{dx} + \frac{xy}{1 - x^2} = x \cdot \frac{y^2}{1 - x^2} d x d y + 1 − x 2 x y = x ⋅ 1 − x 2 y 2
Solution
y ′ + x 1 − x 2 y = x y y' + \frac{x}{1 - x^2} y = x \sqrt{y} y ′ + 1 − x 2 x y = x y
Function y = 0 y = 0 y = 0 y ≠ 0 y \neq 0 y = 0 y \sqrt{y} y
y ′ y + x 1 − x 2 y = x \frac{y'}{\sqrt{y}} + \frac{x}{1 - x^2} \sqrt{y} = x y y ′ + 1 − x 2 x y = x
Let z = y → z ′ = y ′ 2 y z = \sqrt{y} \rightarrow z' = \frac{y'}{2\sqrt{y}} z = y → z ′ = 2 y y ′
2 z ′ + x 1 − x 2 z = x 2z' + \frac{x}{1 - x^2} z = x 2 z ′ + 1 − x 2 x z = x
Method 1
Let z = u v , 2 u ′ v + 2 u v ′ + x 1 − x 2 u v = x z = uv, 2u'v + 2uv' + \frac{x}{1 - x^2} uv = x z = uv , 2 u ′ v + 2 u v ′ + 1 − x 2 x uv = x
{ 2 u ′ v + x 1 − x 2 u v = 0 2 u v ′ = x \left\{ \begin{array}{c} 2u'v + \frac{x}{1 - x^2} uv = 0 \\ 2uv' = x \end{array} \right. { 2 u ′ v + 1 − x 2 x uv = 0 2 u v ′ = x
Let v ≠ 0 v \neq 0 v = 0 2 u ′ v + x 1 − x 2 u v = 0 2u'v + \frac{x}{1 - x^2} uv = 0 2 u ′ v + 1 − x 2 x uv = 0 v v v u u u
2 u ′ + x 1 − x 2 u = 0 2u' + \frac{x}{1 - x^2} u = 0 2 u ′ + 1 − x 2 x u = 0
Solution of the separated homogeneous equation
2 u ′ + x 1 − x 2 u = 0 2u' + \frac{x}{1 - x^2} u = 0 2 u ′ + 1 − x 2 x u = 0 2 d u d x = − x 1 − x 2 u 2\frac{du}{dx} = -\frac{x}{1 - x^2} u 2 d x d u = − 1 − x 2 x u
2 ∫ d u u = − ∫ x d x 1 − x 2 → 2 l n u = 1 2 l n ( c ( 1 − x 2 ) ) → u = c 4 1 − x 2 2 \int \frac {d u}{u} = - \int \frac {x d x}{1 - x ^ {2}} \rightarrow 2 l n u = \frac {1}{2} l n (c (1 - x ^ {2})) \rightarrow u = c ^ {4} \sqrt {1 - x ^ {2}} 2 ∫ u d u = − ∫ 1 − x 2 x d x → 2 l n u = 2 1 l n ( c ( 1 − x 2 )) → u = c 4 1 − x 2
Next, take c = 1 c = 1 c = 1 u = 1 − x 2 4 u = \sqrt[4]{1 - x^2} u = 4 1 − x 2
2 u v ′ = x 2uv' = x 2 u v ′ = x v v v
2 u v ′ = x 2 u v ^ {\prime} = x 2 u v ′ = x 2 1 − x 2 4 v ′ = x 2 \sqrt [ 4 ]{1 - x ^ {2}} v ^ {\prime} = x 2 4 1 − x 2 v ′ = x 2 v ′ = x 1 − x 2 4 2 v ^ {\prime} = \frac {x}{\sqrt [ 4 ]{1 - x ^ {2}}} 2 v ′ = 4 1 − x 2 x 2 v = ∫ x 1 − x 2 4 d x + c 1 2 v = \int \frac {x}{\sqrt [ 4 ]{1 - x ^ {2}}} d x + c _ {1} 2 v = ∫ 4 1 − x 2 x d x + c 1 2 v = − 1 