How unit 1 is examined
First-order equations (linear, Bernoulli, exact, separable) and constant-coefficient higher-order equations carry almost all the marks; Cauchy-Euler and simultaneous equations come next, and first-order higher degree is unasked.
Differential Equations of First Order and First Degree (Leibnitz linear, Bernoulli's, Exact)
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Definition. ==A first-order, first-degree equation has the form $M\,dx + N\,dy = 0$ (or $dy/dx = f(x,y)$); it is solved by first recognising its type: separable, linear, Bernoulli or exact.==
Key points.
- Variables separable: if the equation splits as $f(x)dx + g(y)dy = 0$, integrate each part separately and add one constant.
- Leibnitz linear in $y$: $\dfrac{dy}{dx} + Py = Q$ with $P,Q$ functions of $x$ only; the integrating factor is $I.F. = e^{\int P\,dx}$.
- The linear solution is $y\cdot(I.F.) = \int Q\,(I.F.)\,dx + C$; if $x$ appears only to the first power, treat $x$ as the unknown and use $\dfrac{dx}{dy} + Px = Q$ with $P,Q$ functions of $y$.
- Bernoulli: $\dfrac{dy}{dx} + Py = Qy^n$; divide by $y^n$ and put $v = y^{1-n}$, which turns it into a linear equation in $v$.
- Exact: $M\,dx + N\,dy = 0$ is exact when $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$.
- Exact solution: $\int M\,dx$ (with $y$ constant) $+ \int(\text{terms of } N \text{ free of } x)\,dy = C$.
- A substitution such as $u = x + y$ turns $dy/dx = f(x+y)$ into a separable equation.
- Always add the constant $C$ and substitute back the original variable.
Steps (linear). Write in standard form; find $P$; compute $I.F.$; multiply and integrate.
Example. Solve $(1+y^2)dx = (\tan^{-1}y - x)dy$.
Write $\dfrac{dx}{dy} + \dfrac{x}{1+y^2} = \dfrac{\tan^{-1}y}{1+y^2}$, so $P = \dfrac{1}{1+y^2}$ and $I.F. = e^{\tan^{-1}y}$. Put $t = \tan^{-1}y$:
$$x e^{t} = \int t e^{t}\,dt + C = (t-1)e^{t} + C$$
Answer: $x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$.
| Question | Type and key step | Answer |
|---|---|---|
| $(e^y+1)\cos x\,dx + e^y\sin x\,dy = 0$ | Separable: $\cot x\,dx + \frac{e^y}{e^y+1}dy = 0$ | $(e^y+1)\sin x = C$ |
| $(1+e^{x/y})dx + e^{x/y}(1-\frac xy)dy = 0$ | Exact: both partials $= -\frac{x}{y^2}e^{x/y}$ | $x + ye^{x/y} = C$ |
| $(1+y^2) + (x - e^{-\tan^{-1}y})\frac{dy}{dx} = 0$ | Linear in $x$, $I.F. = e^{\tan^{-1}y}$ | $xe^{\tan^{-1}y} = \tan^{-1}y + C$ |
| $\frac{dy}{dx} + y\tan x = y^2\sec x$ | Bernoulli, $v = 1/y$, $I.F. = \cos x$ | $\frac{\cos x}{y} = C - x$ |
| $\cos x\,dy = y(\sin x - y)dx$ | Bernoulli, $v = 1/y$: $v' + v\tan x = \sec x$, $I.F. = \sec x$ | $\frac{\sec x}{y} = \tan x + C$ |
| $\sin 2x\frac{dy}{dx} - y = \tan x$ | Linear, $P = -\csc 2x$, $I.F. = 1/\sqrt{\tan x}$ | $y = \tan x + C\sqrt{\tan x}$ |
| $(r+\sin\theta-\cos\theta)dr + r(\sin\theta+\cos\theta)d\theta = 0$ | Exact: both partials $= \sin\theta+\cos\theta$ | $\frac{r^2}{2} + r(\sin\theta-\cos\theta) = C$ |
| $\frac{dy}{dx} = \cos(x+y)+\sin(x+y)$ | $u = x+y$, $t = \tan\frac u2$: $dx = \frac{dt}{1+t}$ | $1+\tan\frac{x+y}{2} = Ce^{x}$ |
Answer frame. Open with "The equation is of the form ... so it is solved by the ... method"; state the standard form and the test or substitution; then write $P$, the $I.F.$ (or $M,N$ and the exactness check) line by line; integrate carefully; close with the general solution in the original variables.
