Differential Equations

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 9.

A bank balance that grows continuously

A bank advertises that your balance grows continuously at $5\%$ a year. Notice what that sentence gives you. It does not give a formula for the balance. It gives a rule about how fast the balance changes: the rate of growth is $5\%$ of whatever you have right now. In symbols, $\dfrac{dP}{dt} = \dfrac{P}{20}$. That is a differential equation. Almost every law in physics, biology and economics is stated this way, as a rule about rates. This chapter teaches you to turn the rule back into a formula.

Throughout, $\log$ means the natural logarithm.

The chapter in one line: read the equation (order, degree) → know what a solution is → identify the type (separable, homogeneous, linear) → apply that type's method.

What is a differential equation?

An equation that contains the derivatives of a dependent variable with respect to an independent variable is a differential equation. If there is only one independent variable it is an ordinary differential equation, and this chapter studies only those. The compact notation is $y' = \dfrac{dy}{dx}$, $y'' = \dfrac{d^2y}{dx^2}$ and $y''' = \dfrac{d^3y}{dx^3}$.

Order is the order of the highest derivative in the equation. It is always a positive integer.

Degree is the power of the highest order derivative, after the equation is a polynomial in its derivatives. If any derivative sits inside a function such as $\sin$, $\log$ or an exponential, the equation is not a polynomial in the derivatives, and the degree is not defined.

Worked example: reading order and degree. Find the order and degree of $\left( \dfrac{d^3y}{dx^3} \right)^2 + \left( \dfrac{d^2y}{dx^2} \right)^3 + \left( \dfrac{dy}{dx} \right)^4 + y^5 = 0$.

Common wrong step. Taking the degree as $4$ or $5$, the largest power anywhere in the equation. The degree is the power of the highest order derivative only, and $y^5$ is not a derivative at all.

Pause and predict: what are the order and degree of $y'' + \sin y' = 0$?

Check your answer: order $2$, degree not defined. The highest derivative is $y''$, but $y'$ appears inside $\sin$, so the equation is not a polynomial in its derivatives.

Solutions: general and particular

A solution is a function $y = \varphi(x)$ that satisfies the equation when you substitute it in. Its graph is an integral curve. To check a candidate, differentiate it as often as the equation needs and confirm that both sides match.

Consider the simplest equation $\dfrac{dy}{dx} = 2x$. Integrating gives $y = x^2 + C$, not a single function but a whole family, one curve for every $C$. This is the general solution. It contains as many arbitrary constants as the order of the equation, because getting back from an $n$th-order equation takes $n$ integrations and each one brings its own constant.

A particular solution has no arbitrary constant left. Once the initial condition fixes every constant, nothing is arbitrary.

Method 1: variables separable

If an equation can be rearranged so that all $y$ terms sit with $dy$ on one side and all $x$ terms with $dx$ on the other, it is variables separable. Typically $\dfrac{dy}{dx} = g(x)\,h(y)$, and then:

$$\int \dfrac{dy}{h(y)} = \int g(x)\,dx + C$$

Why this is allowed. Dividing by $h(y)$ gives $\dfrac{1}{h(y)}\dfrac{dy}{dx} = g(x)$. The left side is exactly what the chain rule produces when you differentiate $H(y)$ with $H' = \dfrac{1}{h}$. So integrating both sides with respect to $x$ is legitimate, and in practice you just integrate each side in its own variable.

Put the single constant $C$ on one side, after both integrations. An exponential of a sum such as $e^{x+y}$ is a hidden separable form, because $e^{x+y} = e^x e^y$.

Worked example: a hidden separable equation. Solve $\dfrac{dy}{dx} = e^{x+y}$.

Common wrong step. Integrating $e^{-y}\,dy$ as $e^{-y}$ and dropping the minus sign. The chain rule gives $\int e^{-y}\,dy = -e^{-y}$.

Growth and decay

A great many real laws say "the rate of change is proportional to the amount present". If $P$ is the amount at time $t$, that law is $\dfrac{dP}{dt} = kP$. Separating the variables gives $\dfrac{dP}{P} = k\,dt$, then $\log P = kt + c$, so $P = P_0 e^{kt}$, where $P_0$ is the value at $t = 0$.

$$\dfrac{dP}{dt} = kP \;\Rightarrow\; P = P_0 e^{kt} \qquad (k > 0 \text{ growth}, \ k < 0 \text{ decay})$$

Worked example: the continuously compounding deposit. A bank pays $5\%$ per year, compounded continuously. A deposit of ₹1000 is made. Find the amount after $t$ years and the time to double.

Common wrong step. Writing $P = e^{t/20} + C$. The constant multiplies, because exponentiating $\log P = \dfrac{t}{20} + c$ gives $P = e^{c}e^{t/20}$. Note that the doubling time does not depend on the starting deposit.

Pause and predict: a radioactive substance decays at a rate proportional to the amount present, and its half-life is $10$ years. What fraction remains after $30$ years?

Check your answer: $\dfrac{1}{8}$. In the law $y = y_0 e^{-kt}$ the amount halves every equal period, so three half-lives leave $\left( \dfrac{1}{2} \right)^3 = \dfrac{1}{8}$.