2 ∫ ( 1 − x 2 ) − 1 4 ( − 2 x ) d x + c 1 2 v = - \frac {1}{2} \int (1 - x ^ {2}) ^ {- \frac {1}{4}} (- 2 x) d x + c _ {1} 2 v = − 2 1 ∫ ( 1 − x 2 ) − 4 1 ( − 2 x ) d x + c 1 2 v = − 1 2 ( 1 − x 2 ) − 1 4 + 1 − 1 4 + 1 + c 1 2 v = - \frac {1}{2} \frac {\left(1 - x ^ {2}\right) ^ {- \frac {1}{4} + 1}}{- \frac {1}{4} + 1} + c _ {1} 2 v = − 2 1 − 4 1 + 1 ( 1 − x 2 ) − 4 1 + 1 + c 1 2 v = − 2 3 ( 1 − x 2 ) 3 4 + c 1 2 v = - \frac {2}{3} \sqrt [ 4 ]{(1 - x ^ {2}) ^ {3}} + c _ {1} 2 v = − 3 2 4 ( 1 − x 2 ) 3 + c 1 v = − 1 3 ( 1 − x 2 ) 3 4 + c , c = c 1 2 v = - \frac {1}{3} \sqrt [ 4 ]{(1 - x ^ {2}) ^ {3}} + c, \quad c = \frac {c _ {1}}{2} v = − 3 1 4 ( 1 − x 2 ) 3 + c , c = 2 c 1
Recall the substitution
z = u v = 1 − x 2 4 ( − 1 3 ( 1 − x 2 ) 3 4 + c ) = − 1 − x 2 3 + c 4 1 − x 2 = = x 2 − 1 3 + c 1 − x 2 4 \begin{array}{l} z = u v = \sqrt [ 4 ]{1 - x ^ {2}} \left(- \frac {1}{3} \sqrt [ 4 ]{(1 - x ^ {2}) ^ {3}} + c\right) = - \frac {1 - x ^ {2}}{3} + c ^ {4} \sqrt {1 - x ^ {2}} = \\ = \frac {x ^ {2} - 1}{3} + c \sqrt [ 4 ]{1 - x ^ {2}} \\ \end{array} z = uv = 4 1 − x 2 ( − 3 1 4 ( 1 − x 2 ) 3 + c ) = − 3 1 − x 2 + c 4 1 − x 2 = = 3 x 2 − 1 + c 4 1 − x 2
A particular solution of the nonhomogeneous equation :
z p = 1 3 ( x 2 − 1 ) . z_p = \frac{1}{3}(x^2 - 1). z p = 3 1 ( x 2 − 1 ) .
A general solution of the homogeneous equation :
z c = c 1 − x 2 4 z_c = c\sqrt[4]{1 - x^2} z c = c 4 1 − x 2
So the general solution of the nonhomogeneous equation is
z = c 1 − x 2 4 − 1 3 ( 1 − x 2 ) z = c\sqrt[4]{1 - x^2} - \frac{1}{3}(1 - x^2) z = c 4 1 − x 2 − 3 1 ( 1 − x 2 )
Recall the substitution z = y z = \sqrt{y} z = y
y = c 1 − x 2 4 − 1 3 ( 1 − x 2 ) , \sqrt{y} = c\sqrt[4]{1 - x^2} - \frac{1}{3}(1 - x^2), y = c 4 1 − x 2 − 3 1 ( 1 − x 2 ) ,
where c c c
or y = c 1 − x 2 − 2 c 3 ( 1 − x 2 ) 1 − x 2 4 + 1 9 ( 1 − x 2 ) 2 y = c\sqrt{1 - x^2} - \frac{2c}{3}(1 - x^2)\sqrt[4]{1 - x^2} + \frac{1}{9}(1 - x^2)^2 y = c 1 − x 2 − 3 2 c ( 1 − x 2 ) 4 1 − x 2 + 9 1 ( 1 − x 2 ) 2
Method 2
2 z ′ + x 1 − x 2 z = x 2z' + \frac{x}{1 - x^2}z = x 2 z ′ + 1 − x 2 x z = x
