Meet Your Tutor
Differential equations become systematic when you classify the equation before solving it. I will help you identify order and degree, choose variable separation or a homogeneous method, apply the initial condition and differentiate the result to verify it.
Take a breath before you start this one. Differential Equations has a reputation for looking frightening, and that reputation comes almost entirely from the notation rather than from the ideas. The moment you see d²y/dx² sitting inside a bracket raised to a power, something in the brain says this is not for me. I want to gently disagree with that voice, because here is the honest shape of this chapter: you already know how to differentiate, you already know how to integrate, and this chapter is mostly about deciding which of your integration skills to reach for. That is it.
Think of it like cooking. You already own the knives and the pans. A differential equation just hands you an unfamiliar-looking ingredient and asks, which technique does this one need? Once you can recognise the ingredient — and there are only three recipes on the CBSE menu for 2026-27 — the actual cooking is ordinary integration you have been doing since Class 11. We are going to build that recognition slowly, from absolute zero, with a decision map you can picture during the exam.
Work through this page with a pen in your hand. Reading maths and doing maths are two different activities, and only one of them earns marks. Take it one section at a time, and please do not move on from a section until it genuinely feels comfortable — there is no prize for finishing fast.
What You’ll Learn
- What a Differential Equation Actually Is
- Order of a Differential Equation
- Degree of a Differential Equation
- When the Degree Is Not Defined
- General and Particular Solutions
- Checking Whether a Given Function Is a Solution
- Choosing Your Method: The Decision Map
- Method 1: Variables Separable
- Separable Equations With an Initial Condition
- Method 2: Homogeneous Equations and the y = vx Substitution
- When x = vy Is the Easier Substitution
- Method 3: Linear Equations of the Type dy/dx + Py = Q
- Linear Equations of the Type dx/dy + Px = Q
- Pinning Down the Constant: Particular Solutions
- Good to Know: Forming a Differential Equation From a Family of Curves
- Common Traps and How to Dodge Them
Your Game Plan
Here is the order I would work in if I were sitting next to you. It is deliberately slow at the start, because the first four sections are the ones that make everything afterwards feel easy.
- Spend one sitting on vocabulary. Order, degree, general solution, particular solution. These are free marks in the exam and they take about forty minutes to nail down properly.
- Learn to recognise before you learn to solve. Go to the decision map, and practise only identifying which of the three methods an equation needs. Do not solve anything yet. Ten equations, identification only.
- Master variables separable completely. It is the simplest method and it is hiding inside the other two, so time spent here pays twice.
- Then homogeneous, then linear. One method per sitting. Do not mix them on the day you first meet them — mix them the day after.
- Finally, mixed practice. Cover the method names, shuffle twenty questions, and force yourself to choose. This is the skill the exam actually tests.
- Do the worksheet at the bottom with the answers hidden. Write full solutions on paper first, then check. Checking without writing teaches you almost nothing.
Study Notes
What a Differential Equation Actually Is
Let us start with something you already know. An ordinary equation, like 3x + 5 = 14, asks you to find a number. You solve it and you get x = 3. One number, and you are done.
A differential equation asks a bigger question. It asks you to find a function. And instead of telling you about the function directly, it tells you something about the function’s rate of change, and asks you to work backwards to the function itself.
So an equation like dy/dx = 2x is a differential equation. It is not saying “find a number”. It is saying: I am thinking of a function whose slope at every point equals twice the x-coordinate. Which function am I thinking of? And you, doing the reverse of differentiation, answer: y = x² + C.
Why does anyone care? Because almost every rule in science is naturally written as a statement about rates of change, not about quantities. A cup of tea cools at a rate proportional to how much hotter it is than the room. A population grows at a rate proportional to its own size. Radioactive material decays at a rate proportional to how much is left. In each case you are handed a sentence about the rate, and you want the actual formula. That translation — from a rule about change to a formula for the thing itself — is exactly what this chapter teaches.
Here is that tea example written down, so the abstraction lands. Let T be the temperature of the tea. “Cools at a rate proportional to the excess over room temperature R” becomes dT/dt = −k(T − R). That is a differential equation, and by the end of this chapter you will be able to solve it in about four lines.
Order of a Differential Equation
This is the easiest definition in the chapter, and it appears in the exam almost every year, usually as a one-mark question.
So you scan the equation, you find the highest derivative present, and you read off its order. dy/dx is a first-order derivative. d²y/dx² is second order. d³y/dx³ is third order, and so on.
The single most common mistake here is letting a large power distract you. A power tells you nothing about order. (dy/dx)⁴ is still a first-order derivative; it has just been raised to the fourth power, which is a completely separate piece of information.
Find the order of each equation.
(i) (d³y/dx³)² + 4(dy/dx)⁵ + y = 0
Derivatives present: d³y/dx³ (order 3) and dy/dx (order 1). The highest is third order, so the order is 3. The powers 2 and 5 are irrelevant here.
(ii) d²y/dx² + x dy/dx = sin x
The highest derivative is d²y/dx², so the order is 2.
(iii) dy/dx + 7y = eˣ
Only a first derivative appears, so the order is 1.