Pitfall: When $x$ (not $y$) is the unknown, the standard form must be $dx/dy + Px = Q$; using $dy/dx$ gives a non-linear equation.
Asked: [7 marks] (Dec 2023, Jun 2023, Dec 2024) Solve $(1+y^2)dx = (\tan^{-1}y - x)dy$ (Leibnitz linear method). Asked: [7 marks] (Dec 2023, Jun 2025) Solve $(e^y+1)\cos x\,dx + e^y\sin x\,dy = 0$. Asked: [7 marks] (Jun 2022) Solve $(1+e^{x/y})dx + e^{x/y}(1-x/y)dy = 0$. Asked: [7 marks] (Jun 2022) Solve the linear equation $(1+y^2) + (x - e^{-\tan^{-1}y})\frac{dy}{dx} = 0$. Asked: [7 marks] (Jun 2022) Solve $\frac{dy}{dx} + y\tan x = y^2\sec x$ using Bernoulli's. Asked: [7 marks] (Nov 2022) Solve $\cos x\,dy = y(\sin x - y)dx$ using Bernoulli's. Asked: [7 marks] (Nov 2022) Solve the linear equation $\sin 2x\frac{dy}{dx} - y = \tan x$. Asked: [7 marks] (Nov 2022) Solve $(r+\sin\theta-\cos\theta)dr + r(\sin\theta+\cos\theta)d\theta = 0$. Asked: [7 marks] (Jun 2023) Solve $\frac{dy}{dx} = \cos(x+y)+\sin(x+y)$.
Differential Equations of First Order and Higher Degree
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Definition. An equation of first order and degree $n>1$ in $p = dy/dx$, such as $p^2 + Px + Q = 0$; Clairaut's form is $y = px + f(p)$.
Key points.
- If the equation factorises into $(p - f_1)(p - f_2) = 0$, solve each first-degree factor and combine the solutions.
- Clairaut's equation $y = px + f(p)$ is solved by replacing $p$ with $c$, giving the general solution $y = cx + f(c)$, a family of straight lines.
- Eliminating $p$ between $x = -f'(p)$ and the equation gives the singular solution, the envelope of those lines.
- Equations solvable for $y$ or for $x$ are differentiated with respect to $x$ (or $y$) to get an equation in $p$ and $x$.
Higher order differential equations with constants coefficients
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Definition. ==$f(D)y = X$ with $D = d/dx$ and constant coefficients has the complete solution $y = C.F. + P.I.$, where $C.F.$ solves $f(D)y = 0$ and $P.I. = \dfrac{1}{f(D)}X$.==
Key points.
- Put $y = e^{mx}$ to get the auxiliary equation $f(m) = 0$; its roots decide the $C.F.$
- Real distinct roots $m_1, m_2$ give $C.F. = c_1e^{m_1x} + c_2e^{m_2x}$.
- A repeated root $m$ (twice) gives $(c_1 + c_2x)e^{mx}$.
- Complex roots $\alpha \pm i\beta$ give $e^{\alpha x}(c_1\cos\beta x + c_2\sin\beta x)$.
- For $X = e^{ax}$: $P.I. = \dfrac{e^{ax}}{f(a)}$ if $f(a) \ne 0$; if $f(a) = 0$, multiply by $x$ and differentiate $f$.
- For $X = \sin ax$ or $\cos ax$: replace $D^2$ by $-a^2$; if the result is $pD + q$, rationalise by multiplying by $(pD - q)$.
- For $X = e^{ax}V$: $P.I. = e^{ax}\dfrac{1}{f(D+a)}V$; for $X = x^k$: expand $[f(D)]^{-1}$ as a series in $D$ and operate on $x^k$.
- The constant $5$ is $5e^{0x}$, so its $P.I.$ is $5/f(0)$.
Example. Solve $(D^2 - 4D + 3)y = \cos 2x$.
$m^2 - 4m + 3 = 0$ gives $m = 1, 3$, so $C.F. = c_1e^x + c_2e^{3x}$. With $D^2 = -4$:
$$P.I. = \frac{\cos 2x}{-1-4D} = -\frac{(4D-1)\cos 2x}{16D^2-1} = -\frac{-8\sin 2x - \cos 2x}{-65}$$
Answer: $y = c_1e^x + c_2e^{3x} - \dfrac{8\sin 2x + \cos 2x}{65}$.