2 z ′ + x 1 − x 2 z = 0 2z' + \frac{x}{1 - x^2}z = 0 2 z ′ + 1 − x 2 x z = 0
2 d z d x = − x 1 − x 2 z 2\frac{dz}{dx} = -\frac{x}{1 - x^2}z 2 d x d z = − 1 − x 2 x z
2 ∫ d z z = − ∫ x d x 1 − x 2 → 2 l n z = 1 2 ln ( c ( 1 − x 2 ) ) → → z = c 1 − x 2 4 = C ( x ) 1 − x 2 4 , \begin{array}{l}
2 \int \frac{dz}{z} = - \int \frac{xdx}{1 - x^2} \rightarrow 2lnz = \frac{1}{2}\ln\left(c(1 - x^2)\right) \rightarrow \\
\rightarrow z = c\sqrt[4]{1 - x^2} = C(x)\sqrt[4]{1 - x^2},
\end{array} 2 ∫ z d z = − ∫ 1 − x 2 x d x → 2 l n z = 2 1 ln ( c ( 1 − x 2 ) ) → → z = c 4 1 − x 2 = C ( x ) 4 1 − x 2 ,
substitute it into the initial nonhomogeneous equation 2 z ′ + x 1 − x 2 z = x 2z' + \frac{x}{1 - x^2}z = x 2 z ′ + 1 − x 2 x z = x
2 ( C ( x ) 1 − x 2 4 ) ′ + x 1 − x 2 ( C ( x ) 1 − x 2 4 ) = x 2\left(C(x)\sqrt[4]{1 - x^2}\right)' + \frac{x}{1 - x^2}\left(C(x)\sqrt[4]{1 - x^2}\right) = x 2 ( C ( x ) 4 1 − x 2 ) ′ + 1 − x 2 x ( C ( x ) 4 1 − x 2 ) = x 2 C ′ ( x ) 1 − x 2 4 + 2 C ( x ) 1 4 ( 1 − x 2 ) 1 4 − 1 ( − 2 x ) + x 1 − x 2 1 − x 2 4 ⋅ C ( x ) = x 2C'(x)\sqrt[4]{1 - x^2} + 2C(x)\frac{1}{4}(1 - x^2)^{\frac{1}{4} - 1}(-2x) + \frac{x}{1 - x^2}\sqrt[4]{1 - x^2} \cdot C(x) = x 2 C ′ ( x ) 4 1 − x 2 + 2 C ( x ) 4 1 ( 1 − x 2 ) 4 1 − 1 ( − 2 x ) + 1 − x 2 x 4 1 − x 2 ⋅ C ( x ) = x 2 C ′ ( x ) 4 1 − x 2 + 2 C ( x ) − 2 x 4 ( 1 − x 2 ) 3 4 + x 1 − x 2 1 − x 2 4 ⋅ C ( x ) = x 2 C ^ {\prime} (x) ^ {4} \sqrt {1 - x ^ {2}} + 2 C (x) \frac {- 2 x}{4 \sqrt [ 4 ]{(1 - x ^ {2}) ^ {3}}} + \frac {x}{1 - x ^ {2}} \sqrt [ 4 ]{1 - x ^ {2}} \cdot C (x) = x 2 C ′ ( x ) 4 1 − x 2 + 2 C ( x ) 4 4 ( 1 − x 2 ) 3 − 2 x + 1 − x 2 x 4 1 − x 2 ⋅ C ( x ) = x
divide by 1 − x 2 4 \sqrt[4]{1 - x^2} 4 1 − x 2
2 C ′ ( x ) − C ( x ) x 1 − x 2 + x 1 − x 2 ⋅ C ( x ) = x 1 − x 2 4 2 C ^ {\prime} (x) - C (x) \frac {x}{1 - x ^ {2}} + \frac {x}{1 - x ^ {2}} \cdot C (x) = \frac {x}{\sqrt [ 4 ]{1 - x ^ {2}}} 2 C ′ ( x ) − C ( x ) 1 − x 2 x + 1 − x 2 x ⋅ C ( x ) = 4 1 − x 2 x 2 C ′ ( x ) = x 1 − x 2 4 2 C ^ {\prime} (x) = \frac {x}{\sqrt [ 4 ]{1 - x ^ {2}}} 2 C ′ ( x ) = 4 1 − x 2 x C ( x ) = − 1 4 ∫ ( 1 − x 2 ) − 1 4 ( − 2 x ) d x C (x) = - \frac {1}{4} \int (1 - x ^ {2}) ^ {- \frac {1}{4}} (- 2 x) d x C ( x ) = − 4 1 ∫ ( 1 − x 2 ) − 4 1 ( − 2 x ) d x C ( x ) = − 1 4 ( 1 − x 2 ) − 1 4 + 1 − 1 4 + 1 + c C (x) = - \frac {1}{4} \frac {(1 - x ^ {2}) ^ {- \frac {1}{4} + 1}}{- \frac {1}{4} + 1} + c C ( x ) = − 4 1 − 4 1 + 1 ( 1 − x 2 ) − 4 1 + 1 + c C ( x ) = − 1 4 ( 1 − x 2 ) 3 4 3 4 + c C (x) = - \frac {1}{4} \frac {(1 - x ^ {2}) ^ {\frac {3}{4}}}{\frac {3}{4}} + c C ( x ) = − 4 1 4 3 ( 1 − x 2 ) 4 3 + c z = C ( x ) 4 1 − x 2 4 = ( − 1 4 ( 1 − x 2 ) 3 4 3 4 + c ) 4 1 − x 2 4 = − 1 − x 2 3 + c 4 1 − x 2 4 = = x 2 − 1 3 + c 1 − x 2 4 \begin{array}{l} z = C (x) ^ {4} \sqrt [ 4 ]{1 - x ^ {2}} = \left(- \frac {1}{4} \frac {(1 - x ^ {2}) ^ {\frac {3}{4}}}{\frac {3}{4}} + c\right) ^ {4} \sqrt [ 4 ]{1 - x ^ {2}} = - \frac {1 - x ^ {2}}{3} + c ^ {4} \sqrt [ 4 ]{1 - x ^ {2}} = \\ = \frac {x ^ {2} - 1}{3} + c \sqrt [ 4 ]{1 - x ^ {2}} \\ \end{array} z = C ( x ) 4 4 1 − x 2 = ( − 4 1 4 3 ( 1 − x 2 ) 4 3 + c ) 4 4 1 − x 2 = − 3 1 − x 2 + c 4 4 1 − x 2 = = 3 x 2 − 1 + c 4 1 − x 2
Recall the substitution z = y z = \sqrt{y} z = y
y = c 1 − x 2 4 − 1 3 ( 1 − x 2 ) \sqrt {y} = c \sqrt [ 4 ]{1 - x ^ {2}} - \frac {1}{3} (1 - x ^ {2}) y = c 4 1 − x 2 − 3 1 ( 1 − x 2 ) or y = c 1 − x 2 − 2 c 3 ( 1 − x 2 ) 4 1 − x 2 4 + 1 9 ( 1 − x 2 ) 2 \text {or} \quad y = c \sqrt {1 - x ^ {2}} - \frac {2 c}{3} (1 - x ^ {2}) ^ {4} \sqrt [ 4 ]{1 - x ^ {2}} + \frac {1}{9} (1 - x ^ {2}) ^ {2} or y = c 1 − x 2 − 3 2 c ( 1 − x 2 ) 4 4 1 − x 2 + 9 1 ( 1 − x 2 ) 2 Method 3
2 z ′ + x 1 − x 2 z = x 2z' + \frac{x}{1 - x^2} z = x 2 z ′ + 1 − x 2 x z = x
z ′ + x 2 ( 1 − x 2 ) z = x 2 z' + \frac{x}{2(1 - x^2)} z = \frac{x}{2} z ′ + 2 ( 1 − x 2 ) x z = 2 x
This is a linear first order ordinary differential equation in the dependent variable x x x
d z d x + P ( x ) z = Q ( x ) \frac{dz}{dx} + P(x)z = Q(x) d x d z + P ( x ) z = Q ( x ) d z d x + x 2 ( 1 − x 2 ) z = x 2 \frac{dz}{dx} + \frac{x}{2(1 - x^2)} z = \frac{x}{2} d x d z + 2 ( 1 − x 2 ) x z = 2 x
