Differential Equations

Here is a JEE question that stops most candidates:

$\left(1+y^2\right)dx-\left({\tan }^{-1}y-x\right)dy=0$

Try to separate the variables and you fail. Try it as a homogeneous equation and it isn't one. Try the linear method in $y$ and there is no $\frac{dy}{dx}+Py=Q$ to be found.

It is linear — in $x$. Divide by $dy$ and treat $x$ as the function:

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$

Now it is a standard integrating-factor problem and takes two minutes.

That is this chapter in one example. There are only three methods, and each is mechanical once you have chosen it. The marks are lost in the classification, not in the calculus.

This chapter is worth about 1 question (4 marks) in JEE Main.

1. Order and Degree

Order is the order of the highest derivative present. Degree is the power to which that highest derivative is raised — but only after the equation has been made polynomial in its derivatives.

That qualifier is where the marks are.

Two rules follow.

Clear radicals and fractions in the derivatives first. Squaring the third example gives $1+\frac{dy}{dx}={\left(\frac{d^{2}y}{d{x}^2}\right)}^2$, and the highest derivative appears to the second power — but degree counts the power of the highest-order derivative once the equation is polynomial, which here is $2$. Read the definition carefully rather than the shape.

Degree is undefined when the equation can never be made polynomial in the derivatives — when a derivative sits inside a sine, a logarithm or an exponential. Degree is not "1" and not "0" in that case; there is no degree.

Illustration 1 · Order and degree

State the order and degree of ${\left(\frac{dy}{dx}\right)}^3+2y\frac{d^{2}y}{d{x}^2}=\sin x$, and of $e^{dy/dx}+y=x$.

For the first, the highest derivative is $\frac{d^{2}y}{d{x}^2}$, so the order is 2. The equation is already polynomial in its derivatives, and that highest derivative appears to the first power, so the degree is 1.

The cube on $\frac{dy}{dx}$ is irrelevant. Degree counts the power of the highest-order derivative only.

For the second, the highest derivative is $\frac{dy}{dx}$, so the order is 1. But it sits inside an exponential, and no algebraic manipulation will make the equation polynomial in $\frac{dy}{dx}$.

So the degree is undefined. Writing $1$ here is the standard error.

2. Formation of a Differential Equation

Solving runs one way; forming runs the other. Given a family of curves with $n$ arbitrary constants, differentiate $n$ times and eliminate the constants between the resulting equations.

$n\text{ arbitrary constants}\Rightarrow \text{differentiate }n\text{ times}\Rightarrow \text{an equation of order }n$

The order of the result is forced, not chosen: each differentiation supplies one equation, and $n$ constants need $n$ equations to be eliminated.

Illustration 2 · Forming an equation from a family

Form the differential equation of the family $y=C{x}^2$.

One constant, so differentiate once and then eliminate.

$\frac{dy}{dx}=2Cx\Rightarrow C=\frac{1}{2x}\frac{dy}{dx}$

Substitute back into the original relation, which is the elimination step.

$y=\left(\frac{1}{2x}\frac{dy}{dx}\right)x^2=\frac{x}{2}\frac{dy}{dx}\Rightarrow x\frac{dy}{dx}=2y$

There is a faster route worth knowing. Divide the differentiated equation by the original one, and $C$ cancels without ever being isolated.

$\frac{dy/dx}{y}=\frac{2Cx}{C{x}^2}=\frac{2}{x}$

Dividing beats substituting whenever the constant appears as a single multiplying factor, which is the commonest case.

Illustration 3 · Two constants force order two

Form the differential equation of the family $y=A\sin x+B\cos x$.

Two arbitrary constants, so the result must be of order two and one differentiation will not be enough.

$\frac{dy}{dx}=A\cos x-B\sin x,\frac{d^{2}y}{d{x}^2}=-A\sin x-B\cos x$

The second derivative is the negative of the original expression, so both constants vanish at once.

$\frac{d^{2}y}{d{x}^2}=-y\Rightarrow \frac{d^{2}y}{d{x}^2}+y=0$

No elimination algebra was needed at all, because the second derivative reproduced the family's own shape. That happens for every family built from sines, cosines and exponentials, which is why those families give such clean equations.

