04. Homogeneous Differential Equations
Dated: 08-11-2024
A differential equation of the form
\[\frac{dy}{dx} = f(x, y)\]
is said to be homogeneous if \(f(x, y)\) is homogeneous which means
\[f(tx, ty) = t^n f(x, y)\]
where \(t\) is any number and \(n \in \mathbb R\).
Example
Determine if the following functions are homogeneous.
\[
\begin{cases}
f(x, y) = \frac{xy}{x^2 + y^2} \\
g(x, y) = \ln\left(- \frac{3x^2y}{x^3 + 4xy^2}\right)
\end{cases}
\]
\(f(x, y)\) is homogeneous because
\[f(tx,ty)=\frac{t^{2}xy}{t^{2}(x^{2}+y^{2})}=\frac{xy}{x^{2}+y^{2}}=f(x,y)\]
\(g(x, y)\) is also homogeneous because
\[g(tx,ty)=\ln\left(\frac{-3t^{3}x^{2}y}{t^{3}(x^{3}+4xy^{2})}\right)=\ln\left(\frac{-3x^{2}y}{x^{3}+4xy^{2}}\right)=g(x,y)\]
Therefore, the differential equations
\[
\begin{cases}
\frac{dy}{dx} = f(x, y) \\
\frac{dy}{dx} = g(x, y)
\end{cases}
\]
are also homogeneous differential equations.
Method of Solution
To solve homogeneous differential equations, we we will substitute the following in \(f(x, y)\).
\[v = \frac y x\]
If \(f(x, y)\) is homogeneous of degree zero, we have.
\[f(x, y) = f(1, v) = F(v)\]
\[\because y^\prime = xv^\prime + v\]
\[x\frac{dv}{dx}+v=f(1,v)\]
Summary
- Identify if the equation is
homogeneous or not.
- Use the substitution \(v = \frac y x\).
- Solve \(x\frac{dv}{dx}+v=f(1,v)\) to find \(v\).
- Find \(y\) through \(v = \frac y x\).
- If we have an
initial value problem, we need to use the initial condition to find the constant of integration.
Example
\[\frac{dy}{dx}=\frac{-2x+5y}{2x+y}\]
Step 1
Checking if the \(f(x, y)\) is homogeneous or not.
\[\because f(x, y) = \frac{-2x+5y}{2x+y}\]
\[f(tx,ty) = \frac{-2tx+5ty}{2tx+ty}\]
\[ = \frac {t(-2x + 5y)}{t(2x + y)}\]
\[= \frac{-2x+5y}{2x+y}\]
\[= f(x, y)\]
Hence, \(f(x, y)\) is homogeneous.
Step 2
Substituting \(v = \frac y x\) in the differential equation.
\[xv^{\prime}+v=\frac{-2x+5xv}{2x+xv}=\frac{-2+5v}{2+v}\]
\[\frac{dv}{dx}=\frac{1}{x}\left(\frac{-2+5v}{2+v}-v\right)\]
Step 3
The solutions are
\[-4\ln(|v-2|)+3\ln|v-1|=\ln(|x|)+C\]
Step 4
\[-4\ln|y-2x|+3\ln|y-x|=C\]
\[-4\ln\left|\frac{y-2x}{x}\right|+3\ln\left|\frac{y-x}{x}\right|=\ln|x|+c\]
\[\ln\left|\frac{y-2x}{x}\right|^{-4}+\ln\left|\frac{y-x}{x}\right|^{3}=\ln x+\ln c_{1}, c=\ln c_{1}\]
\[\ln\left|\frac{(y-2x)^{-4}}{x^{-4}}\right|+\ln\left|\frac{(y-x)^{3}}{x^{3}}\right|=\ln c_{1}x\]
\[\ln\left|\frac{(y-2x)^{-4}}{x^{-4}}\cdot\frac{(y-x)^{3}}{x^{3}}\right|=\ln c_{1}x\]
\[\frac{(y-2x)^{-4}}{x^{-4}}\cdot\frac{(y-x)^{3}}{x^{3}}=c_{1}x\]
\[(y-2x)^{-4}(y-x)^{3}=c_{1}\]
The implicit equation can be written as
\[(y - x)^3 = C_1(y - 2x)^4\]
\[\frac{dy}{dx}=\frac{a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\]
This equation is not homogeneous but can be reduced to one.
Case 1
\[\frac {a_1}{b_1} = \frac {a_2}{b_2}\]
First, we will substitute \(z = a_1x + b_1y\) which reduces to separable equation.
After solving the separable equation, put the values of \(z\) back.
