(a) Variable Separable Method

(b) Homogeneous Differential Equations

(c) Linear Differential Equations

\[ \frac{dy}{dx} = g(y)h(x) \]

We can write the above as

$ \frac{1}{g(y)} \, dy = h(x) \, dx $

Integrating on both sides

\[ \int \frac{1}{g(y)} \, dy = \int h(x) \, dx \]

$\frac {dy}{dx} =e^{x-y}$

we can write this as

$e^y dy = e^x dx $

Integration

$e^y =e^x + C$

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 \( f(x, y) = 3x^3 + 7xy^2 \).

\[ f(\lambda x, \lambda y) = 3(\lambda x)^3 + 7(\lambda x)(\lambda y)^2 = \lambda^3(3x^3 + 7xy^2) = \lambda^3 f(x, y) \]

This function is homogeneous of degree 3.

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

\[ \frac{dy}{dx} = \frac{x + y}{x} \]

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 $y^2 -x^2 =Cx$

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

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

$y. e^{\int P(x)dx}= \int Q(x) . e^{\int P(x)dx} dx + C$

(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 \( y \):

\[ y(x) = \cos(x)(\tan(x) + C) \]

**Notes**- Differential equations
- General and Particular Solutions of Differential equations
- Method of solving for Differential equations

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