1. Exact Equations
An ordinary differential equation (ODE) of the form \(M(x,y)dx + N(x,y)dy = 0\) is considered an exact equation if it satisfies the condition: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \quad (M_y = N_x) \]
If this condition is met, there exists a potential function \(F(x,y)\) such that the total differential is: \[ dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy \] The implicit solution to the ODE is always \(F(x,y) = C\).
Example: Finding an Implicit Solution
Solve: \((6xy - y^3)dx + (4y + 3x^2 - 3xy^2)dy = 0\)
- Check for exactness:
\(M = 6xy - y^3 \implies M_y = 6x - 3y^2\)
\(N = 4y + 3x^2 - 3xy^2 \implies N_x = 6x - 3y^2\)
Since \(M_y = N_x\), the equation is exact. - Find \(F(x,y)\):
Integrate \(M\) with respect to \(x\): \[ F = \int (6xy - y^3) dx = 3x^2y - xy^3 + g(y) \] - Solve for \(g(y)\):
Differentiate \(F\) with respect to \(y\) and set it equal to \(N\): \[ \frac{\partial F}{\partial y} = 3x^2 - 3xy^2 + g'(y) = 4y + 3x^2 - 3xy^2 \] \[ g'(y) = 4y \implies g(y) = 2y^2 + C_1 \] - Final Implicit Solution: \[ 3x^2y - xy^3 + 2y^2 = C \]
2. Reducible Second-Order Equations
Some second-order equations can be reduced to first-order equations by substitution. These typically yield solutions with two unknown constants (\(C_1, C_2\)).
Type A: "y" is missing
Equations where \(x, y', \text{ and } y''\) appear, but \(y\) does not.
Substitution: Let \(v = y'\), then \(y'' = v'\).
Solve: \(xy'' + y' = x\) (given \(x > 0\))
Substituting \(v = y'\) gives the 1st order linear equation: \(v' + \frac{1}{x}v = 1\).
Using the integrating factor \(\rho = e^{\int \frac{1}{x} dx} = x\): \[ xv = \int x dx \implies xv = \frac{x^2}{2} + C_1 \implies v = \frac{x}{2} + \frac{C_1}{x} \] Since \(v = y'\), integrate again: \[ y = \int \left( \frac{x}{2} + \frac{C_1}{x} \right) dx = \frac{x^2}{4} + C_1 \ln x + C_2 \]
Type B: "x" is missing
Equations where \(y, y', \text{ and } y''\) appear, but \(x\) does not.
Substitution: Let \(y' = v\). To replace \(y''\), use the chain rule:
\[ y'' = \frac{dv}{dx} = \frac{dv}{dy} \cdot \frac{dy}{dx} = v \frac{dv}{dy} \]Solve: \(y'' = -25y\) (assuming \(y, y' > 0\))
Substitute \(y'' = v \frac{dv}{dy}\): \[ v \frac{dv}{dy} = -25y \implies \int v dv = \int -25y dy \] \[ \frac{1}{2}v^2 = -\frac{25}{2}y^2 + C_1 \implies v = \sqrt{C_1 - 25y^2} \] Since \(v = \frac{dy}{dx}\), separate variables: \[ \int \frac{dy}{\sqrt{C_1 - 25y^2}} = \int dx \] Using the identity \(\int \frac{dx}{\sqrt{k^2 - x^2}} = \sin^{-1}(\frac{x}{k}) + C\): \[ y = C_1 \sin(5x + C_2) \]