Trap. Count the arbitrary constants, not every letter. In the family of circles $x^2+(y-c{)}^2=25$ the radius is fixed, so there is one constant and the equation is of order one.

3. What a Solution Is

A general solution contains as many arbitrary constants as the order of the equation. A particular solution has those constants pinned by given conditions.

So a first-order equation has one arbitrary constant, and one initial condition fixes it. A question giving you a point the curve passes through is handing you that condition.

A solution is anything that satisfies the equation, and the test is always the same: substitute it back. That check costs fifteen seconds and catches sign errors, missing constants and dropped factors, all of which are otherwise invisible.

Trap. A general solution is a family, not a curve. A question asking for "the curve" is asking for a particular solution and must supply a condition; if it supplies none, the answer keeps its constant.

4. The Classification Decision

Run these three tests, in this order, on any first-order equation.

Can you separate? Get all the $y$ terms with $dy$ and all the $x$ terms with $dx$. If the equation factorises as $\frac{dy}{dx}=f(x)g(y)$, it separates.

Is it homogeneous? Every term has the same total degree in $x$ and $y$ together, so the equation can be written as $\frac{dy}{dx}=F\left(\frac{y}{x}\right)$. Substitute $y=vx$.

Is it linear? It has the shape $\frac{dy}{dx}+P(x)y=Q(x)$ — first power of $y$, first power of the derivative, no products of the two.

If none fits, swap the roles of $x$ and $y$ and run the three tests again. That single move solves the question this chapter opened with, and it is the most commonly missed step in the topic.

Illustration 4 · Running the three tests

Classify each equation, without solving any of them.

$\text{(i)}\frac{dy}{dx}=xy+x,\text{(ii)}\frac{dy}{dx}=\frac{x^2+y^2}{xy},\text{(iii)}x\frac{dy}{dx}-3y=x^4$

(i) Factor the right side before deciding anything: $xy+x=x(y+1)$, a function of $x$ times a function of $y$.

$\frac{dy}{y+1}=xdx\Rightarrow \text{separable}$

It is also linear, since it rearranges to $\frac{dy}{dx}-xy=x$. When two methods apply, take the shorter, and separation is shorter here because it needs no integrating factor.

(ii) Nothing separates, since $x$ and $y$ are entangled in both parts. Check the degrees instead: numerator and denominator are both of total degree $2$.

$\frac{dy}{dx}=\frac{x}{y}+\frac{y}{x}=F\left(\frac{y}{x}\right)\Rightarrow \text{homogeneous}$

(iii) The $x^4$ has nothing to do with $y$, so it is not homogeneous, and $y$ cannot be gathered on one side. Divide by $x$ to test linearity.

$\frac{dy}{dx}-\frac{3}{x}y=x^3\Rightarrow \text{linear},P=-\frac{3}{x}$

Each classification took one line, and each was decided by a different feature: a factorisation, a count of degrees, and a rearrangement into standard form. Those are the only three things to look for.

5. Variable Separable

This is the method the syllabus calls separation of variables. Get the two variables onto opposite sides and integrate each:

$\int \frac{dy}{g(y)}=\int f(x)dx+C$

Add the constant once, on one side only. Two constants collapse into one and writing both wastes time.

Illustration 5 · Separable, with a condition

Find the curve through $(1,1)$ with $\frac{dy}{dx}=\frac{2y}{x}$. (JEE previous-year question)

The right side factorises as a function of $x$ times a function of $y$, so it separates:

$\frac{dy}{y}=\frac{2dx}{x}$

Integrating both sides, with one constant:

$\log |y|=2\log |x|+\log C$, so $y=C{x}^2$.

Writing the constant as $\log C$ rather than $C$ is a small trick that turns the sum of logarithms into a product immediately.

Applying $y=1$ at $x=1$ gives $C=1$.

$y=x^2$

Illustration 6 · Separable in disguise

Find the curve through $(1,0)$ satisfying $\left(1+y^2\right)dx-xydy=0$. (JEE previous-year question)

It looks unpromising, but rearranging separates it:

$\frac{dx}{x}=\frac{ydy}{1+y^2}$

Both sides are now standard. The right side is of the form $\frac{f^{\prime }}{f}$ up to a factor of $\frac{1}{2}$:

$\log |x|=\frac{1}{2}\log \left(1+y^2\right)+\log C$

Doubling and exponentiating: $x^2=C^2\left(1+y^2\right)$.

