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Splitting a first-order equation into two integrals, fixing the constant with an initial condition, and saying which interval the answer holds on.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to recognise a separable equation, split it into two integrals and evaluate them, use an initial condition to fix the arbitrary constant, recover the constant solutions that dividing removed, and state the interval on which the solution you found actually exists.
You can integrate the standard functions and you have met substitution. You also know from the previous lesson that a first-order equation hands you a slope at every point. This lesson is the first method that turns that slope rule into a formula, and it works exactly when the rule splits cleanly into an $x$ part and a $y$ part.
| Term | What it means |
|---|---|
| Separable | An equation whose right-hand side factors into a function of $x$ times a function of $y$: $y' = g(x)h(y)$. |
| Implicit solution | An answer left as a relation between $x$ and $y$, such as $y^{2} = x^{2} + C$. |
| Explicit solution | An answer solved for $y$ as a formula in $x$. |
| Singular solution | A constant solution that the general formula misses, because dividing by $h(y)$ removed it. |
| Interval of existence | The largest interval containing the initial point on which the solution actually holds. |
| Half-life | For a quantity decaying as $e^{-\lambda t}$, the time $\frac{\ln 2}{\lambda}$ in which it halves. |
An equation is separable when it can be written
$$y' = g(x)\,h(y).$$
Then divide by $h(y)$ and multiply by $dx$:
$$\int \frac{dy}{h(y)} = \int g(x)\,dx,$$
two ordinary integrals with one arbitrary constant between them. The result is usually an implicit solution — a relation between $x$ and $y$ — and it is solved for $y$ afterwards when that is possible.
The initial condition is applied after integrating. Before that there is no constant to fix.
Two things deserve care. Dividing by $h(y)$ assumes $h(y) \ne 0$, and every root of $h$ is itself a constant solution — sometimes one the general formula misses. And the formula you end up with may fail somewhere: the answer is the branch that contains the initial point, and only that branch.
Another way: picture
Think of the slope rule as a product of two dials, one set by $x$ and one by $y$. Separating is unhooking the dials so each can be integrated on its own. When the rule cannot be unhooked — when $x$ and $y$ are added rather than multiplied, as in $y' = x + y$ — no amount of algebra separates it, and the next lesson's method is the one that answers it.
Another way: steps
Dividing by $h(y)$ is only legal where $h(y) \ne 0$, so every root of $h$ has to be checked separately. For $y' = y(3 - y)$ the roots are $y = 0$ and $y = 3$, and both constant functions solve the equation exactly. Neither comes out of the integrated formula for any finite value of the constant.
These are called singular solutions, and losing them is the standard way a correct calculation gives an incomplete answer. In a population model the two lost solutions are extinct and at carrying capacity, which is not a footnote.
The habit worth building is small: before dividing, write down what you are assuming is non-zero, and come back to it at the end.
$y' = \dfrac{x}{y}$ separates to $y\,dy = x\,dx$ and integrates to $y^{2} = x^{2} + C$. Solving for $y$ needs a square root, and the sign of that root is decided by the initial condition: through $(0, 4)$ the answer is $y = \sqrt{x^{2} + 16}$, and through $(0, -4)$ it is the negative root.
That is the general pattern. The implicit relation describes a whole curve — here a hyperbola with two branches — and the initial condition names the branch. Writing $y = \pm\sqrt{x^{2} + C}$ as the answer to an initial value problem is giving two answers where the problem has one.
The whole method rests on one question: does the right-hand side factor into a function of $x$ times a function of $y$? Answering it quickly is a skill of its own, and the usual trap is an equation that factors after a little algebra.
| Equation | Separable? | Why |
|---|---|---|
| $y' = xy + x$ | yes | $xy + x = x(y + 1)$ |
| $y' = e^{x + y}$ | yes | $e^{x + y} = e^{x}e^{y}$ |
| $y' = x + y$ | no | a sum, and no factoring makes it a product |
| $y' = \sin(xy)$ | no | $x$ and $y$ are locked inside one function |
| $y' = \frac{y}{x}$ | yes | $\frac{1}{x}$ times $y$ |
When the answer is yes, the method is the same five moves every time. Factor the right side as $g(x)h(y)$. Divide by $h(y)$ and multiply by $dx$, writing down the roots of $h$ as you do. Integrate both sides and write one constant, at that moment. Solve for $y$ if it can be done, renaming constants as they absorb signs and exponentials. Apply the initial condition and state the interval that contains the initial point.
