First-Order ODEs

A first-order ODE $\frac{dy}{dx} = f(x, y)$ is separable if $f(x, y) = g(x)h(y)$.

$\frac{dy}{dx} = g(x)h(y) \implies \frac{dy}{h(y)} = g(x)\, dx$

Integrating both sides: $\int \frac{dy}{h(y)} = \int g(x)\, dx + C$.

Example. Solve $\frac{dy}{dx} = xy$.

Separating: $\frac{dy}{y} = x\, dx$. Integrating: $\ln|y| = \frac{x^2}{2} + C$. Thus $y = Ce^{x^2/2}$ where $C \neq 0$Plus the trivial solution $y = 0$.

A linear first-order ODE has the form

$\frac{dy}{dx} + P(x)y = Q(x)$

Theorem 2.1 (Integrating Factor). The solution is

$y(x) = e^{-\int P(x)\, dx}\left(\int Q(x) e^{\int P(x)\, dx}\, dx + C\right)$

Proof. Multiply both sides by $\mu(x) = e^{\int P(x)\, dx}$:

$\frac{d}{dx}(\mu y) = \mu \frac{dy}{dx} + \mu P y = \mu \frac{dy}{dx} + \mu" y = \mu\left(\frac{dy}{dx} + Py\right) = \mu Q$

Integrating: $\mu y = \int \mu Q\, dx + C$. Solving for $y$ gives the result. $\blacksquare$

Problem. Solve $y' + \frac{2}{x}y = x^2$ for $x > 0$.

Solution. $P(x) = 2/x$, $Q(x) = x^2$.

$\mu(x) = e^{\int 2/x\, dx} = e^{2\ln x} = x^2$.

$y = x^{-2}\left(\int x^2 \cdot x^2\, dx + C\right) = x^{-2}\left(\frac{x^5}{5} + C\right) = \frac{x^3}{5} + \frac{C}{x^2}$. $\blacksquare$

The ODE $M(x, y)\, dx + N(x, y)\, dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

When exact, there exists $\Psi(x, y)$ such that $\frac{\partial \Psi}{\partial x} = M$ and $\frac{\partial \Psi}{\partial y} = N$And the solution is $\Psi(x, y) = C$.

Theorem 2.2. If $M$ and $N$ have continuous partial derivatives on a connected domain $D$Then $M\, dx + N\, dy = 0$ is exact if and only if $M_y = N_x$.

Proof. If exact, $M = \Psi_x$ and $N = \Psi_y$So $M_y = \Psi_{xy} = \Psi_{yx} = N_x$ by Clairaut. Conversely, if $M_y = N_x$Define $\Psi(x, y) = \int_{x_0}^x M(t, y)\, dt + \int_{y_0}^y N(x_0, s)\, ds$. Then $\Psi_x = M(x, y)$ and $\Psi_y = \int_{x_0}^x M_y(t, y)\, dt + N(x_0, y) = \int_{x_0}^x N_x(t, y)\, dt + N(x_0, y) = N(x, y) - N(x_0, y) + N(x_0, y) = N(x, y)$. $\blacksquare$

Problem. Solve $(2xy + 3)\, dx + (x^2 - 1)\, dy = 0$.

Solution. $M = 2xy + 3$, $N = x^2 - 1$. Check: $M_y = 2x = N_x$. Exact.

$\Psi_x = 2xy + 3 \implies \Psi = x^2 y + 3x + h(y)$.

$\Psi_y = x^2 + h'(y) = x^2 - 1 \implies h'(y) = -1 \implies h(y) = -y$.

Solution: $x^2 y + 3x - y = C$. $\blacksquare$

If $M_y \neq N_x$One can sometimes find an integrating factor $\mu(x, y)$ such that $(\mu M)_y = (\mu N)_x$.

Case 1: If $\frac{M_y - N_x}{N}$ depends only on $x$Then $\mu(x) = e^{\int \frac{M_y - N_x}{N}\, dx}$.

Case 2: If $\frac{N_x - M_y}{M}$ depends only on $y$Then $\mu(y) = e^{\int \frac{N_x - M_y}{M}\, dy}$.