At $(1,0)$: $1=C^2$, so

$x^2-y^2=1$

The curve is a rectangular hyperbola. Checking a second point is worthwhile: at $y=1$, $x^2=2$, and substituting back into the original equation confirms it.

6. Homogeneous Equations

Substitute $y=vx$, so that

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

The result always separates in $v$ and $x$. That is the entire point of the substitution — it converts a type you cannot separate into one you can.

After integrating, substitute $v=\frac{y}{x}$ back. Leaving the answer in $v$ is an incomplete solution.

If the equation is simpler in the other direction, substitute $x=vy$ instead. Choose whichever makes the algebra shorter.

Illustration 7 · Homogeneous

Solve $2xy\frac{dy}{dx}=x^2+3y^2$. (JEE previous-year question)

Every term is of total degree $2$, so it is homogeneous. Substitute $y=vx$, giving $\frac{dy}{dx}=v+x\frac{dv}{dx}$:

$2x(vx)\left(v+x\frac{dv}{dx}\right)=x^2+3v^2{x}^2$

Divide through by $x^2$:

$2v\left(v+x\frac{dv}{dx}\right)=1+3v^2$, so $2v^2+2vx\frac{dv}{dx}=1+3v^2$.

The $v^2$ terms partly cancel, leaving $2vx\frac{dv}{dx}=1+v^2$, which separates:

$\frac{2vdv}{1+v^2}=\frac{dx}{x}$

The left side is again $\frac{f^{\prime }}{f}$, so integrating gives $\log \left(1+v^2\right)=\log |x|+\log C$, hence $1+v^2=Cx$.

Substituting $v=\frac{y}{x}$ back and multiplying by $x^2$:

$x^2+y^2=C{x}^3$

Illustration 8 · Homogeneous with a condition

Given $\left(x^2+y^2\right)dy=xydx$ with $y(1)=1$, find $x_0$ such that $y(x_0)=e$. (JEE previous-year question)

Homogeneous again. With $y=vx$:

$v+x\frac{dv}{dx}=\frac{x\cdot vx}{x^2+v^2{x}^2}=\frac{v}{1+v^2}$

So $x\frac{dv}{dx}=\frac{v}{1+v^2}-v=\frac{-v^3}{1+v^2}$, which separates as

$-\frac{\left(1+v^2\right)dv}{v^3}=\frac{dx}{x}$

Splitting the left side into $v^{-3}+v^{-1}$ and integrating:

$\frac{1}{2v^2}-\log |v|=\log |x|+C$

Now substitute $v=\frac{y}{x}$. The term $-\log \frac{y}{x}$ becomes $-\log y+\log x$, and that $\log x$ cancels the one on the right:

$\frac{x^2}{2y^2}-\log y=C$

Apply $y=1$ at $x=1$: $\frac{1}{2}-0=C$, so $C=\frac{1}{2}$.

At $y=e$: $\frac{x^2}{2e^2}-1=\frac{1}{2}$, so $x^2=3e^2$.

So $x_0=\pm \sqrt{3}e$.

Illustration 9 · Homogeneous with a tidy integral

Solve $\frac{dy}{dx}=\frac{x-y}{x+y}$. (JEE previous-year question)

Numerator and denominator are both of degree $1$, so it is homogeneous. With $y=vx$:

$v+x\frac{dv}{dx}=\frac{1-v}{1+v}$

So $x\frac{dv}{dx}=\frac{1-v}{1+v}-v=\frac{1-2v-v^2}{1+v}$, giving

$\frac{(1+v)dv}{1-2v-v^2}=\frac{dx}{x}$

Now notice that $\frac{d}{dv}\left(1-2v-v^2\right)=-2(1+v)$, so the left side is $-\frac{1}{2}$ of the $\frac{f^{\prime }}{f}$ form:

$-\frac{1}{2}\log \left|1-2v-v^2\right|=\log |x|+C$

Rearranging gives $1-2v-v^2=\frac{K}{x^2}$, and substituting $v=\frac{y}{x}$ then multiplying by $x^2$:

$x^2-2xy-y^2=K$

Spotting that the numerator was proportional to the derivative of the denominator is what kept this to three lines.