Every move has a reason. Dividing by $h(y)$ is legal only where $h(y) \ne 0$, which is why its roots are written down. The single constant is there because the two constants from the two integrals would combine into one anyway. The condition comes last because before integrating there is nothing for it to fix. And the interval comes from the formula: a denominator that reaches zero, a logarithm of something that reaches zero, a square root of something that turns negative, or a tangent whose angle reaches a right angle all end the solution.
How to check the answer. Differentiate your $y$ and substitute it back into the original equation; both sides must agree for every $x$ in the interval. Then put the initial point into the formula and confirm the value. Finally, go back to the roots of $h$ you wrote down: each is a constant solution, and you should be able to say whether your formula already contains it (as $y = 0$ is contained in $y = Ae^{x^{3}}$ at $A = 0$) or whether it has to be listed on its own. If the check fails, the most common causes are a constant added after exponentiating instead of before, and a sign lost when an absolute value was removed.
Every radioactive nucleus decays at random, but a large sample decays at a rate proportional to how much is left: $N' = -\lambda N$. That equation is separable, and separating it is how carbon dating works.
Divide by $N$ and integrate: $\ln N = -\lambda t + C$, so $N = N_0e^{-\lambda t}$ with $N_0$ the amount at $t = 0$. The half-life $t_{1/2}$ is the time for $N$ to halve: $\frac{1}{2} = e^{-\lambda t_{1/2}}$, so $t_{1/2} = \frac{\ln 2}{\lambda}$. For carbon-14, $t_{1/2} = 5730$ years, which gives $\lambda = \frac{0.6931}{5730} \approx 1.21 \times 10^{-4}$ per year.
A piece of charcoal from a hearth holds a quarter of the carbon-14 a living tree holds. Solve $\frac{1}{4} = e^{-\lambda t}$: taking logarithms, $t = \frac{\ln 4}{\lambda} = \frac{2\ln 2}{\lambda} = 2t_{1/2} = 11\,460$ years. Two halvings, as the quarter suggests. A sample with $0.1$ of the original is $\frac{\ln 10}{\lambda} \approx 19\,000$ years old, and beyond about ten half-lives so little is left that the method stops being usable, which is why carbon dating reaches back about $50\,000$ years and no further.
Nothing in the dating used the constant solution $N = 0$ that dividing by $N$ set aside, but it is there: a sample with no carbon-14 has none forever, and it is the one sample the method cannot date.
The same separable equation, $Q' = -\lambda Q$, reappears wherever something disappears at a rate proportional to what is left: a medicine cleared by the liver (a half-life of about six hours for caffeine), the charge draining from a capacitor through a resistor, the light absorbed as it passes through water, the value of a machine written down by a fixed percentage each year. Each has its own $\lambda$ and its own half-life $\frac{\ln 2}{\lambda}$, and each is answered by the three moves above: separate, integrate, exponentiate. Recognising that a new situation is this equation in disguise is most of the work of modelling it.
Water leaving a tank through a hole at the bottom flows at a speed of $\sqrt{2gh}$ (Torricelli's law), so the depth $h$ of water in a straight-sided tank falls at a rate proportional to $\sqrt{h}$: $h' = -k\sqrt{h}$, with $k$ set by the sizes of the hole and the tank.
Separate: $h^{-1/2}\,dh = -k\,dt$. Integrate: $2\sqrt{h} = -kt + C$. With $h(0) = h_0$, $C = 2\sqrt{h_0}$, so $\sqrt{h} = \sqrt{h_0} - \frac{k}{2}t$. The tank is empty when $\sqrt{h} = 0$, at $t = \frac{2\sqrt{h_0}}{k}$.
A tank filled to $4$ metres with $k = 0.2$ per minute (in the units of $\text{m}^{1/2}$) empties at $t = \frac{2 \times 2}{0.2} = 20$ minutes. Half full, at $2$ metres, it takes $\frac{2\sqrt{2}}{0.2} \approx 14.1$ minutes, not ten: a tank drains faster when it is full, because the pressure at the hole is higher, so the last half of the water takes longer than the first.
The equation also shows something the formula alone hides. At $h = 0$ the right-hand side $-k\sqrt{h}$ is zero, so $h = 0$ is a constant solution, and separating lost it. A tank that is empty now could have emptied at any earlier time, so the equation cannot be run backwards from an empty tank. The next lesson but one explains why: $\sqrt{h}$ has no derivative at $0$.
