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JEE Mains Maths · Formula sheet

Differential Equations formulas

21 formulas and 21 common traps for JEE Mains Maths Differential Equations, grouped by subtopic.

Full notes with worked examples

Forming a Differential Equation

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Eliminating the constants of a family

Order = number of constants

y=f(x;c1,…,cn) ⇒ F(x,y,y′,…,y(n))=0y=f(x;c_1,\dots,c_n)\ \Rightarrow\ F\big(x,y,y',\dots,y^{(n)}\big)=0

Order, degree, and the equation a solution satisfies

Degree after clearing radicals

y=xy′+1+y′2 ⇒ (y−xy′)2=1+y′2 : order 1, degree 2y=x y'+\sqrt{1+y'^2}\ \Rightarrow\ (y-xy')^2=1+y'^2\ :\ \text{order }1,\ \text{degree }2

Common traps

Use the conditions first

A general circle has three constants. Conditions such as 'through the origin' or 'centre on y=xy=x' remove some of them before you differentiate; skipping that step gives an equation of too high an order.

Clear the radical before counting

y=xy′+1+y′2y=xy'+\sqrt{1+y'^2} looks like degree 1, but squaring to remove the root gives y′2y'^2 terms, so the degree is 2. The degree is read only after the radical is gone.

Separating the Variables

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Separate, integrate, fix the constant

Separable form

dydx=f(x) g(y) ⇒ ∫dyg(y)=∫f(x) dx+C\frac{dy}{dx}=f(x)\,g(y)\ \Rightarrow\ \int\frac{dy}{g(y)}=\int f(x)\,dx+C

When the slope depends on x alone

Direct integration

dydx=f(x) ⇒ y=∫f(x) dx+C\frac{dy}{dx}=f(x)\ \Rightarrow\ y=\int f(x)\,dx+C

Substituting for x + y, and exact differentials

Substitution for a linear combination

t=ax+by+c ⇒ dtdx=a+b dydxt=ax+by+c\ \Rightarrow\ \frac{dt}{dx}=a+b\,\frac{dy}{dx}

Common traps

Constant before exponentiating

From ln⁡y=x2+C\ln y=x^2+C, the solution is y=Aex2y=Ae^{x^2}, not ex2+Ce^{x^2}+C. Fix the constant in whichever form you keep, and never add it after exponentiating.

Two conditions for a second-order equation

Integrating twice brings two constants, so it needs two conditions — often one on the derivative and one on the value.

Differentiate the substitution fully

With t=2x+3yt=2x+3y, dtdx=2+3dydx\frac{dt}{dx}=2+3\frac{dy}{dx}, not 3dydx3\frac{dy}{dx}. Dropping the 2 gives a separable equation with the wrong answer.

Homogeneous Equations

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Putting y = vx

The substitution

y=vx:v+xdvdx=F(v) ⇒ ∫dvF(v)−v=ln⁡∣x∣+Cy=vx:\quad v+x\frac{dv}{dx}=F(v)\ \Rightarrow\ \int\frac{dv}{F(v)-v}=\ln|x|+C

When y = vx is not the first move

Shift of origin

x=X+h, y=Y+k:dYdX=a1X+b1Ya2X+b2Yx=X+h,\ y=Y+k:\quad \frac{dY}{dX}=\frac{a_1X+b_1Y}{a_2X+b_2Y}

Common traps

Return to y before using the point

The condition is given in xx and yy. Either convert it to v=yxv=\frac yx at that point, or substitute back first; mixing the two gives the wrong constant.

Parallel lines need a different substitution

If a1x+b1ya_1x+b_1y and a2x+b2ya_2x+b_2y are proportional, the two lines never meet and no shift exists. Put t=a1x+b1yt=a_1x+b_1y instead; the equation then separates.

Linear Equations: The Integrating Factor

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Standard form and the integrating factor

Linear equation

dydx+Py=Q ⇒ y e∫P dx=∫Q e∫P dx dx+C\frac{dy}{dx}+Py=Q\ \Rightarrow\ y\,e^{\int P\,dx}=\int Q\,e^{\int P\,dx}\,dx+C

Integrating factors from trigonometric P

Common trigonometric factors

P=ktan⁡x ⇒ μ=sec⁡kx,P=11+x2 ⇒ μ=etan⁡−1xP=k\tan x\ \Rightarrow\ \mu=\sec^kx,\qquad P=\frac{1}{1+x^2}\ \Rightarrow\ \mu=e^{\tan^{-1}x}

Common traps

Standard form before the factor

The integrating factor comes from PP only after the coefficient of y′y' is 1. In x y′+2y=x2x\,y'+2y=x^2, PP is 2x\frac2x, not 2.

Sign of the tangent term

∫tan⁡x dx=ln⁡sec⁡x\int\tan x\,dx=\ln\sec x, so +tan⁡x+\tan x gives sec⁡x\sec x and −tan⁡x-\tan x gives cos⁡x\cos x. Swapping them is the commonest slip on this page.

Integrating Factors in Disguise

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The left side is already a derivative

Product rule in reverse

u dydx+dudx y=ddx(uy)u\,\frac{dy}{dx}+\frac{du}{dx}\,y=\frac{d}{dx}(uy)

When P is the derivative of a logarithm

Log-derivative coefficient

P=g′(x)g(x) ⇒ e∫P dx=g(x)P=\frac{g'(x)}{g(x)}\ \Rightarrow\ e^{\int P\,dx}=g(x)

Common traps

Divide only if it helps

Dividing x4y′+4x3y=fx^4y'+4x^3y=f by x4x^4 and computing e∫4/x=x4e^{\int4/x}=x^4 just multiplies back by x4x^4. If the left side is already exact, integrate it directly.