Degree of a Differential Equation
Degree is slightly fussier than order, but only because of one condition that people skip past. Here it is in full.
Two things in that rule deserve emphasis. First, the tidying-up step comes before you read anything off. Second, when you do read it off, you look at the power of the highest-order derivative only — the powers on the lower derivatives are decoys.
Find the order and degree of (d²y/dx²)² + (dy/dx)³ − 5y = 0.
Order. The highest derivative is d²y/dx², so the order is 2.
Degree. The equation is already a polynomial in the derivatives — no roots, nothing trapped inside a function — so we may read it off. The highest-order derivative d²y/dx² is raised to the power 2, so the degree is 2.
Notice that the cube on dy/dx played no part at all. It is a lower-order derivative, so its power is simply not what “degree” is asking about.
Find the order and degree of (1 + (dy/dx)²)3/2 = d²y/dx².
The order is easy: the highest derivative is d²y/dx², so the order is 2.
Degree, however, cannot be read off yet — the left-hand side has a fractional power of 3/2, so this is not a polynomial in the derivatives. Square both sides to clear it:
(1 + (dy/dx)²)³ = (d²y/dx²)²
Now everything is a whole-number power. The highest-order derivative d²y/dx² appears to the power 2, so the degree is 2.
If you had skipped the squaring step and stared at the original, you would have had nothing sensible to say. That is the whole reason the tidying-up step exists.
When the Degree Is Not Defined
This is a small topic that carries a surprisingly reliable one-mark question, and it is the kind of mark that students hand back simply because nobody warned them. So consider yourself warned.
Sometimes an equation can never be turned into a polynomial in its derivatives, no matter how much algebra you throw at it. That happens when a derivative is trapped inside another function — sin, cos, tan, log, or an exponential. In that situation the degree is not defined. Note carefully: the degree is not zero, and it is not one. It simply does not exist.
(i) sin(dy/dx) + y = 0
The only derivative is dy/dx, so the order is 1. But dy/dx is locked inside a sine. No amount of rearranging will free it into a polynomial, so the degree is not defined.
(ii) d²y/dx² + edy/dx = 0
Highest derivative is d²y/dx², so the order is 2. But dy/dx sits in the exponent, so again the degree is not defined.
(iii) d²y/dx² + sin x × dy/dx = 0
Careful — this one is fine! The sine contains x, not a derivative. Both derivatives appear to the first power, so the order is 2 and the degree is 1.
General and Particular Solutions
When you solve dy/dx = 2x you write y = x² + C, and you have probably never thought hard about that C. It deserves a moment, because it is the whole reason this chapter has two kinds of answer.
That C is not decoration. Different values of C give genuinely different functions — y = x², y = x² + 7, y = x² − 3 — and every single one of them has slope 2x at every point. So the differential equation on its own does not pin down one curve. It pins down an entire family of curves, all parallel in shape, sliding up and down.
How many arbitrary constants should the general solution have? Exactly as many as the order of the equation. A first-order equation gives one constant, a second-order equation gives two. That is a useful sanity check: if you solve a first-order equation and end up with two constants, something has gone wrong.
The extra information that selects one member is called an initial condition (or boundary condition). It usually arrives as a sentence like “given that y = 3 when x = 0”, or in shorthand as y(0) = 3.
Show that y = A cos x + B sin x satisfies d²y/dx² + y = 0, and then find the particular solution for which y = 1 and dy/dx = 2 when x = 0.
Verifying. Differentiating, dy/dx = −A sin x + B cos x, and d²y/dx² = −A cos x − B sin x = −y. So d²y/dx² + y = 0 for every A and B. It is a solution, and since the equation has order 2 and the answer carries two constants, this is the general solution.
Applying the conditions. Put x = 0 into y: 1 = A cos 0 + B sin 0 = A, so A = 1.
Put x = 0 into dy/dx: 2 = −A sin 0 + B cos 0 = B, so B = 2.
The particular solution is y = cos x + 2 sin x. Quick check: at x = 0 it gives y = 1, and its derivative −sin x + 2 cos x gives 2 at x = 0. Both conditions hold.
Checking Whether a Given Function Is a Solution
Some questions do not ask you to solve anything. They hand you a function and an equation and say “verify that this function is a solution”. These are among the friendliest marks in the paper, because there is no cleverness required — just careful differentiation.
The single best habit here is to work on one side only. Do not manipulate both sides at once and hope they meet in the middle — take the left-hand side, substitute, simplify, and show it turning into the right-hand side. Examiners want to see that one-directional flow.
Verify that y = e−3x is a solution of d²y/dx² + 6dy/dx + 9y = 0.
Differentiate once: dy/dx = −3e−3x.
Differentiate again: d²y/dx² = 9e−3x.
Now substitute into the left-hand side:
LHS = 9e−3x + 6(−3e−3x) + 9(e−3x) = (9 − 18 + 9)e−3x = 0 = RHS.
Since the left-hand side reduces to zero, y = e−3x is a solution.
As a bonus, the same equation is also satisfied by y = x e−3x. If you would like the practice, differentiate it twice using the product rule and watch the terms cancel in exactly the same way.
