There are various method to solve Differential equations depending on the order and degree. we will concentrate on Methods of Solving First Order, First Degree Differential Equations in this page.There are three method to solve First Order, First Degree Differential Equations
(a) Variable Separable Method
(b) Homogeneous Differential Equations
(c) Linear Differential Equations
Variable Separable Method
If the differential equation is of the form
\[ \frac{dy}{dx} = g(y)h(x) \]
We can write the above as
$ \frac{1}{g(y)} \, dy = h(x) \, dx $
Integrating on both sides
\[ \int \frac{1}{g(y)} \, dy = \int h(x) \, dx \]
Example
$\frac {dy}{dx} =e^{x-y}$
we can write this as
$e^y dy = e^x dx $
Integration
$e^y =e^x + C$
Homogeneous Differential Equations
homogeneous Function
A function is termed as homogeneous if it exhibits consistent behavior when all of its variables are scaled by a common factor. More formally, a function \( f(x, y, z, ...) \) is said to be homogeneous of degree \( n \) if, for any scalar \( \lambda \), the following condition holds:
\[ f(\lambda x, \lambda y, \lambda z, ...) = \lambda^n f(x, y, z, ...) \]
Here, \( n \) can be any real number, positive, negative, or even non-integer.
Consider the function \( f(x, y) = 3x^3 + 7xy^2 \).
Further we can see that
\( f(x, y) = x^3 (3+ 7 \frac {y^2}{x^2} ) \).
A Homogeneous Differential Equations is of type
\[ \frac{dy}{dx} = F(x,y) = f\left(\frac{y}{x}\right) \] -(1)
Where F(x,y) is homogeneous function of degree 0
Substitute \(v = \frac{y}{x} \Rightarrow y = vx \) and solve.
$y= vx$
$\frac {dy}{dx} = v + x \frac {dv}{dx}$
From (1)
$f(v) = v + x \frac {dv}{dx}$
$f(v) -v = x \frac {dv}{dx}$
$\frac {dx}{x} = \frac {dv}{f(v) -v} $
Now we can integrate and solve it
Example
(1)
\[ \frac{dy}{dx} = \frac{x + y}{x} \]
This can be written as
\[ \frac{dy}{dx} = 1 + \frac{y}{x} \]
So this is Homogeneous Differential Equations
Now, let \( v = \frac{y}{x} \) which implies \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute these values back into the equation:
\[ v + x\frac{dv}{dx} = 1 + v \]
\[ x\frac{dv}{dx} = 1 \]
\[ \int dx = \int \frac{dv}{x} \]
\[ \ln|x| = \ln|v| + C_1 \]
\[ |x| = C|v| \]
\[ x = C(y/x) \]
\[ x - Cy = 0 \]
\[ y = \frac{x}{C} \]
where \( C \) and \( C_1 \) are constants of integration, and \( C = e^{C_1} \).
(2)
\[ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \]
This can be written as
\[ \frac{dy}{dx} = \frac {1 + (y/x)^2}{2(y/x)} \]
So this is Homogeneous Differential Equations
Now, let \( v = \frac{y}{x} \), which implies \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute these values back into the equation:
\[ v + x\frac{dv}{dx} = \frac{1 + v^2}{2v} \]
\[ x\frac{dv}{dx} = \frac{1 - v^2}{2v} \]
Now, separate and integrate:
$ln |v^2 -1| = - ln |x| + C$
$ln |(v^2-1)x| = C$
or
$(v^2 -1)x = C_1$
Substituting the value of v
$y^2 -x^2 =C_1 x$
or
$y^2 -x^2 =Cx$
Linear Differential Equations
A Differential Equations of the form where P and Q are either function of x or constant is called the Linear Differential Equations
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
Lets multiply both the side by the function g(x),then
$ g(x) \frac{dy}{dx} + P(x). g(x) y = Q(x) g(x)$
Let Function g(x) be such that the RHS side becomes of derivative y. g(x)
$ g(x) \frac{dy}{dx} + P(x). g(x) y= \frac {d}{dx} y. g(x)$
or
$g(x) \frac{dy}{dx} + P(x). g(x) y=g(x) \frac{dy}{dx} y g'(x)$
$ P(x). g(x)= g'(x)$
$P(x) = \frac {g'(x)}{g(x)}$
Integrating both side
$\int P(x) dx= ln |g(x) |$
or
$g(x) =e^{ \int P(x) dx}$
This is called the integrating factor
So Multiply through by the integrating factor \( e^{\int P(x)dx} \) , we get the differential equation in the form
$\frac {d}{dx} (y. e^{\int P(x)dx}) = Q(x) . e^{\int P(x)dx} $
Integrating both side
$y. e^{\int P(x)dx}= \int Q(x) . e^{\int P(x)dx} dx + C$
Examples
(1)
\[ \frac{dy}{dx} - 2y = e^{3x} \]
To solve this equation, we can use an integrating factor, \( g(x) \), which is given by:
\[ g(x) = e^{\int -2 dx} = e^{-2x} \]
Multiply the entire equation by \( g(x) \):
\[ e^{-2x}\frac{dy}{dx} - 2e^{-2x}y = 1 \]
\[ \frac{d}{dx}(e^{-2x}y) = 1 \]
Now integrate both sides:
\[ \int \frac{d}{dx}(e^{-2x}y) dx = \int 1 dx \]
\[ e^{-2x}y = x + C \]
Solve for \( y \):
\[ y(x) = e^{2x}(x + C) \]
where \( C \) is the constant of integration.
(2)
\[ \frac{dy}{dx} + y\tan(x) = \cos(x) \]
The integrating factor, \( g(x) \), is calculated as follows:
\[ g(x) = e^{\int \tan(x) dx} = e^{-\ln|\cos(x)|} = \frac{1}{\cos(x)} \]
Multiply the entire equation by \( g(x) \):
\[ \frac{1}{\cos(x)}\frac{dy}{dx} + \frac{\sin(x)}{\cos^2(x)}y = \frac{1}{\cos^2(x)} \]
\[ \frac{d}{dx}\left(\frac{y}{\cos(x)}\right) = \frac{1}{\cos^2(x)} \]
Integrate both sides:
\[ \int \frac{d}{dx}\left(\frac{y}{\cos(x)}\right) dx = \int \frac{1}{\cos^2(x)} dx \]
\[ \frac{y}{\cos(x)} = \tan(x) + C \]
Solve for \( y \):
\[ y(x) = \cos(x)(\tan(x) + C) \]