7. Linear Equations and the Integrating Factor

For $\frac{dy}{dx}+P(x)y=Q(x)$, multiply through by

$\text{IF}=e^{\int Pdx}$

The point of that multiplier is that it makes the left side an exact derivative. After multiplying, the left side is precisely $\frac{d}{dx}\left(y\cdot \text{IF}\right)$, so integrating both sides gives

$y\cdot \text{IF}=\int Q\cdot \text{IF}dx+C$

Why it works. Multiplying gives $e^{\int P}\frac{dy}{dx}+P{e}^{\int P}y$. By the product rule, $\frac{d}{dx}\left(y{e}^{\int P}\right)=e^{\int P}\frac{dy}{dx}+y{e}^{\int P}\cdot P$, which is the same expression. The integrating factor is chosen to make that match.

Two cautions. Put the equation in standard form first — the coefficient of $\frac{dy}{dx}$ must be $1$, or $P$ is wrong. And no constant is needed inside the integrating factor; it would cancel.

For an equation linear in $x$, everything mirrors: $\frac{dx}{dy}+P(y)x=Q(y)$, with $\text{IF}=e^{\int Pdy}$.

Illustration 10 · Linear in y

Solve $\frac{dy}{dx}+2y=e^{-x}$.

It is already in standard form with $P=2$ and $Q=e^{-x}$.

$\text{IF}=e^{\int 2dx}=e^{2x}$

Multiplying through, the left side becomes $\frac{d}{dx}\left(y{e}^{2x}\right)$, so

$y{e}^{2x}=\int e^{-x}{e}^{2x}dx=\int e^{x}dx=e^x+C$

$y=e^{-x}+C{e}^{-2x}$

Check by substituting: $\frac{dy}{dx}=-e^{-x}-2C{e}^{-2x}$, and adding $2y$ gives $e^{-x}$, as required.

Illustration 11 · Linear in x, not y

Find the integrating factor for $\left(1+y^2\right)dx-\left({\tan }^{-1}y-x\right)dy=0$. (JEE previous-year question)

No separation, and it is not homogeneous. Test linearity in $y$: there is no clean $\frac{dy}{dx}+Py=Q$.

Now swap the roles. Divide throughout by $dy$ and by $1+y^2$:

$\frac{dx}{dy}=\frac{{\tan }^{-1}y-x}{1+y^2}$, which rearranges to $\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$.

This is linear in $x$, with $P(y)=\frac{1}{1+y^2}$.

$\text{IF}=e^{\int \frac{dy}{1+y^2}}=e^{{\tan }^{-1}y}$

The whole difficulty was classification. Once $x$ is treated as the dependent variable, the method is the standard one.

Illustration 12 · Linear, but not yet in standard form

Solve $x\frac{dy}{dx}+2y=x^2$.

It is linear, but $P$ cannot be read off until the coefficient of the derivative is $1$. Divide throughout by $x$:

$\frac{dy}{dx}+\frac{2}{x}y=x$

Now $P=\frac{2}{x}$ and $Q=x$. Reading $P=2$ from the original form is the standard error here, and it produces a completely different answer.

$\text{IF}=e^{\int \frac{2}{x}dx}=e^{2\log x}=x^2$

Multiplying, the left side becomes $\frac{d}{dx}\left(y{x}^2\right)$, so

$y{x}^2=\int x\cdot x^{2}dx=\frac{x^4}{4}+C$

$y=\frac{x^2}{4}+\frac{C}{x^2}$

Checking: $x\frac{dy}{dx}=\frac{x^2}{2}-\frac{2C}{x^2}$ and $2y=\frac{x^2}{2}+\frac{2C}{x^2}$, which sum to $x^2$.

8. Exact Differentials Worth Recognising

Some equations that fail all three tests collapse instantly if you recognise a differential you already know.

Whenever you see $xdy-ydx$ in a question, look for one of these. Dividing by $x^2$, $y^2$ or $x^2+y^2$ usually finishes the problem in one line.

Illustration 13 · An exact differential

Solve $xdy=y\left(dx+ydy\right)$ with $y>0$ and $y(1)=1$, then find $y$ when $x=-3$. (JEE previous-year question)

Expanding: $xdy=ydx+y^{2}dy$, so $xdy-ydx=y^{2}dy$.