A Bernoulli equation has the form

$\frac{dy}{dx} + P(x)y = Q(x)y^n$

Where $n \neq 0, 1$. The substitution $v = y^{1-n}$ transforms it into a linear equation:

$\frac{dv}{dx} + (1 - n)P(x)v = (1 - n)Q(x)$

Example. Solve $y' + y = y^2 e^x$.

Here $n = 2$So set $v = y^{-1}$. Then $v' = -y^{-2}y'$And the equation becomes $v' - v = -e^x$. Integrating factor: $e^{-x}$. So $(v e^{-x})' = -1$Giving $ve^{-x} = -x + C$, $v = -xe^x + Ce^x$ And $y = 1/(C - x)e^x$.

Theorem 2.3 (Picard-Lindelöf). If $f$ and $\partial f/\partial y$ are continuous on a rectangle Containing $(x_0, y_0)$Then the IVP $y' = f(x, y)$, $y(x_0) = y_0$ has a unique solution in some Neighbourhood of $x_0$.

Several substitutions reduce specific equations to separable or linear form:

1. Homogeneous equations: $y' = f(y/x)$. Set $v = y/x$.
2. Equations of the form $y' = f(ax + by + c)$: set $v = ax + by + c$.

An ODE of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ is called homogeneous (not to be Confused with the linearity sense). The substitution $v = y/x$I.e., $y = vx$Gives $y' = v + xv'$ So the equation becomes:

$v + x\frac{dv}{dx} = F(v) \implies x\frac{dv}{dx} = F(v) - v$

This is separable: $\frac{dv}{F(v) - v} = \frac{dx}{x}$.

Problem. Solve $y' = \frac{x^2 + y^2}{xy}$.

A Riccati equation has the form

$\frac{dy}{dx} = q_0(x) + q_1(x)y + q_2(x)y^2$

If a particular solution $y_1(x)$ is known, the substitution $y = y_1 + \frac{1}{v}$ reduces the Riccati equation to a linear first-order equation in $v$:

$\frac{dv}{dx} = -(q_1 + 2q_2 y_1)v - q_2$

Example. Solve $y' = 1 + x^2 - 2xy + y^2$ given that $y_1 = x$ is a particular solution.

Substituting $y = x + 1/v$: $y' = 1 - v'/v^2$. The equation becomes

$1 - v'/v^2 = 1 + x^2 - 2x(x + 1/v) + (x + 1/v)^2$

$1 - v'/v^2 = 1 + x^2 - 2x^2 - 2x/v + x^2 + 2x/v + 1/v^2$

$1 - v'/v^2 = 1 + 1/v^2$

$-v'/v^2 = 1/v^2 \implies v' = -1$

So $v = -x + C$And $y = x + \frac{1}{C - x}$.

Problem. A body at $90\degree\mathrm{C}$ is placed in a room at $20\degree\mathrm{C}$. After 10 Minutes, its temperature is $60\degree\mathrm{C}$. When will it reach $30\degree\mathrm{C}$?

Problem. A tank contains 100 L of brine with 20 kg of salt. Fresh water enters at 3 L/min and The mixture leaves at 3 L/min. Find the amount of salt after 30 minutes.

Problem. Solve $(3xy + 2y^2)\, dx + (x^2 + 2xy)\, dy = 0$.

Given a one-parameter family of curves $F(x, y, C) = 0$The orthogonal trajectories are curves That intersect every member of the family at right angles. To find them:

1. Find the differential equation $\frac{dy}{dx} = f(x, y)$ of the given family.
2. Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ (equivalently, negate the slope).
3. Solve the new ODE.

Example. Find the orthogonal trajectories of $y = Cx^2$.

$\frac{dy}{dx} = 2Cx = \frac{2y}{x}$.

Orthogonal trajectories satisfy $\frac{dy}{dx} = -\frac{x}{2y}$.

Separating: $2y\, dy = -x\, dx$. Integrating: $y^2 = -\frac{x^2}{2} + C$Or $\frac{x^2}{2} + y^2 = C$. These are ellipses.

:::caution Common Pitfall When separating variables, dividing by $h(y)$ can lose solutions where $h(y) = 0$. Always check whether $h(y) = 0$ yields valid solutions before dividing.

::: :::caution Common Pitfall Not every first-order ODE falls into a standard category. Equations like $y' = e^{x^2} + \sin(y^2)$ cannot be solved by elementary methods and require numerical Techniques.

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