Check the derivative, do not integrate blindly

Differentiate the denominator and compare with the numerator before attempting the integral. When they match, the factor is the denominator itself; attempting a long integral wastes the time the question is testing.

Equations Reducible to Linear

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Linear in x: dx/dy + P(y)x = Q(y)

Linear in x

dxdy+P(y) x=Q(y) ⇒ x e∫P dy=∫Q e∫P dy dy+C\frac{dx}{dy}+P(y)\,x=Q(y)\ \Rightarrow\ x\,e^{\int P\,dy}=\int Q\,e^{\int P\,dy}\,dy+C

Bernoulli equations

Bernoulli substitution

y′+Py=Qyn, z=y1−n ⇒ z′+(1−n)P z=(1−n)Qy'+Py=Qy^n,\ z=y^{1-n}\ \Rightarrow\ z'+(1-n)P\,z=(1-n)Q

Substituting for a function of y

Substitution

f′(y) dydx+P(x) f(y)=Q(x), u=f(y) ⇒ dudx+Pu=Qf'(y)\,\frac{dy}{dx}+P(x)\,f(y)=Q(x),\ u=f(y)\ \Rightarrow\ \frac{du}{dx}+Pu=Q

Common traps

Integrate in y throughout

Once xx is the unknown, every integral is with respect to yy, including the integrating factor. Writing e∫P dxe^{\int P\,dx} out of habit mixes the variables.

The factor 1 - n

With z=y−1z=y^{-1}, z′=−y−2y′z'=-y^{-2}y': the minus sign flips both PP and QQ. Forgetting it gives an integrating factor of the wrong sign.

sin 2y becomes 2 tan y

After dividing by cos⁡2y\cos^2y, sin⁡2y\sin2y becomes 2tan⁡y2\tan y, not tan⁡y\tan y. The factor 2 goes straight into the integrating factor.

Using a Linear Solution

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Limits that fix the constant

General solution with constant coefficient

y′+ky=Q ⇒ y=yp(x)+Ce−kxy'+ky=Q\ \Rightarrow\ y=y_p(x)+Ce^{-kx}

Maxima, critical points and derivatives of the solution

Derivative from the equation

dydx=Q(x)−P(x) y  (no need to differentiate the solution)\frac{dy}{dx}=Q(x)-P(x)\,y\ \ \text{(no need to differentiate the solution)}

Integrating the solution: odd parts and areas

Symmetric interval

∫−aa(godd+geven) dx=2∫0ageven dx\int_{-a}^{a}\big(g_{\text{odd}}+g_{\text{even}}\big)\,dx=2\int_0^{a}g_{\text{even}}\,dx

Common traps

Which infinity

eαx→0e^{\alpha x}\to0 as x→−∞x\to-\infty only when α>0\alpha>0. Read which end the limit is taken at before deciding which sign of α\alpha the condition forces.

Check the endpoint and the domain

A quadratic in t=cos⁡xt=\cos x has its vertex inside the range only if that value of tt is attainable on the given interval. Otherwise the extreme value is at an end.

Split before integrating

Integrating the whole solution term by term on [−a,a][-a,a] wastes time on parts that cancel. Identify the odd terms first and drop them.

Equations Hidden in Integrals and Limits

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Differentiating an integral equation

Fundamental theorem

f(x)=h(x)+∫axg(t,f(t)) dt ⇒ f′(x)=h′(x)+g(x,f(x)), f(a)=h(a)f(x)=h(x)+\int_a^xg\big(t,f(t)\big)\,dt\ \Rightarrow\ f'(x)=h'(x)+g\big(x,f(x)\big),\ f(a)=h(a)

Equations from limits and derivative rules

The limit as a derivative

lim⁡t→xt2f(x)−x2f(t)t−x=2xf(x)−x2f′(x)\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=2xf(x)-x^2f'(x)

Common traps

The starting value is free

Differentiating throws away the constant, so the equation alone cannot give ff. Put the lower limit into the original relation to get f(a)f(a) — the question almost never states it.

Watch the order in the denominator

…x−t\frac{\dots}{x-t} and …t−x\frac{\dots}{t-x} differ by a sign. Differentiate in tt and divide by the derivative of the denominator in tt, which is −1-1 for x−tx-t.

Curves and Growth from Rates

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Curves from tangent, normal and area conditions

Tangent at (x, y)

Y−y=dydx(X−x):X-int=x−yy′,  Y-int=y−xy′Y-y=\frac{dy}{dx}(X-x):\quad X\text{-int}=x-\frac{y}{y'},\ \ Y\text{-int}=y-xy'

Growth, decay and cooling

Newton's law of cooling

dTdt=−k(T−A) ⇒ T−A=(T0−A) e−kt\frac{dT}{dt}=-k(T-A)\ \Rightarrow\ T-A=(T_0-A)\,e^{-kt}

Common traps

Midpoint and intercept are not the same

'The yy-axis bisects PQPQ' means the midpoint has x=0x=0, i.e. x+(x−yy′)=0x+\left(x-\frac{y}{y'}\right)=0. Setting the intercept itself to zero is a different condition.

The difference decays, not the temperature

In cooling, T−AT-A is multiplied by e−kte^{-kt}, not TT itself. Halving the temperature instead of its excess over the room gives a wrong answer.

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