Case 2
\[\frac {a_1}{b_1} \ne \frac {a_2}{b_2}\]
In this case, we will substitute
\[x = X + h, \quad y = Y + k\]
Then the equation becomes
\[
\frac{dY}{dX} = \frac{a_{1}X + b_{1}Y + a_{1}h + b_{1}k + c_{1}}{a_{2}X + b_{2}Y + a_{2}h + b_{2}k + c_{2}}
\]
Choose \(h\) and \(k\) such that
\[
\begin{cases}
a_{1}h + b_{1}k + c_{1} &= 0 \\
a_{2}h + b_{2}k + c_{2} &= 0
\end{cases}
\]
This reduces to
\[
\frac{dY}{dX} = \frac{a_{1}X + b_{1}Y}{a_{2}X + b_{2}Y}
\]
This one is homogeneous and can be solved accordingly.
Example
\[\frac{dy}{dx} = -\frac{2x + 3y - 1}{2x + 3y + 2}\]
\[\because \frac{a_1}{b_1} = \frac{a_2}{b_2} = 1\]
Substitute
\[z = 2x + 3y\]
Original differential equation becomes
\[
\frac{dy}{dx} = \frac{1}{3}\left(\frac{dz}{dx} - 2\right)
\]
\[
\frac{1}{3}\left(\frac{dz}{dx} - 2\right) = -\frac{z - 1}{z + 2}
\]
\[
\frac{dz}{dx} = \frac{-z + 7}{z + 2}
\]
\[
\left(\frac{z + 2}{-z + 7}\right) dz = dx
\]
Integrating both sides, we will get
\[
-z - 9\ln(z - 7) = x + A
\]
Put values of \(z\) back
\[
-ln(2x+3y-7)^{9} = 3x+3y+A
\]
\[
(2x+3y-7)^{-9} = ce^{3(x+y)}, \quad c = e^{A}
\]
Example
\[
\frac{dy}{dx} = \frac{(x + 2y - 4)}{2x + y - 5}
\]
Substituting
\[
x = X + h, \quad y = Y + k
\]
Original different equation becomes
\[
\frac{dY}{dX} = \frac{(X + 2Y) + (h + 2k - 4)}{(2X + Y) + (2h + k - 5)}
\]
We choose \(h\) and \(k\) such that
\[
h + 2k - 4 = 0, \quad 2h + k - 5 = 0
\]
This gives us
\[h = 2, k = 1\]
\[
\frac{dY}{dX} = \frac{X + 2Y}{2X + Y}
\]
\[
X \frac{dV}{dX} = \frac{1 - V^2}{2 + V}
\]
\[
\left[\frac{2 + V}{1 - V^2}\right] dV = \frac{dX}{X}
\]
Integrating both sides and we get
\[
\int\left[\frac{3}{2(1-V)} + \frac{1}{2(1+V)}\right] dV = \int\frac{dX}{X}
\]
\[
-\frac{3}{2}\ln(1-V) + \frac{1}{2}\ln(1+V) = \ln X + \ln A
\]
\[
\frac{(1-V)^3}{(1+V)} = CX^{-2}, \quad C = A^{-2}
\]
\[
-\frac{3}{2}ln(1-V)+\frac{1}{2}ln(1+V)=ln~X+ln~A
\]
\[
ln(1-V)^{-\frac{3}{2}}+ln(1+V)^{\frac{1}{2}}=ln~XA
\]
\[
ln(1-V)^{-\frac{3}{2}}(1+V)^{\frac{1}{2}}=ln~XA
\]
\[
(1-V)^{-\frac{3}{2}}(1+V)^{\frac{1}{2}}=XA
\]
Raising both sides to power of \(-2\).
\[
(1-V)^3(1+V)^{-1} = X^{-2}A^{-2}
\]
Putting \(V = \frac Y X\) back
\[
\left(1-\frac{Y}{X}\right)^{3}\left(1+\frac{Y}{X}\right)^{-1} = X^{-2}A^{-2}
\]
\[
\left(\frac{X-Y}{X}\right)^{3}\left(\frac{X+Y}{X}\right)^{-1} = X^{-2}A^{-2}
\]
\[
\frac{(X-Y)^{3}}{X+Y}X^{-3+1} = X^{-2}A^{-2}
\]
\[\because c = A^{-2}\]
\[c = \frac{(X - Y)^3}{X + Y}\]
Put \(X = x - 2\) and \(Y = y - 1\)
\[
\frac {(x+y-1)^{3}}{x+y}-3=c
\]
References
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