First-Order Differential Equations

Every equation so far has constrained a number. $x^2-3x-10=0$ asks which values of $x$ make it true, and the answer is a list.

${\displaystyle \frac{dy}{dx}}=2xy$ asks something different. It constrains a function by relating it to its own rate of change, and the answer is a function, here $y=3e^{x^2}$, which can be checked: differentiating gives $y^{\prime }=6x{e}^{x^2}$, and $2xy=2x\cdot 3e^{x^2}$ is the same thing.

That shift matters because rates are how most quantitative laws are stated. "Grows at a rate proportional to its size", "cools at a rate proportional to the temperature difference", "the current adjusts until the voltage balances". Each is a sentence about a derivative, and writing it down produces a differential equation rather than an algebraic one.

The answer is a family rather than a single function, because knowing how fast something changes does not say where it started. The general solution carries an arbitrary constant, and an initial condition picks out one member.

Two methods handle the first-order cases. If the rate factors into an $x$-part and a $y$-part, the variables separate onto opposite sides and each integrates on its own. If the equation is linear in $y$, multiplying by a chosen factor turns the left side into a product's derivative, which collapses under integration.

Both reduce the problem to integration, so the techniques from that unit do the real work, and every answer can be verified by differentiating it back.

Start from the simplest case: ${\displaystyle \frac{dy}{dx}}=2x$. This asks which functions have derivative $2x$, and the integration unit already answered it: $y=x^2+C$, a family with one arbitrary constant.

That constant is not a loose end. The equation specifies the slope at every point and says nothing about height, so translating a solution vertically produces another solution. The family is the set of all those translates, and picking one requires information the equation does not contain.

An initial condition supplies it. Requiring $y(0)=3$ forces $C=3$, selecting one curve from infinitely many. One arbitrary constant, one condition, which is why a first-order equation needs exactly one.

Why the general case is harder. ${\displaystyle \frac{dy}{dx}}=2xy$ cannot be integrated directly, because the right side involves the unknown $y$. Antidifferentiating requires knowing $y$, which is what is being sought.

Separation is the move that breaks the circle. Dividing by $y$ puts every $y$ on one side and every $x$ on the other:

and now each side integrates in its own variable, giving $\ln |y|=x^2+C$ and hence $y=A{e}^{x^2}$. With $y(0)=3$ the constant is $A=3$, and the solution checks: $y^{\prime }/y$ evaluated numerically at $x=0,0.5,1$ gives $0$, $1.000000$, $2.000000$, exactly $2x$.

What separation costs. Dividing by $y$ assumed $y\ne 0$. But $y=0$ is itself a solution, since the zero function has zero derivative and $2x\cdot 0=0$, and the division discarded it. The general form $y=A{e}^{x^2}$ recovers it at $A=0$ in this case, but that is luck rather than method: the constant solutions where the divisor vanishes must be checked separately.

Why the second method exists. ${\displaystyle \frac{dy}{dx}}=x+y$ resists separation entirely, since $x+y$ cannot be written as $f(x)g(y)$. Rearranged as $y^{\prime }-y=x$ it is linear, and multiplying by $e^{-x}$ makes the left side $(e^{-x}y{)}^{\prime }$, a single derivative, which integration undoes. The two methods answer two different shapes, and neither subsumes the other.

Why separation needs a product $f(x)g(y)$. The step that makes the method work is writing $\frac{dy}{g(y)}=f(x)dx$, with every $y$ on the left and every $x$ on the right. That is possible exactly when the right-hand side factors. A sum such as $x+y$ cannot be split this way, because dividing by $y$ leaves $\frac{x}{y}+1$, which still mixes the variables.

Testing for separability is therefore a question about factoring, not about difficulty: $2xy$ factors as $(2x)(y)$, $y/x$ factors as $(1/x)(y)$, $y^2$ factors as $(1)(y^2)$, and $x+y$ does not.

Why the integrating factor is $e^{\int P}$ and nothing else. Multiplying $y^{\prime }+Py=Q$ by $\mu$ gives $\mu y^{\prime }+\mu Py$, and the aim is for that to be $(\mu y{)}^{\prime }=\mu y^{\prime }+{\mu }^{\prime }y$. Matching the two requires

which is itself a separable equation with solution $\mu =e^{\int Pdx}$. The factor is not guessed. It is forced by the requirement that the product rule apply.

The constant of integration in $\int P$ is omitted deliberately: including it multiplies $\mu$ by a constant, which cancels when both sides are later divided by $\mu$.

Why standard form matters. $\mu$ is built from $P$, the coefficient of $y$ when the coefficient of $y^{\prime }$ is 1. For $2y^{\prime }+4y=8$ the correct $P$ is 2, obtained after dividing through, using 4 gives $\mu =e^{4x}$, which does not satisfy ${\mu }^{\prime }=\mu P$ and collapses nothing.

Why the constant enters at integration. Writing $\ln |y|=x^2$ and appending a constant afterwards gives $y=e^{x^2}+C$, which does not solve the equation: differentiating gives $2x{e}^{x^2}$, while $2xy=2x{e}^{x^2}+2xC$. The constant must be introduced where it arises, inside the logarithm's argument after exponentiating, yielding $y=A{e}^{x^2}$.

Why constant solutions can be lost. Separation divides by $g(y)$, which is invalid where $g(y)=0$. Each root $y=c$ of $g$ gives a constant function with $y^{\prime }=0=f(x)g(c)$. A genuine solution that the division removed. For $y^{\prime }=y(1-y)$ the roots $y=0$ and $y=1$ are both solutions, and neither is produced by the separated integral.

