Differential Equations Introduction Homogeneous Functions
Definition. A function f(x,y) is said to be a homogeneous function of degree n in x and y if \(f(k x, k y)=k^n f(x, y)\) for all values of k where n is a real number.
Introduction To Homogeneous Differential Equations
Example 1.
Given
A function f(x,y) is said to be a homogeneous function of degree n in x and y if \(f(k x, k y)=k^n f(x, y)\) for all values of k where n is a real number.
Let \(f(x, y)=\left(x^2+y^2\right) /\left(x^3+y^3\right)\)
Now \(f(k x, k y)=\left(k^2 x^2+k^2 y^2\right) /\left(k^3 x^3+k^3 y^3\right)=\frac{1}{k}\left(\frac{x^2+y^2}{x^3+y^3}\right)=k^{-1} f(x, y)\)
∴ f(x,y) is a homogeneous function of degree -1.
Illustrations Of Homogeneous Differential Equations
Example. 2. Let \(f(x, y)=(\sqrt[3]{x}+\sqrt[3]{y}) /(x+y)\)
Now \(f(k x, k y)=\frac{\sqrt[3]{(k x)}+\sqrt[3]{(k y)}}{k x+k y}=\frac{k^{1 / 3}(\sqrt[3]{x}+\sqrt[3]{y})}{k(x+y)}=k^{-2 / 3}\left(\frac{\sqrt[3]{x}+\sqrt[3]{y}}{x+y}\right)=k^{-2 / 3} f(x, y)\)
∴ f(x,y) is a homogeneous function of degree -2/3.
Example 3. Let \(f(x, y)=\frac{y^2+x^2 e^{-x / y}}{x+y}\)
Now \(f(k x, k y)=\frac{k^2 y^2+k^2 x^2 e^{-k x / k y}}{k x+k y}=k \frac{y^2+x^2 e^{-x / y}}{x+y}=k f(x, y)\)
∴ f(x,y) is a homogeneous function of degree 1.
Example 4. Let f(x,y) = cos x + tan y
Now \(f(k x, k y)=\cos k x+\tan k y \neq k^n f(x, y)\) for any value of n
∴ f(x,y) is not a homogeneous function.
Further a homogeneous function of degree n in x and y can be expressed as , \(x^n f(x / y) \text { or } y^n f(x / y)\)
Example 5. Let \(f(x, y)=a x^3+3 b x^2 y+c y^3\)
Now \(f(x ; y)=x^3\left[a+3 b(y / x)+c(y / x)^3\right]=x^3 f(y / x)\)
Also \(f(x, y)=y^3\left[c+3 b(x / y)^2+a(x / y)^3\right]=y^3 f(x / y)\)
∴ f(x,y) is a homogeneous function of degree 3.
Methods To Convert Differential Equations To Homogeneous Form
Example 6. Let \(f(x, y)=(\sqrt{x}+\sqrt{y}) /(\sqrt{x}-\sqrt{y}) .\)
Now \(f(x, y)=\frac{x^{1 / 2}\left[1+(y+x)^{1 / 2}\right]}{x^{1 / 2}\left[1-(y / x)^{1 / 2}\right]}=x^0 f(y / x)\)
∴ f(x, y) is a homogeneous function of degree 0.
Note. If f(x,y) is a homogeneous function of degree zero, then f(x,y) is a function of y/x or x/y alone.
Differential Equations Introduction Homogeneous Differential Equation
Definition. A differential equation \(\frac{d y}{d x}=f(x, y)\) of first order and first degree is called homogeneous in x and y if the function f(x,y) is a homogeneous function of degree zero in x and y.
Differential Equations Introduction Working rule to solve a Homogeneous Differential Equation
1. Let \(\frac{d y}{d x}=f(x, y)\) be the homogeneous equation.
Then express it in the form \(\frac{d y}{d x}=\phi\left(\frac{y}{x}\right)\) ……………………….(1)
2. To solve (z), put \(y / x=v \Rightarrow y=v x\) …………………….(2)
3. Differentiating (2) w.r.t. x: \(x: \frac{d y}{d x}=v+x \frac{d v}{d x}\) ……………………….(3)
4. Now (1), (2), (3) => \(v+x \frac{d v}{d x}=\phi(v) \Rightarrow x \frac{d v}{d x}=\phi(v)-v\)
Separating the variables : \(\frac{d x}{x}=\frac{d v}{\phi(v)-v}\)
Integrating: \(\int \frac{d x}{x}=\int \frac{d v}{\phi(v)-v}+c\) where c is an arbitrary constant.