Two habits cause most of the lost marks here, and both are about when. The first is adding $+C$ after solving for $y$ rather than at the moment of integrating: $\ln|y| = x^{3}$ then $y = e^{x^{3}} + C$ is a different and wrong family, because the constant has to go through the exponential with everything else. The second is substituting the initial condition early, into a line that still contains an integral sign — there is nothing to determine yet, and the substitution quietly throws the condition away. A solution is a function together with an interval it is a solution on. The existence theorem promises only a neighbourhood, and a formula that blows up at a finite point stops being an answer there — so a solution written without its interval is an answer to a question nobody asked.
Solve $y' = 3x^{2}y$ with $y(0) = 2$. The right-hand side is $3x^{2}$ times $y$, so separate: divide by $y$ and multiply by $dx$.
$\dfrac{dy}{y} = 3x^{2}\,dx \qquad (y \ne 0)$
Split first, and record what was assumed non-zero: $y = 0$ is a solution that dividing has just set aside.
Integrate both sides.
$\displaystyle\int \frac{dy}{y} = \int 3x^{2}\,dx \quad\Rightarrow\quad \ln|y| = x^{3} + C$
One constant covers both sides: two constants would only combine into one.
Exponentiate both sides.
$|y| = e^{x^{3} + C} = e^{C}e^{x^{3}}$
$e^{\ln u} = u$, and $e^{a + b} = e^{a}e^{b}$ splits the constant off.
Remove the absolute value and rename the constant.
$y = \pm e^{C}e^{x^{3}} = Ae^{x^{3}}, \qquad A = \pm e^{C}$
The sign from the absolute value is absorbed into the new constant $A$. $A = 0$ gives back the solution $y = 0$ that was divided away.
Apply the initial condition $y(0) = 2$.
$2 = Ae^{0} = A \quad\Rightarrow\quad y = 2e^{x^{3}}$
The condition is applied last, once the constant exists.
Check it in the equation and state the interval.
$y' = 2e^{x^{3}} \cdot 3x^{2} = 3x^{2} \cdot 2e^{x^{3}} = 3x^{2}y, \qquad x \in \mathbb{R}$
The chain rule brings down $3x^{2}$, which is exactly the right-hand side. Nothing in the formula breaks, so it holds on the whole line.
Solve $y' = y^{2}$ with $y(0) = 1$. Divide both sides by $y^{2}$ and multiply by $dx$.
$\dfrac{dy}{y^{2}} = dx \qquad (y \ne 0)$
The right-hand side is a function of $y$ alone, so it separates with $g(x) = 1$.
Integrate both sides.
$\displaystyle\int y^{-2}\,dy = \int dx \quad\Rightarrow\quad -\frac{1}{y} = x + C$
The power rule with exponent $-2$: $\int y^{-2}\,dy = \frac{y^{-1}}{-1}$.
Apply the initial condition $y(0) = 1$.
$-\dfrac{1}{1} = 0 + C \quad\Rightarrow\quad C = -1$
At $x = 0$ only the constant is left on the right.
Multiply both sides by $-1$, then take reciprocals.
$\dfrac{1}{y} = 1 - x \quad\Rightarrow\quad y = \dfrac{1}{1 - x}$
Isolating $\frac{1}{y}$ first keeps the reciprocal step clean.
Find where it exists.
$1 - x = 0 \text{ at } x = 1 \quad\Rightarrow\quad (-\infty, 1)$
At $x = 1$ the solution is infinite. Beyond it the formula belongs to a different branch and solves nothing that starts at $(0, 1)$.
Check it in the equation.
$y' = \dfrac{1}{(1 - x)^{2}} = \left(\dfrac{1}{1 - x}\right)^{2} = y^{2}$
Differentiating $(1 - x)^{-1}$ gives $(1 - x)^{-2}$, the chain rule's $-1$ cancelling the power rule's $-1$.
Solve $y' = 1 + y^{2}$ with $y(0) = 1$. The right side is $1$ times $1 + y^{2}$, and $1 + y^{2}$ is never zero.
$g(x) = 1, \qquad h(y) = 1 + y^{2} \ge 1$
Because $h$ has no roots, dividing by it loses no constant solutions.
Divide both sides by $1 + y^{2}$ and multiply by $dx$.
$\dfrac{dy}{1 + y^{2}} = dx$
Each variable sits with its own differential.
Integrate both sides.
$\displaystyle\int \frac{dy}{1 + y^{2}} = \int dx \quad\Rightarrow\quad \arctan y = x + C$
$\frac{1}{1 + y^{2}}$ is the derivative of $\arctan y$.
Apply the initial condition $y(0) = 1$.