| Question | Roots and C.F. | P.I. and answer |
|---|---|---|
| $(D^2+3D+2)y = \sin 3x$ | $m=-1,-2$ | $P.I. = -\frac{7\sin 3x + 9\cos 3x}{130}$ (put $D^2 = -9$: $\frac{1}{3D-7}$) |
| $(D^2-5D+6)y = 4e^x+5$ | $m=2,3$ | $2e^x + \frac56$; $y = c_1e^{2x}+c_2e^{3x}+2e^x+\frac56$ |
| $(D^2-2D-3)y = x^3e^{-3x}$ | $m=3,-1$ | $e^{-3x}\frac{1}{D^2-8D+12}x^3 = e^{-3x}\left(\frac{x^3}{12}+\frac{x^2}{6}+\frac{13x}{72}+\frac{5}{54}\right)$ |
| $(D^3-7D^2+14D-8)y = e^x\cos 2x$ | $m=1,2,4$ | $e^x\frac{1}{D^3-4D^2+3D}\cos 2x$; $P.I. = e^x\frac{8\cos 2x - \sin 2x}{130}$ |
| $y''+y' = (1+e^x)^{-1}$ | $m=0,-1$ | Variation of parameters: $y = c_1 + c_2e^{-x} + x - (1+e^{-x})\ln(1+e^x)$ |
Answer frame. Open with "Let $y = e^{mx}$ be a trial solution"; write the auxiliary equation, roots and $C.F.$; state the $P.I.$ rule for the given $X$; simplify step by step; close with $y = C.F. + P.I.$ written out in full.
Pitfall: For $\sin ax$ or $\cos ax$ replace only $D^2$ by $-a^2$, never $D$ itself; forgetting to rationalise the leftover $D$ is the commonest lost mark.
Asked: [7 marks] (Dec 2023, Dec 2024) Solve $(D^2-4D+3)y = \cos 2x$; also $(D^2+3D+2)y = \sin 3x$. Asked: [7 marks] (Jun 2022) Solve $(D^2-2D-3)y = x^3e^{-3x}$. Asked: [7 marks] (Nov 2022) Solve $(D^3-7D^2+14D-8)y = e^x\cos 2x$. Asked: [7 marks] (Jun 2023) Solve $\frac{d^2y}{dx^2}+\frac{dy}{dx} = (1+e^x)^{-1}$. Asked: [7 marks] (Jun 2025) Solve $(D^2-5D+6)y = 4e^x+5$.
Homogeneous Linear Differential equations
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Definition. The Cauchy-Euler (homogeneous linear) equation $x^2y'' + axy' + by = X$ has each term $x^ky^{(k)}$; the substitution $x = e^z$ makes its coefficients constant.
Key points.
- Put $x = e^z$, $z = \log x$, and $\theta = d/dz$; then $x\dfrac{dy}{dx} = \theta y$ and $x^2\dfrac{d^2y}{dx^2} = \theta(\theta-1)y$.
- For the Legendre form $(a+bx)^2y'' + \dots$, put $a + bx = e^z$; here $(1+x)^2y'' = \theta(\theta-1)y$ and $(1+x)y' = \theta y$.
- The transformed equation $f(\theta)y = Z(z)$ is solved exactly as a constant-coefficient equation in $z$.
- Finally replace $z$ by $\log x$ (or $\log(1+x)$) and $e^z$ by $x$.
Example. $(1+x)^2y'' + (1+x)y' + y = \cos\log(1+x)$ becomes $[\theta(\theta-1)+\theta+1]y = \cos z$, i.e. $(\theta^2+1)y = \cos z$. Then $C.F. = c_1\cos z + c_2\sin z$; since $\cos z$ is a solution of the homogeneous part, $P.I. = \frac z2\sin z$.
Answer: $y = c_1\cos\log(1+x) + c_2\sin\log(1+x) + \frac12\log(1+x)\sin\log(1+x)$.
For $x^2y''+5xy'+4y = x\log x$: $(\theta+2)^2y = ze^z$, $y = (c_1+c_2\log x)x^{-2} + \frac x9\left(\log x - \frac23\right)$.
Answer frame. Open with "This is a homogeneous (Cauchy-Euler) equation, so put $1+x = e^z$"; write the operator form; solve $C.F.$ and $P.I.$ in $z$; close by back-substituting $z = \log(1+x)$.
Asked: [7 marks] (Dec 2024, Jun 2025) Solve $(1+x)^2\frac{d^2y}{dx^2}+(1+x)\frac{dy}{dx}+y = \cos\log(1+x)$; also $x^2\frac{d^2y}{dx^2}+5x\frac{dy}{dx}+4y = x\log x$.
Simultaneous Differential Equations
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Definition. A set of linear equations in two unknowns $x(t), y(t)$ with one independent variable $t$ is solved by writing it in operator form ($D = d/dt$) and eliminating one unknown.
Key points.
