\( v(x) = \frac{y(x)}{x} \), find \( y' \)
\[ y = x v \]
\[ y' = (x v)' = 1 \cdot v + x \cdot v' \]
\( v(x) = (y(x))^n \), find \( y' \)
\( n \neq 0 \)
\[ v'(x) = n \cdot y^{1-n} y'(x) \]
\[ y'(x) = \frac{1}{n \, y^{1-n}} v' \]
\[ \frac{dy}{dx} = f(x) \rightsquigarrow y(x) = \int f(x)\,dx + C \]
\[ \frac{dy}{dx} = \frac{f(x)}{g(y)} \rightsquigarrow g(y)\frac{dy}{dx} = f(x) \] \[ \int g(y)\,dy = \int f(x)\,dx + C \]
\[ \frac{dy}{dx} + p(x)y = q(x) \] \[ \mu(x) = e^{\int p(x)\,dx} \rightsquigarrow \mu'(x) = \mu(x)p(x) \] \[ \mu(x)y' + \mu(x)p(x)y = \mu(x)q(x) \] \[ (\mu(x)p(x))' = \mu(x)q(x) \]
\[ \rightsquigarrow \text{integrate } \& \text{ find } y(x) \]
\[ \frac{dy}{dx} = f(x, y) \]
Not linear or seperable.
but we can do a substitution and convert to seperable or linear
\[ \frac{dy}{dx} = (x+y)^2 \]
Let
\[ v(x) = x+y \]
\[ v'(x) = 1 + y' \rightsquigarrow y' = v' - 1 \]
\[ v' - 1 = v^2 \]
\[ \left. \frac{dv}{dx} = v^2 + 1 \right\} \underline{\text{Separable}} \]
\[ \frac{1}{v^2 + 1} \, dv = 1 \cdot dx \]
\[ \int \frac{1}{v^2 + 1} \, dv = \int 1 \, dx + c \rightsquigarrow \tan^{-1}(v) = x + c \]
\[ v(x) = x + y(x) = \tan(x+c) \]
\[ y(x) = \tan(x+c) - x \]
Homogeneous function: \( f(x,y) = \phi\left(\frac{y}{x}\right) \)
Any function such that \( f(kx, ky) = f(x,y) \) is called a Homogeneous function.
\[ y' = \phi(y/x) \]
Let \( v = y/x \), \( y' = (xv)' = v + xv' \)
New equation:
\[ v + xv' = \phi(v) \] \[ xv' = \phi(v) - v \] \[ \frac{1}{\phi(v) - v} \frac{dv}{dx} = \frac{1}{x} \]
a Seperable equation.
eg 1: \( y' = \frac{x^2 + xy + y^2}{x^2} \)
\[ f(x, y) = \frac{x^2 + xy + y^2}{x^2} \]
try writing \( f \) as a function of 1 variable: \( y/x \)
\[ \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 \]
Let \( \phi(v) = 1 + v + v^2 \) then \( f(x, y) = \phi\left(\frac{y}{x}\right) \)
\[ \frac{dy}{dx} = 1 + \left(\frac{y}{x}\right) + \left(\frac{y}{x}\right)^2 \leadsto \text{Homogeneous} \]
Let \( v = \frac{y}{x} \implies y' = (xv)' = v + xv' \)
\[ \cancel{v} + xv' = 1 + \cancel{v} + v^2 \] \[ xv' = 1 + v^2 \quad \Big\} \text{ separable equation} \] \[ \frac{1}{1+v^2} \frac{dv}{dx} = \frac{1}{x} \] \[ \int \frac{1}{1+v^2} \, dv = \int \frac{1}{x} \, dx + C \] \[ \tan^{-1}(v) = \ln|x| + C \] \[ v(x) = \tan(\ln|x| + C) \] \[ y(x) = x v(x) = x \tan(\ln|x| + C) \]
Substitute: \( v = \frac{y}{x}, \quad y' = v + x v' \)