So P ( x ) = x 2 ( 1 − x 2 ) , Q ( x ) = x 2 P(x) = \frac{x}{2(1 - x^2)}, Q(x) = \frac{x}{2} P ( x ) = 2 ( 1 − x 2 ) x , Q ( x ) = 2 x
The integrating factor is
μ ( x ) = e ∫ P ( x ) d x \mu(x) = e^{\int P(x)dx} μ ( x ) = e ∫ P ( x ) d x
that is,
μ ( x ) = e ∫ x 2 ( 1 − x 2 ) d x \mu(x) = e^{\int \frac{x}{2(1 - x^2)} dx} μ ( x ) = e ∫ 2 ( 1 − x 2 ) x d x ∫ x 2 ( 1 − x 2 ) d x = − 1 4 ∫ − 2 x 1 − x 2 d x = − 1 4 ln ( 1 − x 2 ) \int \frac{x}{2(1 - x^2)} dx = -\frac{1}{4} \int \frac{-2x}{1 - x^2} dx = -\frac{1}{4} \ln(1 - x^2) ∫ 2 ( 1 − x 2 ) x d x = − 4 1 ∫ 1 − x 2 − 2 x d x = − 4 1 ln ( 1 − x 2 ) μ ( x ) = e − 1 4 ln ( 1 − x 2 ) = 1 1 − x 2 4 \mu(x) = e^{-\frac{1}{4} \ln(1 - x^2)} = \frac{1}{\sqrt[4]{1 - x^2}} μ ( x ) = e − 4 1 l n ( 1 − x 2 ) = 4 1 − x 2 1
Consider a differential equation
d z d x + x 2 ( 1 − x 2 ) z = x 2 \frac{dz}{dx} + \frac{x}{2(1 - x^2)} z = \frac{x}{2} d x d z + 2 ( 1 − x 2 ) x z = 2 x
multiplying by 1 1 − x 2 4 \frac{1}{\sqrt[4]{1 - x^2}} 4 1 − x 2 1
1 1 − x 2 4 d z d x + x 2 ( 1 − x 2 ) 1 − x 2 4 z = x 2 1 − x 2 4 \frac {1}{\sqrt [ 4 ]{1 - x ^ {2}}} \frac {d z}{d x} + \frac {x}{2 (1 - x ^ {2}) ^ {\sqrt [ 4 ]{1 - x ^ {2}}}} z = \frac {x}{2 ^ {\sqrt [ 4 ]{1 - x ^ {2}}}} 4 1 − x 2 1 d x d z + 2 ( 1 − x 2 ) 4 1 − x 2 x z = 2 4 1 − x 2 x
So, ( z 1 − x 2 4 ) ′ = x 2 1 − x 2 4 \left(\frac{z}{\sqrt[4]{1 - x^2}}\right)' = \frac{x}{2\sqrt[4]{1 - x^2}} ( 4 1 − x 2 z ) ′ = 2 4 1 − x 2 x
( z ( 1 − x 2 ) − 1 / 4 ) ′ = x 2 1 − x 2 4 \left(z (1 - x ^ {2}) ^ {- 1 / 4}\right) ^ {\prime} = \frac {x}{2 ^ {\sqrt [ 4 ]{1 - x ^ {2}}}} ( z ( 1 − x 2 ) − 1/4 ) ′ = 2 4 1 − x 2 x
integrating both sides with respect to x x x
z ( 1 − x 2 ) − 1 / 4 = ∫ x d x 2 1 − x 2 4 z (1 - x ^ {2}) ^ {- 1 / 4} = \int \frac {x d x}{2 ^ {\sqrt [ 4 ]{1 - x ^ {2}}}} z ( 1 − x 2 ) − 1/4 = ∫ 2 4 1 − x 2 x d x z ( 1 − x 2 ) − 1 / 4 = − 1 4 ∫ ( 1 − x 2 ) − 1 4 ( − 2 x ) d x z (1 - x ^ {2}) ^ {- 1 / 4} = - \frac {1}{4} \int (1 - x ^ {2}) ^ {- \frac {1}{4}} (- 2 x) d x z ( 1 − x 2 ) − 1/4 = − 4 1 ∫ ( 1 − x 2 ) − 4 1 ( − 2 x ) d x z ( 1 − x 2 ) − 1 / 4 = − 1 4 ( 1 − x 2 ) − 1 4 + 1 − 1 4 + 1 + c , z (1 - x ^ {2}) ^ {- 1 / 4} = - \frac {1}{4} \frac {\left(1 - x ^ {2}\right) ^ {- \frac {1}{4} + 1}}{- \frac {1}{4} + 1} + c, z ( 1 − x 2 ) − 1/4 = − 4 1 − 4 1 + 1 ( 1 − x 2 ) − 4 1 + 1 + c ,