5. After integration replaced by (y / x) to get the general solution of the given homogeneous equation.
Note: If the given homogeneous equation is reduced to the form \(\frac{d x}{d y}=\mathrm{F}\left(\frac{x}{y}\right)\) then put \(x=v y \Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}\). This gives a differential equation in v and y where the variables are separable. After integration replace v by (x/y) to obtain the required general solution.
Differential Equations Introduction Equations Reduced to Homogeneous Form
Non-homogeneous equations of the first degree in x and y:
If the equation \(\frac{d y}{d x}=f(x, y)\) is of the form \(\left(a_2 x+b_2 y+c_2\right) \frac{d y}{d x}=a_1 x+b_1 y+c_1\) where \(a_1, b_1, c_1, a_2, b_2, c_2\) are real numbers and \(c_1 \neq 0 \text { or } c_2 \neq 0\), then it is called a non-homogeneous differential equation of the first degree in x and y.
2. General solution of non-homogeneous equation \(\left(a_2 x+b_2 y+c_2\right) \frac{d y}{d x}=a_1 x+b_1 y+c_1\)…… (1)
Equation (1) can be reduced to a homogeneous form or variables separable form by some transformation.
Differential Equations Introduction Working rule to solve the equation \(\left(a_2 x+b_2 y+c_2\right) \frac{d y}{d x}=a_1 x+b_1 y+c_1\)
Case 1.
(1) Let \(a_1 b_2-a_2 b_1=0\)
Let \(\frac{a_2}{a_1}=\frac{b_2}{b_1}=t\) (non-zero real number) \(\Rightarrow a_2=t a_1, b_2=t b_1\)
(2) Substitute the value \(a_2=t a_1, b_2=t b_1\) in (1)
(1) \(\Rightarrow\left[t\left(a_1 x+b_1 y\right)+c_2\right] \frac{d y}{d x}=a_1 x+b_1 y+c_1\) …………………….(2)
(3) If \(c_2=t c_1\), then (2) reduces to \(\frac{d y}{d x}=\frac{1}{t}\)
∴ G. S. of (1) is ty = x + c
(4) If \(\) then put \(a_1 x+b_1 y=u \Rightarrow \frac{d u}{d x}=a_1+b_1 \frac{d y}{d x}\) ……………………….(3)
(2) and (3) => \(\left(t u+c_2\right)\left[\frac{1}{b_1}\left(\frac{d u}{d x}-a_1\right)\right]=u+c_1 \Rightarrow \frac{d u}{d x}=\frac{\left(b_1+t a_1\right) u+\left(c_2+c_1 b_1\right)}{t u+c_2}\) ……………………..(4)
(4) is in variables separable form hence it can be solved by separating the variables.
∴ The G. S. of (4) is \(\phi(u, x, c)=0\)
Hence the G. S. of (1) is \(\phi\left(a_1 x+b_1 y, x, c\right)=0\)
Case 2. : Let \(a_1 b_2-a_2 b_1 \neq 0 \Rightarrow \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
1. (1) \(\Rightarrow \frac{d y}{d x}=\frac{a_1 x+b_1 y+c_1}{a_2 x+b_2 y+c_2}\) ………………………..(2)
2. Put x = X + h and y = Y + k in (1) => \(\frac{d x}{d \mathrm{X}}=1, \frac{d y}{d \mathrm{Y}}=1 \Rightarrow \frac{d y}{d x}=\frac{d \mathrm{Y}}{d \mathrm{X}}\) ………………………….(3)
3. (2) and (3) \(\frac{d Y}{d X}=\frac{\left(a_1 X+b_1 Y\right)+\left(a_1 h+b_1 k+c_1\right)}{\left(a_2 X+b_2 Y\right)+\left(a_2 h+b_2 k+c_2\right)}\) …………………..(4)
4. Choose h and k such that \(a_1 h+b_1 k+c_1=0\) and \(a_2 h+b_2 k+c_2=0\)
Solve these equations to get h and k. ………………………..(5)
5. (4) and (5) \(\frac{d \mathrm{Y}}{d \mathrm{X}}=\frac{a_1 \mathrm{X}+b_1 \mathrm{Y}}{a_2 \mathrm{X}+b_2 \mathrm{Y}}\)
This is clearly a homogeneous equation which can be solved by putting Y = VX ……………………….(6)
6. The G. S. of (6) is F (X,Y,c) = 0 …………….(7)
7. The G.S. of (1) is obtained by replacing X and Y with x – h and y – k in (7)
∴ General solution of (1) is f(x-h,y-k,c) = 0
Note: In case 2 of the above method geometrically (A, k) is the point of intersection of two lines \(a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0\)