Example 1: Solving via Substitution
Solve the ODE: $$\frac{dy}{dx} = (4x + y)^2$$
Step 1: Substitution
Set $v = 4x + y$. By isolating $y$, we get $y = v - 4x$.
Step 2: Implicit Differentiation
Differentiating $y$ with respect to $x$: $$\frac{dy}{dx} = \frac{dv}{dx} - 4$$
Step 3: Substitute back into original ODE
$$\frac{dv}{dx} - 4 = v^2 \implies \frac{dv}{dx} = v^2 + 4$$
This is now a separable equation.
Step 4: Solve for $v$
$$\int \frac{dv}{v^2 + 4} = \int dx$$
Using the arctan integral formula: $$\frac{1}{2} \tan^{-1}\left(\frac{v}{2}\right) = x + C$$
$$\tan^{-1}\left(\frac{v}{2}\right) = 2x + C \implies v = 2 \tan(2x + C)$$
Step 5: "Un-substitute"
Since $y = v - 4x$: $$y = 2 \tan(2x + C) - 4x$$
General Rule: For equations of the form $y' = f(ax + by + c)$, use the substitution $v = ax + by + c$ to make the ODE separable.
Homogeneous Equations
Definition: An equation is "homogeneous" if changing $x \to \epsilon x$ and $y \to \epsilon y$ and simplifying yields the original equation.
Example 2: Implicit Solution
Find an implicit solution for: $$\frac{dy}{dx} = \frac{2x - y}{x + 7y}$$
Verification: $$\frac{d(\epsilon y)}{d(\epsilon x)} = \frac{2(\epsilon x) - \epsilon y}{\epsilon x + 7\epsilon y} = \frac{\epsilon(2x - y)}{\epsilon(x + 7y)} = \frac{2x - y}{x + 7y}$$ Since the $\epsilon$ cancels, it is homogeneous.
Method: For homogeneous equations, set $v = \frac{y}{x}$ (or $y = vx$).
Using implicit differentiation: $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$
Substitution: $$v + x\frac{dv}{dx} = \frac{2x - vx}{x + 7vx} = \frac{2 - v}{1 + 7v}$$ $$x\frac{dv}{dx} = \frac{2 - v}{1 + 7v} - v = \frac{2 - v - v(1 + 7v)}{1 + 7v}$$ $$x\frac{dv}{dx} = \frac{2 - 2v - 7v^2}{1 + 7v}$$
Continuing Example 2 (Solving the Separable Equation):
$$\int \frac{1 + 7v}{2 - 2v - 7v^2} dv = \int \frac{1}{x} dx$$
Let $u = 2 - 2v - 7v^2$, then $du = (-2 - 14v)dv = -2(1 + 7v)dv$.
The integral becomes: $$-\frac{1}{2} \ln|u| = \ln|x| + C$$
$$-\frac{1}{2} \ln|2 - 2v - 7v^2| = \ln|x| + C$$
Multiplying by -2 and exponentiating: $$|2 - 2v - 7v^2| = |x|^{-2} \cdot e^C \implies 2 - 2v - 7v^2 = \frac{C}{x^2}$$
Final Step (Replace $v = y/x$):
$$2 - \frac{2y}{x} - \frac{7y^2}{x^2} = \frac{C}{x^2}$$
Multiplying by $x^2$ to get an implicit solution $F(x,y) = C$: $$2x^2 - 2xy - 7y^2 = C$$
Bernoulli Equations
Form: $y' + P(x)y = Q(x)y^n$ where $n \in \mathbb{R}$.
- If $n=0$, it is a 1st order linear equation.
- If $n=1$, it is linear and separable.
- Otherwise, use substitution: $v = y^{1-n}$. This transforms the ODE into a 1st order linear equation.
Example 3: Bernoulli ODE
Solve: $x^2y' + xy = y^2$
Standard form ($y' + P y = Q y^n$): $$y' + \frac{1}{x}y = \frac{1}{x^2}y^2$$
Here $n=2$, so $v = y^{1-2} = y^{-1}$. This implies $y = v^{-1}$ and $y' = -v^{-2}v'$.
Substitute: $$-v^{-2}v' + \frac{1}{x}v^{-1} = \frac{1}{x^2}v^{-2}$$
Multiply by $-v^2$: $$v' - \frac{1}{x}v = -\frac{1}{x^2}$$
Solve using Integrating Factor ($\rho$): $$\rho = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$$ $$\frac{1}{x}v = \int \left(\frac{1}{x} \cdot -\frac{1}{x^2}\right) dx = \int -x^{-3} dx = \frac{1}{2}x^{-2} + C$$ $$v = \frac{1}{2x} + Cx$$
Un-substitute: $$y = v^{-1} = \frac{1}{\frac{1}{2x} + Cx} = \frac{2x}{1 + 2Cx^2}$$
Example 4: General Solution (Homogeneous)
Solve: $x^2y' = xy + x^2e^{y/x}$
Isolate $y'$: $y' = \frac{y}{x} + e^{y/x}$
Set $v = y/x \implies y = vx$ and $y' = v + x\frac{dv}{dx}$.
Sub: $$v + x\frac{dv}{dx} = v + e^v \implies x\frac{dv}{dx} = e^v$$
Solve: $$\int e^{-v} dv = \int \frac{1}{x} dx \implies -e^{-v} = \ln|x| + C$$
$$e^{-v} = C - \ln|x| \implies -v = \ln(C - \ln|x|)$$ $$v = -\ln(C - \ln|x|)$$
Final Solution: $$y = -x \ln(C - \ln|x|)$$