Calculus
11th
New unit on applications of integrals starts with integrating differential equations with two variables.
Method: Separation of Variables.
Move all y's to the left side and all x's to the right side.
Structure: y's and dy's on the left; x's and dx's on the right.
Integrate left side with respect to y and right side with respect to x.
Isolate y by performing necessary algebraic manipulations.
Start with ( y \frac{dy}{dx} = 2x ).
Multiplying both sides by dx: ( y \frac{dy}{dx} dx = 2x dx ).
Resulting in ( y dy = 2x dx ).
Integrate both sides:
( \int y dy = \int 2x dx )
Left side: ( \frac{y^2}{2} + C_1 )
Right side: ( x^2 + C_2 )
Combine constants to get ( \frac{y^2}{2} = x^2 + C ).
Final form: ( y = \sqrt{2x^2 + C} ).
Rearranging: ( dy = y^2 dx )
Dividing by y^2: ( \frac{1}{y^2} dy = dx ).
Integrating gives: ( -\frac{1}{y} = x + C ).
Isolating y leads to: ( y = -\frac{1}{x + C} ).
Equation: ( 3y^2 \frac{dy}{dx} = x + \sin(x) ).
Rearranging: ( 3y^2 dy = (x + \sin(x)) dx ).
When integrated:
Left: ( y^3 + C_1 )
Right: ( \frac{1}{2}x^2 - \cos(x) + C_2 )
Isolating y results in: ( y = \sqrt[3]{\frac{1}{2}x^2 - \cos(x) + C} ).
Given: ( dy = xy dx ) leads to ( \frac{1}{y} dy = x dx ).
Integration yields: ( \ln(y) = \frac{1}{2}x^2 + C ).
To isolate y, exponentiate both sides:
( y = e^{\frac{1}{2}x^2 + C} ).
Result: ( y = C e^{\frac{1}{2}x^2} ) where C is a constant.