First-Order Differential Equations

${\displaystyle \frac{dy}{dt}}=ky$: exponential growth and decay. Separable, with solution $y=y_0{e}^{kt}$. Positive $k$ grows, negative $k$ decays. For a half-life of 5, $k=-\frac{\ln 2}{5}=-0.13862944$, and the ratio $y/y_0$ reads $1$, $0.5$, $0.25$ at $t=0,5,10$.

Radioactive decay, continuously compounded interest, unconstrained population growth and first-order chemical kinetics are all this equation.

${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y}{x}}$: solutions are lines through the origin. $y=Ax$, since the equation says the slope equals the ratio $y/x$, which is exactly the property of a line through the origin. With $y(1)=4$, the solution $y=4x$ has $y^{\prime }=4=y/x$ at $x=1,2,5$.

$y^{\prime }+2y=6$: steady state plus transient. Solution $y=3+C{e}^{-2x}$. Whatever the initial value, the solution approaches 3: at $x=3$ it is $2.995$. The constant 3 is where $y^{\prime }=0$, the equilibrium the system settles into, and $C{e}^{-2x}$ is the memory of the start, decaying away.

Newton's law of cooling has this shape, with 3 replaced by the ambient temperature.

${\displaystyle \frac{dy}{dx}}=x+y$: linear, not separable. Solution $y=C{e}^x-x-1$. The particular part $-x-1$ tracks the forcing term and the $C{e}^x$ is the homogeneous solution. Verified at $x=0,1,2$: the derivative and $x+y$ agree at $0$, $1.718282$.

$y^{\prime }=y(1-y)$: the logistic equation. Separable, with general solution $y={\displaystyle \frac{1}{1+A{e}^{-x}}}$. Checked at $A=1$ and $A=4$ across $x=0,2,5$, the derivative matches $y(1-y)$ to eight decimals.

It has two constant solutions, $y=0$ and $y=1$, both found by inspection rather than by the method: separation divides by $y(1-y)$, which vanishes at both. The general form recovers $y=1$ at $A=0$ but never reaches $y=0$ for any finite $A$, so that solution must be restored by hand.

This is constrained population growth, where 1 is the carrying capacity.

$y^{\prime }=y^2$: a solution that escapes. Separable, giving $y={\displaystyle \frac{1}{C-x}}$; with $y(0)=1$, $y={\displaystyle \frac{1}{1-x}}$. Checked at $x=0,0.5,0.9,0.99$, the derivative matches $y^2$, and the values climb $1$, $2$, $10$, $100$.

The solution blows up as $x\to 1$, in finite $x$, even though the right-hand side $y^2$ is perfectly well behaved everywhere. A solution need not exist for all $x$.

Five of the six are separable and three are also linear. Equilibrium, carrying capacity and finite-time blow-up are visible in the solution rather than in the equation.

A solution is a function, not a number. Asked to solve $\frac{dy}{dx}=2xy$, reporting " $x=0$" or " $y=3$" answers a question that was not posed. The unknown is the function itself, and $y=3e^{x^2}$ is a complete answer, verified by differentiating: $y^{\prime }=6x{e}^{x^2}$ and $2xy=6x{e}^{x^2}$.

Attempting separation on a sum. $\frac{dy}{dx}=x+y$ cannot be separated, because $x+y$ does not factor as $f(x)g(y)$. Dividing by $y$ gives $\frac{x}{y}+1$, which still contains both variables. The equation is linear and must be solved that way, giving $y=C{e}^x-x-1$ at $x=0,1,2$.

The test is factoring, not effort: $2xy$ factors, $y/x$ factors, $y^2$ factors, $x+y$ does not.

Computing $\mu$ from the wrong $P$. For $2y^{\prime }+4y=8$, the standard form is $y^{\prime }+2y=4$, so $P=2$ and $\mu =e^{2x}$. Taking $P=4$ from the undivided equation gives $\mu =e^{4x}$, and the left side then fails to collapse, $(e^{4x}y{)}^{\prime }=e^{4x}{y}^{\prime }+4e^{4x}y$, which is not what multiplying produced. Dividing to standard form first is not tidying; it is what makes the method work.

Adding the constant after solving for $y$. From $\ln |y|=x^2$, writing $y=e^{x^2}+C$ gives a function that does not solve $y^{\prime }=2xy$: its derivative is $2x{e}^{x^2}$, while $2xy=2x{e}^{x^2}+2xC$. The constant must enter at the integration and be carried through the exponentiation, giving $y=A{e}^{x^2}$.

Losing the constant solutions. For the logistic equation $y^{\prime }=y(1-y)$, separation divides by $y(1-y)$, which vanishes at $y=0$ and $y=1$. Both are genuine solutions: each is constant, so $y^{\prime }=0$, and $y(1-y)$ is 0 at both.

The general solution $y=\frac{1}{1+A{e}^{-x}}$ recovers $y=1$ at $A=0$, but no finite $A$ gives $y=0$. That solution exists and the method cannot produce it, which is why the roots of the divisor are noted before dividing.

Assuming a solution exists for all $x$. $y^{\prime }=y^2$ with $y(0)=1$ has solution $y=\frac{1}{1-x}$, whose values run $1$, $2$, $10$, $100$ at $x=0,0.5,0.9,0.99$ and which ceases to exist at $x=1$. The equation is smooth everywhere; the solution still escapes in finite $x$. Existence is local, and "solve the equation" does not come with a guarantee of a global answer.

Assuming uniqueness. $y^{\prime }=y^{1/3}$ with $y(0)=0$ has both $y=0$ and $y=(2x/3{)}^{3/2}$ passing through the origin. A method that returns one of them has not found them all, and uniqueness requires the extra hypothesis on $\partial F/\partial y$ that fails here.

Applying the initial condition too early. Substituting $y(0)=3$ into $\ln |y|=x^2+C$ before exponentiating gives $C=\ln 3$, which is correct, but doing the same to an intermediate expression that is not yet the general solution fixes a constant in the wrong place. The condition belongs at the end, applied to the finished general solution.

Solving a differential equation means finding a value of $x$ or $y$, as when solving an algebraic equation, so an answer such as $y=3e^{x^2}$ is incomplete.