- Write the system as $f_1(D)x + g_1(D)y = T_1$ and $f_2(D)x + g_2(D)y = T_2$.
- Eliminate one unknown by treating $D$ like a number, giving a single higher-order equation such as $(D^2+1)x = \dots$
- Solve that equation as $C.F. + P.I.$ to get $x(t)$.
- Get the other unknown $y$ from one original equation (no new constants), not by solving a second equation.
- Initial conditions are applied last to fix the constants.
Example. $Dx - y = e^t$, $x + Dy = \sin t$, $x(0)=1$, $y(0)=0$.
From the first, $y = x' - e^t$, so $y' = x'' - e^t$; the second gives $(D^2+1)x = e^t + \sin t$. Then $x = c_1\cos t + c_2\sin t + \frac{e^t}{2} - \frac t2\cos t$. $x(0)=1$ gives $c_1 = \frac12$; $y(0)=x'(0)-1=0$ gives $c_2 = 1$.
Answer: $x = \frac12\cos t + \sin t + \frac12e^t - \frac t2\cos t$, $y = \frac12\cos t - \frac12\sin t - \frac12e^t + \frac t2\sin t$.
For $(D-7)x + y = 0$, $-2x + (D-5)y = 0$: $y = -(D-7)x$ gives $(D^2-12D+37)x = 0$, roots $6\pm i$. So $x = e^{6t}(c_1\cos t + c_2\sin t)$ and $y = 7x - x' = e^{6t}[(c_1-c_2)\cos t + (c_1+c_2)\sin t]$.
Answer frame. Open with the operator form; eliminate one variable and show the resulting equation; solve it; recover the other variable; close with the pair of solutions and the constants found.
Pitfall: Recover the second unknown from an original equation, not by solving a second higher-order equation, or extra constants appear.
Asked: [7 marks] (Jun 2023) Solve $\frac{dx}{dt} - y = e^t$, $\frac{dy}{dt} + x = \sin t$; $x(0)=1$, $y(0)=0$. Asked: [7 marks] (Dec 2024) Solve $\frac{dx}{dt}-7x+y=0$ and $\frac{dy}{dt}-2x-5y=0$.
Last-minute revision
- Linear: $y' + Py = Q$, $I.F. = e^{\int P dx}$, $y\cdot I.F. = \int Q\,I.F.\,dx + C$.
- Bernoulli: $y' + Py = Qy^n$; put $v = y^{1-n}$.
- Exact: $M_y = N_x$; solution $\int M dx + \int(N \text{ without } x)dy = C$.
- Clairaut: $y = px + f(p)$ has solution $y = cx + f(c)$.
- Complete solution $= C.F. + P.I.$; auxiliary equation $f(m) = 0$.
- Repeated root: $(c_1+c_2x)e^{mx}$; complex root $\alpha\pm i\beta$: $e^{\alpha x}(c_1\cos\beta x + c_2\sin\beta x)$.
- $P.I.$ of $e^{ax}$: $e^{ax}/f(a)$; of $\sin ax$: put $D^2 = -a^2$; of $e^{ax}V$: $e^{ax}\frac{1}{f(D+a)}V$.
- Cauchy-Euler: $x = e^z$, $x^2D^2 = \theta(\theta-1)$, $xD = \theta$.
- Simultaneous: eliminate one variable, solve, get the other from an original equation.
- Repeated paper question: $(1+y^2)dx = (\tan^{-1}y - x)dy$ (three sessions).
Memory hooks
- "Left-over $y$ friend": $I.F.$ is $e$ to the integral of the $P$ next to $y$.
- Bernoulli: the power on the right-hand side is $n$, so $v = y^{1-n}$.
- Exact = "cross partials agree".
- $D^2$ turns into $-a^2$ for sine and cosine only.
- Cauchy-Euler: $x$ powers match derivative orders, so use $e^z$.
Coverage checklist
- Differential Equations of First Order and First Degree (Leibnitz linear, Bernoulli's, Exact): all nine questions (Jun 2022, Nov 2022, Jun 2023, Dec 2023, Dec 2024, Jun 2025).
- Differential Equations of First Order and Higher Degree: no past questions; Clairaut covered.
- Higher order differential equations with constants coefficients: $\cos 2x$, $\sin 3x$, $x^3e^{-3x}$, $e^x\cos 2x$, $(1+e^x)^{-1}$, $4e^x+5$.
- Homogeneous Linear Differential equations: $(1+x)^2$ and $x^2y''+5xy'+4y$.
- Simultaneous Differential Equations: Jun 2023 initial-value system and Dec 2024 system.