Why verification is always available. A solution is a function making the equation an identity, so substituting it and differentiating settles the matter without reference to how it was found. For $y=3-2e^{-2x}$ and $y^{\prime }+2y=6$, evaluating at $x=0,0.5,1,3$ gives $6.000000$ each time. No other kind of equation offers so direct a check.

Step 1 — classify before solving. Look at the right-hand side of $\frac{dy}{dx}=F(x,y)$:

An equation can be both: $2xy$ is $(2x)(y)$ and also $y^{\prime }-2xy=0$. It can be one and not the other: $x+y$ is linear only; $y^2$ is separable only.

Step 2 — separate and integrate.

1. Write ${\displaystyle \frac{dy}{g(y)}}=f(x)dx$.
2. Note the roots of $g$ before dividing. Each $y=c$ with $g(c)=0$ is a constant solution.
3. Integrate both sides, adding one constant on the right.
4. Solve for $y$, absorbing the constant into its final form, so that $e^{x^2+C}$ becomes $A{e}^{x^2}$.
5. Restore any constant solutions from step 2 that the general form does not already include.

Step 3 — integrating factor.

1. Put it in standard form: divide until the coefficient of $y^{\prime }$ is 1, giving $y^{\prime }+P(x)y=Q(x)$.
2. Compute $\mu =e^{\int Pdx}$, omitting the constant of integration.
3. Multiply through. The left side is now $(\mu y{)}^{\prime }$; verify this if in doubt by differentiating the product.
4. Integrate: $\mu y=\int \mu Qdx+C$.
5. Divide by $\mu$ to isolate $y$.

Step 4 — apply the initial condition. Only after the general solution is complete. Substitute $x_0$ and $y_0$, solve for the constant, and write the particular solution.

Applying the condition to an intermediate expression fixes the wrong constant.

Step 5 — verify. Differentiate the answer and substitute it into the original equation; it must reduce to an identity. Check the initial condition separately by evaluating at $x_0$.

This check is decisive and cheap. For $y=3-2e^{-2x}$ solving $y^{\prime }+2y=6$, the left side evaluates to $6.000000$ at $x=0,0.5,1,3$, and $y(0)=1$.

For a linear equation with constant coefficients, the solution splits into a steady state and a transient that decays. In $y=3+C{e}^{-2x}$ the 3 is what remains as $x$ grows, reaching $2.995$ by $x=3$, and the exponential is what the initial condition controls.

1. Separable: ${\displaystyle \frac{dy}{dx}}=2xy$ with $y(0)=3$.

Classify. The right side factors as $(2x)(y)$, so it is separable. (It is also linear, as $y^{\prime }-2xy=0$, either method works.)

Note the constant solution. $g(y)=y$ vanishes at $y=0$, so $y=0$ is a solution that the division will discard.

Separate and integrate.

The constant entered at the integration and was absorbed as $A=\pm e^C$. Note $A=0$ recovers the discarded $y=0$, so nothing is lost here.

Apply the condition. $y(0)=A=3$, giving $y=3e^{x^2}$.

Verify. Numerically $y^{\prime }/y$ at $x=0,0.5,1$ gives $0$, $1.000000$, $2.000000$, exactly $2x$. And $y(0)=3$.

2. Separable: ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y}{x}}$ with $y(1)=4$.

The right side factors as $\left(\frac{1}{x}\right)(y)$.

With $y(1)=4$: $A=4$, so $y=4x$.

Verify. $y^{\prime }=4$ and $\frac{y}{x}=\frac{4x}{x}=4$ at $x=1,2,5$. The solutions are lines through the origin, which the equation predicts, since the slope equals the ratio $y/x$.

3. Linear: $y^{\prime }+2y=6$ with $y(0)=1$.

Classify. Already in standard form, with $P=2$ and $Q=6$. Not separable as written, but $6-2y$ does factor, so it happens to be both; the linear route is shown.

Integrating factor. $\mu =e^{\int 2dx}=e^{2x}$. Multiplying through:

Integrate and divide. $e^{2x}y=3e^{2x}+C$, so $y=3+C{e}^{-2x}$.

Apply the condition. $y(0)=3+C=1$ gives $C=-2$, so $y=3-2e^{-2x}$.

Verify. $y^{\prime }+2y$ evaluates to $6.000000$ at $x=0,0.5,1,3$, and $y(0)=1$.

Read it. The solution is a steady state 3 plus a transient $-2e^{-2x}$ that decays: at $x=3$ the value is already $2.995$, and as $x$ grows it approaches 3 exactly.

4. Linear but not separable: ${\displaystyle \frac{dy}{dx}}=x+y$ with $y(0)=0$.

Classify. $x+y$ does not factor as $f(x)g(y)$, separation is unavailable. Rearranged, $y^{\prime }-y=x$, which is linear with $P=-1$.

Integrating factor. $\mu =e^{\int (-1)dx}=e^{-x}$:

Integrate. The right side needs parts, with $u=x$ and $dv=e^{-x}dx$:

so $e^{-x}y=-x{e}^{-x}-e^{-x}+C$ and $y=C{e}^x-x-1$.

Apply the condition. $y(0)=C-1=0$ gives $C=1$, so $y=e^x-x-1$.

Verify. At $x=0,1,2$ the derivative is $0$, $1.718282$, $6.389056$, and $x+y$ is $0$, $1.718282$.

This is the case that shows why two methods are needed: no amount of rearranging makes $x+y$ separable.

${\displaystyle \frac{dy}{dt}}=ky$: 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.