where c c c
z ( 1 − x 2 ) − 1 / 4 = − 1 3 ( 1 − x 2 ) 3 4 + c , z (1 - x ^ {2}) ^ {- 1 / 4} = - \frac {1}{3} (1 - x ^ {2}) ^ {\frac {3}{4}} + c, z ( 1 − x 2 ) − 1/4 = − 3 1 ( 1 − x 2 ) 4 3 + c ,
whence
z = − 1 − x 2 3 + c 1 − x 2 4 , z = - \frac {1 - x ^ {2}}{3} + c \sqrt [ 4 ]{\mathbf {1} - \mathbf {x} ^ {2}}, z = − 3 1 − x 2 + c 4 1 − x 2 , z = x 2 − 1 3 + c 1 − x 2 4 , z = \frac {x ^ {2} - 1}{3} + c \sqrt [ 4 ]{\mathbf {1} - \mathbf {x} ^ {2}}, z = 3 x 2 − 1 + c 4 1 − x 2 ,
where c c c
The solution to a linear first order differential equation is then
z ( x ) = ∫ μ ( x ) Q ( x ) d x + C μ ( x ) z (x) = \frac {\int \mu (x) Q (x) d x + C}{\mu (x)} z ( x ) = μ ( x ) ∫ μ ( x ) Q ( x ) d x + C z ( x ) = ∫ − x 2 1 − x 2 4 d x + C ∫ − x 1 − x 2 1 d x + C = − 1 3 ( 1 − x 2 ) 3 4 + c 1 4 / 1 − x 2 = x 2 − 1 3 + c 1 − x 2 4 , z (x) = \frac {\int_ {- \frac {x}{2} \sqrt [ 4 ]{1 - x ^ {2}}} d x + C}{\int_ {- \frac {x}{1 - x ^ {2}}} ^ {1} d x + C} = \frac {\frac {- 1}{3} (1 - x ^ {2}) ^ {\frac {3}{4}} + c}{\frac {1}{4 / 1 - x ^ {2}}} = \frac {x ^ {2} - 1}{3} + c \sqrt [ 4 ]{\mathbf {1} - \mathbf {x} ^ {2}}, z ( x ) = ∫ − 1 − x 2 x 1 d x + C ∫ − 2 x 4 1 − x 2 d x + C = 4/1 − x 2 1 3 − 1 ( 1 − x 2 ) 4 3 + c = 3 x 2 − 1 + c 4 1 − x 2 ,
where c c c
Recall the substitution z = y z = \sqrt{y} z = y
y = c 1 − x 2 4 − 1 3 ( 1 − x 2 ) , \sqrt{y} = c \sqrt[4]{1 - x^2} - \frac{1}{3}(1 - x^2), y = c 4 1 − x 2 − 3 1 ( 1 − x 2 ) ,
where c c c
or y = c 1 − x 2 − 2 c 3 ( 1 − x 2 ) 1 − x 2 4 + 1 9 ( 1 − x 2 ) 2 y = c \sqrt{1 - x^2} - \frac{2c}{3}(1 - x^2) \sqrt[4]{1 - x^2} + \frac{1}{9}(1 - x^2)^2 y = c 1 − x 2 − 3 2 c ( 1 − x 2 ) 4 1 − x 2 + 9 1 ( 1 − x 2 ) 2
Answer: y = c 1 − x 2 4 − 1 3 ( 1 − x 2 ) , y = 0. \sqrt{y} = c \sqrt[4]{1 - x^2} - \frac{1}{3}(1 - x^2), \ y = 0. y = c 4 1 − x 2 − 3 1 ( 1 − x 2 ) , y = 0.
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#340153 on Dec 2023