02 Exact DE
Family of Curves¶
\[ \begin{aligned} f(x, y) &= c \\ d( \ f(x, y) \ ) &= d(c) \\ f_x dx + f_y dy &= 0 \end{aligned} \]
This last step
- looks like Homogeneous Equation
- is the exact differential equation of the given curve
The solution is the given equation, and from that we derived the exact differential equation.
Solution of a DE¶
Consider a first-order differential equation of the form
\[ M dx + N dy = 0 \]
if there happens to be a function \(f(x, y)\) such that
\[ \begin{aligned} f_x = M(x, y), \quad f_y &= N(x, y) \\ \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy &= 0 \\ d( f(x, y) ) &= 0 \\ f(x, y) &= c \end{aligned} \]
The final step is the general solution of the given differential equation.
Exact DE¶
is a differential equation where \(M_y = N_x\)
Shortcut Method for Exact DE¶
This is only for exact DE
Consider this DE
\[ (\ y + y \cos(xy) \ )dx + (\ x + x \cos(xy) \ ) = 0 \]
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Check if the given DE is exact
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Put integration sign for both sides
\[ \int (\ y + y \cos(xy) \ ) dx + \int(\ x + x \cos(xy) \ ) dx = \int 0 \]
- Simplifications
- Treat \(y\) as a constant in the \(dx\) integral
- Drop all terms containing \(x\) in the \(dy\) integral think like this: drop your ex example
- \(y \cos(x) \to 0\)
- \(x \cos(x) \to 0\)
- \(y + y \cos(x) \to y\)
\[ \int (\ y + y \cos(xy) \ ) dx + \int (0 + 0) dy = c \]
- Integrate
\[ \begin{aligned} yx + y \left( \frac{ \sin xy }{ y } \right) &= c \\ yx + \sin(xy) &= c \end{aligned} \]
Exact DE Formulae¶
\[ \begin{aligned} d(xy) &= xdy + y dx \\ d(x^2 + y^2) &= 2x dx + 2y dy \\ d \left(\frac{x^2 + y^2}{2} \right) &= x dx + y dy \end{aligned} \]
\[ \begin{aligned} d\left(\frac{x}{y}\right) &= \frac{ydx - xdy}{y^2} \quad \left(\frac{u}{v} \right)' \text{ formula}\\ &= \frac{1}{y}dx - \frac{x}{y^2} dy \\ d\left(\frac{y}{x}\right) &= \frac{xdy - ydx}{x^2} \\ &= \frac{1}{x}dy - \frac{y}{x^2} dx \end{aligned} \]
\[ \begin{aligned} d\left(\log{ |\frac{x}{y}| }\right) &= \frac{1}{\frac{x}{y}} \left( \frac{y dx - x dy}{y^2} \right) \\ &= \frac{y dx - x dy}{xy} \\ \end{aligned} \]
\[ \begin{aligned} d\left( \log \vert \frac{y}{x} \vert \right) &= \frac{x dy - y dx}{xy} \end{aligned} \]
\[ \begin{aligned} d \left( \tan^{-1} \frac{x}{y} \right) &= \frac{1}{1 + \frac{x^2}{y^2} } \left( \frac{y dx - x dy}{y^2} \right) \\ &= \frac{y dx - x dy}{x^2 + y^2} \\ d \left( \tan^{-1} \frac{y}{x} \right) &= \frac{x dy - ydx}{x^2 + y^2} \end{aligned} \]
IDK¶
You cannot integrate \(\int f(x,y) \ dx\) wrt to \(dx\) alone
it is only possible for something sir said and double integration (there \(dy\) will also be there in outer integral)