$\arctan 1 = 0 + C \quad\Rightarrow\quad C = \dfrac{\pi}{4}$
$\tan\frac{\pi}{4} = 1$, so $\arctan 1 = \frac{\pi}{4}$.
Take the tangent of both sides.
$y = \tan\left(x + \dfrac{\pi}{4}\right)$
$\tan(\arctan y) = y$ undoes the inverse function.
Find the interval: the tangent is finite only while its angle stays strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.
$-\dfrac{\pi}{2} < x + \dfrac{\pi}{4} < \dfrac{\pi}{2} \quad\Rightarrow\quad -\dfrac{3\pi}{4} < x < \dfrac{\pi}{4}$
Subtract $\frac{\pi}{4}$ from all three parts. The branch containing $x = 0$ is the answer, and it blows up at both ends.
Check it in the equation.
$y' = \sec^{2}\left(x + \dfrac{\pi}{4}\right) = 1 + \tan^{2}\left(x + \dfrac{\pi}{4}\right) = 1 + y^{2}$
The identity $\sec^{2}\theta = 1 + \tan^{2}\theta$ closes the check.
Separate the variables.
$y\,dy = 2x\,dx$
Multiply both sides by $y\,dx$. Here $y = 0$ is not a solution anyway, because the equation is undefined there.
Integrate both sides.
Apply $y(1) = 3$ and choose the root.
Put the steps of solving $y' = 6xy$ with $y(0) = 1$ into the order they must be done.
Number the steps in order (write the number in the box):
Complete the worked solution of $y' = 12x$ with $y(0) = 9$, and find $y(2)$.
Separate the variables: multiply both sides by $dx$.
$\dfrac{dy}{dx} = 12x \quad\Rightarrow\quad dy = 12x\,dx$
The right-hand side depends on $x$ alone, so the variables are already apart.
Integrate both sides.
$\displaystyle\int dy = \int 12x\,dx \quad\Rightarrow\quad y = 12 \cdot \frac{x^{2}}{2} + C$
The power rule: $\int x\,dx = \frac{x^{2}}{2}$, with one constant for a first-order equation.
Simplify the coefficient of $x^{2}$.
$12 \cdot \dfrac{1}{2} =$ a
Multiplying by one half halves the coefficient.
Substitute $x = 0$ and $y = 9$ to find $C$.
$9 = (\text{coefficient}) \cdot 0^{2} + C \quad\Rightarrow\quad C = 9$
At $x = 0$ the square term vanishes, so the constant is the starting value.
Substitute $x = 2$ into the particular solution.
$y(2) = (\text{coefficient}) \cdot 2^{2} + 9 =$ v
Square first, multiply by the coefficient, then add the constant.
Which method is $y' = \dfrac{x}{y}$ set up for?
Solve $y' = 10x$ with $y(0) = 4$. Write $y$ as a formula in $x$.
Answer:
$y' = 2y^{2}$ with $y(0) = 2$. At which value of $x$ does the solution become infinite?
Answer:
Across a species, the size $y$ of an organ grows with body size $x$ so that its relative growth rate is $4$ times the body's: $xy' = 4y$ (allometric growth). An animal of body size $1$ has organ size $4$. Solve $xy' = 4y$ with $y(1) = 4$, for $x > 0$. Write $y$ as a formula in $x$.
Answer:
The solution of $y' = 5y^{2}$, $y(0) = 4$, is $y = \dfrac{1}{\frac{1}{4} - 5x}$. On which interval containing $x = 0$ is it actually a solution?
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Across a species, the size $y$ of an organ grows with body size $x$ so that its relative growth rate is $4$ times the body's: $xy' = 4y$ (allometric growth). An animal of body size $1$ has organ size $8$. Solve $xy' = 4y$ with $y(1) = 8$, for $x > 0$. Write $y$ as a formula in $x$.
Answer:
You can separate, integrate, fix the constant from the initial condition and name the interval the answer holds on. Say in your own words which solutions dividing by a factor of $y$ can lose.
14. Your turn: solve $y' = \dfrac{2x}{y}$ with $y(1) = 3$, step 2
$\dfrac{y^{2}}{2} = x^{2} + C \quad\Rightarrow\quad y^{2} = 2x^{2} + K$
Doubling both sides turns $2C$ into a new constant $K$.
14. Your turn: solve $y' = \dfrac{2x}{y}$ with $y(1) = 3$, step 3
$9 = 2 + K \;\Rightarrow\; K = 7, \qquad y = \sqrt{2x^{2} + 7}$
The positive root, because the initial value is positive. It holds for every real $x$, since $2x^{2} + 7$ is never zero.