The left side is the signature of a known differential. Divide both sides by $y^2$:

$\frac{xdy-ydx}{y^2}=dy$

The left side is exactly $-d\left(\frac{x}{y}\right)$. So $-d\left(\frac{x}{y}\right)=dy$, and integrating:

$-\frac{x}{y}=y+C$, that is $x=-y^2-Cy$.

At $(1,1)$: $1=-1-C$, so $C=-2$ and $x=-y^2+2y$.

Setting $x=-3$: $y^2-2y-3=0$, so $(y-3)(y+1)=0$.

Since $y>0$ is given, $y=-1$ is rejected.

$y=3$

Attempting this by separation or substitution goes nowhere. Recognising $xdy-ydx$ is the whole solution.

Illustration 14 · Spotting a product differential

Find the general solution of $(x+y)dx+xdy=0$. (JEE previous-year question)

It is homogeneous, so the substitution would work. But grouping is faster.

Split the first bracket and regroup:

$xdx+\left(ydx+xdy\right)=0$

The bracket is exactly $d(xy)$, and the first term is $d\left(\frac{x^2}{2}\right)$. So the whole equation reads

$d\left(\frac{x^2}{2}+xy\right)=0$

An expression whose differential vanishes is constant, so $\frac{x^2}{2}+xy=c$, and doubling:

$x^2+2xy=c$

No integration was performed at all. Whenever you see $ydx+xdy$ sitting inside an equation, group it rather than substituting.

9. A Note on Syllabus Emphasis

The unit text names ordinary differential equations, their order and degree, formation of differential equations, separation of variables, and homogeneous and linear equations of the type $\frac{dy}{dx}+P(x)y=Q(x)$. That is the whole unit.

The four exact differentials in the previous section are worth keeping anyway, because they rescue equations that appear to fail all three named tests and take one line each. Bernoulli's equation is worth a glance for Advanced sitters, since it reduces to the linear type by a substitution.

Illustration 15 · Beyond JEE Main: reducing to the linear type

Solve $\frac{dy}{dx}+\frac{y}{x}=y^2$.

The $y^2$ on the right breaks linearity, so none of the three tests applies as written. Divide throughout by $y^2$ to isolate a combination that can be substituted.

$y^{-2}\frac{dy}{dx}+\frac{1}{x}{y}^{-1}=1$

Now put $v=y^{-1}$, so that $\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}$, which is exactly the first term with its sign flipped.

$-\frac{dv}{dx}+\frac{v}{x}=1\Rightarrow \frac{dv}{dx}-\frac{v}{x}=-1$

This is linear in $v$ with $P=-1/x$, so the integrating factor is $e^{-\log x}=1/x$.

$\frac{v}{x}=\int -\frac{1}{x}dx=-\log |x|+C\Rightarrow \frac{1}{xy}=C-\log |x|$

The whole trick is that dividing by the offending power produces a derivative you already have. That is the same instinct as swapping $x$ and $y$: reshape until one of the three named methods applies.

Summary

Solving is a classification problem: three methods, each mechanical once chosen.

Order is the highest derivative; degree is its power after the equation is made polynomial in the derivatives, and is undefined when that is impossible.

Forming runs the other way: $n$ arbitrary constants need $n$ differentiations, and the order of the result is forced rather than chosen. Divide rather than substitute when the constant is a single multiplying factor.

A general solution has as many constants as the order; a point on the curve is the condition that fixes one.

A general solution is a family of curves; one condition selects one member, and substituting the answer back is the check that costs fifteen seconds.

Test in order: separable, then homogeneous, then linear. If none fits, swap $x$ and $y$ and test again.

Separable: split and integrate, adding one constant on one side only. Writing it as $\log C$ turns sums of logarithms into products.

Homogeneous: every term the same total degree; substitute $y=vx$, which always separates, then substitute back.

Linear: force the coefficient of the derivative to $1$, then multiply by $e^{\int Pdx}$, which makes the left side an exact derivative by construction.

The integrating factor needs no constant of its own, and mirrors exactly for equations linear in $x$.

Learn the four standard exact differentials. Seeing $xdy-ydx$ should immediately suggest dividing by $x^2$, $y^2$ or $x^2+y^2$.