齐次微分方程

在数学中，微分方程是与一个或多个函数及其导数相关的方程。在本文中，我们将讨论齐次方程，但在转到本主题之前，让我们首先了解齐次函数。

同质函数

x和y中的函数f(x，y)被认为是每个项的阶数为p的齐次函数。例如： f(x，y)=(x ² + y ² – xy)是^2级的齐次函数，其中p =2。类似地，g(x，y)=(x ³ – 3xy ² + 3x ² y + y ³ )是阶数为3的齐次函数，其中p =3。通常，阶数为n的齐次函数ƒ(x，y)可表示为：

ƒ(x, y) = xⁿ ƒ(y/x)

齐次微分方程

dy / dx = f(x，y)/ g(x，y)形式的方程，其中f(x，y)和g(x，y)都是简单词中n度的齐次函数，两个函数具有相同的度数，称为齐次微分方程。例如： dy / dx =(x ² -y ² )/ xy是齐次微分方程。

求解齐次微分方程

令dy / dx = f(x，y)/ g(x，y)是齐次微分方程。现在将y = vx和dy / dx =(v + x dv / dx)放在给定的方程式中，我们得到

v + x dy / dx = F(v)

=>∫dv/ {F(v)– v} =∫dx/ x

=>∫dv/ {F(v)– v} = log | x | + C

现在，将v替换为(y / x)以获取所需的解决方案。让我们看一些例子。

示例1：求解dy / dx = y ² – x ² / 2xy?

解决方案：

Clearly, since each of the functions (y² – x²) and 2xy is a homogenous function of degree 2, the given equation is homogeneous.

Putting y = vx and dy/dx = v + x dy/dx, the given equation becomes

v + x dv/dx = (v²x² – x²)/2vx²

=> v + x dv/dx = v² – 1/2v   [after dividing (v²x²/2vx² – x²/2vx²)]

=> x dv/dx = ((v² – 1/2v) – v)

=> x dv/dx = -(1 + v²)/2v

=> 2v/(1 + v²)dv = -1/x dx

=> ∫2v/(1 + v²)dv = -∫1/x dx [Integrating both the sides]

=> log | 1 + v² | = -log | x | + log C

=> log | 1 + v² | + log | x | = log C

=> log | x(1 + v²) | = log C

=> x(1 + v²) = ±C

=> x(1 + v²) = C₁

=> x(1 + y²/x²) = C₁  [Putting the original value of v = y/x]

=> (x² + y²) = xC₁, which is the required solution  

示例2：求解(x√(x ² + y ² )– y ² )dx + xy dy = 0?

解决方案：

The given equation may be written as 

dy/dx = y² – x√(x² + y²)/xy, which is clearly homogeneous 

Putting y = vx and dy/dx = v + x dv/dx in it, we get

v + x dv/dx = {v²x² – x√(x² + v²y²)}/vx²  

=> x dv/dx = [{v² – √(1 + v²)}/v – v]

=> x dv/dx = -√(1 + v²)/v

=> ∫v/√(1 + v²)dv = -∫dx/xc [Integrating both the sides]

=> √(1 + v²) = -log | x | + C

=> √(x² + y²) + x log | x | = Cx, which is the required solution after putting the value of v = y/x.

示例3：求解x dy / dx – y =√(x ² + y ² )?

解决方案：

The given equation may be written as dy/dx = {y + √(x² + y²)}/x ,which is clearly homogeneous.

Putting y = vx and dy/dx = v + x dv/dx in it, we get

v + x dv/dx = {vx + √(x² + v²x²)}/x 

=> v + x dv/dx = v + √(1+v²) [After dividing the {vx + √(x² + v²x²)}/x]

=> x dv/dx = √(1 + v²) [v on the both sides gets cancelled]

=> dv/√(1+v²) = 1/x dx [after rearranging]

=> ∫dv/√(1+v²) = ∫1/x dx [integrating both sides]

=> log | v | + √(1 + v²) | = log | x | + log C

=> log | {v + √(1 + v²)}/x | = log | C |

=> {v + √(1 + v²)}/x = ±C

=> v + √(1 + v²) = C₁x, where C₁ = ±C

=> y + √(x² + y²) = C₁x², which is the required solution after putting the value of v = y/x

示例4：求解(x cos(y / x))(y dx + x dy)= y sin(y / x)(x dy – y dx)?

解决方案：

The given equation may be written as

(x cos(y/x) + y sin(y/x))y – (y sin(y/x) – x cos (y/x)) x . dy/dx = 0

=> dy/dx = {x cos (y/x) + y sin(y/x)}y / {y sin(y/x) – x cos(y/x)}x

=> dy/dx = {cos (y/x) + (y/x)sin(y/x)}(y/x) / {(y/x)sin(y/x) – cos(y/x)} [Dividing numerator and denominator by x²], which is clearly homogeneous ,being a function of (y/x).

Putting y = vx and dy/dx = (v + x dv/dx) in it, we get

v + x dv/dx = v(cos v + sin v)/(v sin v – cos v)

=> x dv/dx = [v(cos v + sin v)/(v sin v – cos v) -v]

=> x dv/dx = 2vcos v/(v sin v – cos v)

=>∫{(v sin v – cos v)/2vcos v}dv = ∫x dx  [Integrating both sides]

=> ∫tan v dv – ∫ dv/v = ∫ 2/x dx

=> -log | cos v | – log | v | + log C = 2 log | x |

=> log | cos v | + log | v | + 2log | x | = log | C |

=> log | x²v cos v | = log | C |

=> | x²v cos v | = C [After cancelling log on the both sides]

=> x²v cos v = ± C

=> x²v cos v = C₁ [here we taking ±C = C₁]

=> xy cos(y/x) = C_1, which is the required solution after putting the actual value of v = y/x