\[ \cancel{v} + x v' = \cos^2(v) + \cancel{v} \] \[ x v' = \cos^2(v) \] \[ \frac{1}{\cos^2(v)} v' = \frac{1}{x} \] \[ \int \sec^2 v \, dv = \int \frac{1}{x} \, dx + C \] \[ \tan(v(x)) = \ln|x| + C \] \[ v(x) = \tan^{-1}(\ln|x| + C) \] \[ y(x) = x \tan^{-1}(\ln|x| + C) \]
\[ \frac{x-y}{x+y} \]
divide with \(x\) both in Numerator & Denominator
\[ \frac{\frac{x-y}{x}}{\frac{x+y}{x}} = \frac{1-\frac{y}{x}}{1+\frac{y}{x}} \]
\[ \frac{x^6+x^2 y^4+x^3 y^3}{x^2 y^4+y^6} \]
\(\rightarrow\) divide with \(x^6\)
\[ \frac{\frac{x^6+x^2 y^4+x^3 y^3}{x^6}}{\frac{x^2 y^4+y^6}{x^6}} \]
\[ = \frac{1+\left(\frac{y}{x}\right)^4+\left(\frac{y}{x}\right)^3}{\left(\frac{y}{x}\right)^4+\left(\frac{y}{x}\right)^6} \]
\[ \ln y - \ln x \]
\[ = \ln(y/x) \]
\[ y' + p(x)y = q(x)y^n, \quad n \neq 0, 1 \]
try to do a substitution and convert to linear
divide with \( y^n \)
\[ y^{-n}y' + p(x)y^{1-n} = q(x) \]
Let \( v = y^{1-n} \leadsto v' = (1-n)y^{-n}y' \leadsto y^{-n}y' = \frac{1}{1-n}v' \)
\[ \frac{1}{1-n}v' + p(x)v = q(x), \quad (n \neq 0, 1) \] \[ v' + (1-n)p(x)v = (1-n)q(x) \qquad \text{Linear ODE} \]
eg:
\[ y' + 2y = e^x y^2 \]
divide with \( y^2 \)
\[ y^{-2} y' + 2 y^{-1} = e^x \]
Let \( v = y^{-1} \implies v' = -1 y^{-2} y' \implies y^{-2} y' = -v' \)
\[ -v' + 2v = e^x \] \[ v'(x) - 2v(x) = -e^x \quad \text{linear equation} \]
integrating factor, \( e^{\int -2 \, dx} = e^{-2x} \)
Multiply with \( \mu(x) = e^{-2x} \) on both sides
\[ e^{-2x} v' - 2e^{-2x} v = -e^{-2x} \cdot e^x \] \[ (e^{-2x} v)' = -e^{-x} \]
integrate both sides
\[ e^{-2x} \cdot v = \int -e^{-x} \, dx + c \] \[ e^{-2x} v(x) = +e^{-x} + c \] \[ v(x) = +e^x + c e^{2x} \] \[ y(x) = \frac{1}{v(x)} = \frac{1}{e^x + c e^{2x}} \]
Eg:
\[ y' - \frac{3}{2} x y = 2 x y^{-1} \]
Multiply with \( y \) on both sides
\[ y y' - \frac{3}{2} x y^2 = 2x \]
\( v = y^2 \implies v' = 2 y y' \implies y y' = \frac{1}{2} v' \)
\[ \frac{1}{2} v' - \frac{3}{2} x v = 2x \] \[ \implies v' - 3x v = 4x \]
integrating factor, \( \mu(x) = e^{-\int 3x \, dx} = e^{\frac{-3x^2}{2}} \)
\[ v' e^{-3x^2/2} - 3x e^{-3x^2/2} v = 4x e^{-3x^2/2} \]
\[ \left(e^{-3x^2/2} v\right)' = 4x e^{-3x^2/2} \]
integrate both sides
\[ e^{-3x^2/2} v = \int 4x e^{-3x^2/2} \, dx + C \] \[ = -\frac{4}{3} e^{-3x^2/2} + C \] \[ v(x) = -\frac{4}{3} + C e^{3x^2/2} \] \[ v = y^2 \implies y(x) = \pm \sqrt{-\frac{4}{3} + C e^{3x^2/2}} \]