Beyond the textbook. For the exam: growth and decay follow $\dfrac{dP}{dt} = kP$ and the answer is $P_0 e^{kt}$. In reality: the same equation, with only the sign and meaning of $k$ changed, describes radioactive decay and the early growth of a population. A close relative is Newton's law of cooling, $\dfrac{dT}{dt} = -k(T - T_0)$, where the temperature $T$ of a cooling object approaches the surrounding temperature $T_0$ at a rate proportional to the difference. Why it matters: one differential equation covers many unrelated phenomena, which is the reason this chapter is worth learning once and applying everywhere.

Beyond the textbook. For the exam: separate the variables, integrate both sides and add $C$. In reality: dividing by $h(y)$ can silently drop solutions where $h(y) = 0$. For $\dfrac{dy}{dx} = y^2$ the separated solution is $y = -\dfrac{1}{x + C}$, yet $y = 0$ also satisfies the equation and is missing from that family. Why it matters: a complete answer to a "find all solutions" question has to check the zeros of $h(y)$ separately, though a CUET question normally asks only for the general solution you get by separating.

Method 2: homogeneous equations

A function $F(x, y)$ is homogeneous of degree $n$ if $F(\lambda x, \lambda y) = \lambda^n F(x, y)$ for every non-zero $\lambda$. To test it, replace $x$ by $\lambda x$ and $y$ by $\lambda y$ and see whether a clean power of $\lambda$ comes out.

The equation $\dfrac{dy}{dx} = F(x, y)$ is a homogeneous differential equation when $F$ is homogeneous of degree zero, which means it depends only on the ratio $\dfrac{y}{x}$: $\dfrac{dy}{dx} = g\left( \dfrac{y}{x} \right)$. In the form $M\,dx + N\,dy = 0$, both $M$ and $N$ must be homogeneous of the same degree.

Why the substitution $y = vx$ works. Since the equation only cares about the ratio $v = \dfrac{y}{x}$, make that ratio the new variable. Then $y = vx$ and, by the product rule, $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. The equation becomes $x\dfrac{dv}{dx} = g(v) - v$, and the variables now separate:

$$\int \dfrac{dv}{g(v) - v} = \int \dfrac{dx}{x} + C$$

After integrating, put back $v = \dfrac{y}{x}$. If the equation is written naturally as $\dfrac{dx}{dy} = h\left( \dfrac{x}{y} \right)$, use $x = vy$ instead, so that $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$.

Worked example: a homogeneous equation. Solve $\dfrac{dy}{dx} = \dfrac{x + y}{x}$.

Common wrong step. Writing $\dfrac{dy}{dx} = v$ and forgetting that $v$ itself depends on $x$. The product rule gives the extra term: $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. A second common slip is stopping at $v = \log\lvert x\rvert + C$ and not substituting back.

Pause and predict: is $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$ homogeneous, and which substitution would you try?

Check your answer: yes. The numerator and the denominator both have degree $2$, so the right side has degree $0$. Put $y = vx$.

Method 3: linear equations and the integrating factor

A linear differential equation in $y$ has the standard form

$$\dfrac{dy}{dx} + P\,y = Q$$

where $P$ and $Q$ are constants or functions of $x$ only. The coefficient of $\dfrac{dy}{dx}$ must be $1$. The unknown $y$ and its derivative must appear only to the first power and never multiplied together or wrapped in another function.

Why an integrating factor works. Multiply the equation by $e^{\int P\,dx}$, called the integrating factor (I.F.). By the product rule, the left side becomes $\dfrac{d}{dx}\left( y\cdot e^{\int P\,dx} \right)$, a perfect derivative, because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$. The right side is just $Q\,e^{\int P\,dx}$, so integrating gives the solution:

$$y\cdot \text{I.F.} = \int Q\cdot \text{I.F.}\,dx + C, \qquad \text{I.F.} = e^{\int P\,dx}$$

The recipe: (1) divide to make the coefficient of $\dfrac{dy}{dx}$ equal to $1$; (2) read off $P$ and $Q$; (3) find I.F. $= e^{\int P\,dx}$; (4) write $y\cdot\text{I.F.} = \int Q\cdot\text{I.F.}\,dx + C$.

When the equation is naturally linear in $x$, with the form $\dfrac{dx}{dy} + P_1x = Q_1$ ($P_1$, $Q_1$ functions of $y$ only), the same idea gives I.F. $= e^{\int P_1\,dy}$ and $x\cdot\text{I.F.} = \int Q_1\cdot\text{I.F.}\,dy + C$.

Worked example: dividing first. Solve $x\dfrac{dy}{dx} + 2y = x^2$ for $x > 0$.

Common wrong step. Reading $P = 2$ from the equation as written and getting I.F. $= e^{2x}$. $P$ is the coefficient of $y$ only after $\dfrac{dy}{dx}$ has coefficient $1$.

Worked example: watching the sign. Solve $\dfrac{dy}{dx} - y = \cos x$.

Common wrong step. Using I.F. $= e^{x}$ because the equation shows "$-y$" and "minus times minus" feels like a plus. $P$ is $-1$, so the integrating factor is $e^{-x}$. A wrong sign here propagates through the whole calculation.

Choosing the method

Linear is more than first degree. The equation $\dfrac{dy}{dx} = y^2$ has degree $1$ but is not linear, because it contains $y^2$. It is separable, though. An equation can be linear and separable at the same time: $\dfrac{dy}{dx} = y$ is both.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Application of Integrals completes the integration toolkit. Every method here ends in an integral from Integrals.
• Earlier ideas reused: Application of Derivatives introduced the derivative as a rate of change, which is where these equations come from.
• Next: Vector Algebra introduces vectors, the next topic in the syllabus.