Show that y = C/x is a solution of xdy/dx + y = 0 for every constant C.
Write y = Cx−1. Then dy/dx = −Cx−2 = −C/x².
LHS = x(−C/x²) + C/x = −C/x + C/x = 0 = RHS.
So y = C/x satisfies the equation whatever value C takes — which is precisely what makes it the general solution of this first-order equation.
Choosing Your Method: The Decision Map
This section contains no new mathematics. It contains something more valuable: a habit. Before you write a single line of working on a first-order differential equation, ask three questions in a fixed order. That order matters, and here is why.
Ask about separability first, because it is the fastest method and because some equations that look homogeneous or linear are quietly separable. Spotting that saves you five minutes. Ask about homogeneity second. Ask about linearity last — not because it is hardest, but because if the first two fail, linear is almost certainly the answer, so you barely need to check.
| Method | How to spot it | What to do |
|---|---|---|
| Variables separable | dy/dx can be written as f(x) × g(y) — every x can be herded to one side and every y to the other. | Rewrite as dy/g(y) = f(x) dx, integrate both sides, add a single constant C. |
| Homogeneous | All terms have the same total degree, so dy/dx depends only on the ratio y/x (for example (x² + y²)/(2xy)). | Put y = vx, so dy/dx = v + x dv/dx. The equation becomes separable in v and x. Solve, then replace v by y/x. |
| Linear | Fits dy/dx + Py = Q with P, Q in x only (or dx/dy + Px = Q with P, Q in y only). The dependent variable and its derivative both appear to the first power, and never multiplied together. | Compute IF = e∫P dx, then use y × IF = ∫(Q × IF) dx + C. |
Method 1: Variables Separable
This is the foundation method, and it is beautifully simple in principle: get all the y-stuff (including dy) onto one side, all the x-stuff (including dx) onto the other, then integrate each side with respect to its own variable.
dy / g(y) = f(x) dx
and integrate both sides. You need only one arbitrary constant on the whole equation, conventionally written on the right.
Two practical points before we begin. First, treating dy and dx as though they were ordinary quantities you can multiply and divide is not fully rigorous, but it is exactly what CBSE expects and it always gives the right answer for the equations in this syllabus. Do it without guilt. Second, when you divide by g(y) you are quietly assuming g(y) is not zero — that is standard at this level and you are not expected to comment on it.
Solve dy/dx = (1 + y²) / (1 + x²).
Separate. Everything with y goes left, everything with x goes right:
dy / (1 + y²) = dx / (1 + x²)
Integrate both sides. Both are standard forms you know:
tan−1y = tan−1x + C
That is the general solution. You could rearrange it as y = tan(tan−1x + C), but there is no need — an implicit answer is perfectly acceptable, and often tidier.
Solve sec²x tan y dx + sec²y tan x dy = 0.
It does not look separable at a glance, but look again: the first term has only x-things multiplied by only y-things, and so does the second. Move one term across and divide by tan x tan y:
(sec²y / tan y) dy = − (sec²x / tan x) dx
Each side is now of the form (derivative of the bottom) / (the bottom), which integrates to a logarithm:
log|tan y| = − log|tan x| + log|C|
Writing the constant as log|C| lets us combine the logs immediately:
log|tan y| + log|tan x| = log|C|, so tan x tan y = C.
Solve dy/dx = (x + 1) / (2 − y), where y ≠ 2.
Separate: (2 − y) dy = (x + 1) dx.
Integrate both sides: 2y − y²/2 = x²/2 + x + C′.
Multiplying through by 2 to clear the fractions, and renaming the constant:
4y − y² = x² + 2x + C
You cannot make y the subject here without a square root and a sign ambiguity, so leave it exactly like this. An implicit general solution is a complete answer.
Separable Equations With an Initial Condition
Everything you just learned, plus one final step: use the given point to find C. The important discipline is when to do it — find the general solution first, all the way, and only then substitute. Substituting too early is a reliable way to lose marks.
Solve dy/dx = ex − y given that y = 1 when x = 0.
First rewrite the right-hand side so the separation is obvious: ex − y = ex × e−y. Now it is visibly f(x) × g(y).
Separate: ey dy = ex dx.
Integrate: ey = ex + C. That is the general solution.
Now, and only now, apply the condition. Put x = 0, y = 1: e1 = e0 + C, so e = 1 + C, giving C = e − 1.
The particular solution is ey = ex + e − 1, or explicitly y = log(ex + e − 1).
Check: at x = 0 this gives y = log(1 + e − 1) = log e = 1. The condition is satisfied.
Solve dy/dx + 2xy = 0 given y(0) = 3.
It is tempting to call this linear — it does fit dy/dx + Py = Q with Q = 0 — but run the decision map properly and you will notice it separates in one step, which is far quicker.
Rearrange: dy/dx = −2xy, so dy/y = −2x dx.
Integrate: log|y| = −x² + C′.
Exponentiate both sides: y = Ae−x², where A = eC′ is just a new constant.
Apply y(0) = 3: 3 = Ae0 = A. So the particular solution is y = 3e−x².
This is exactly the shape of a decay curve — and notice how much less work it was than computing an integrating factor would have been.
Method 2: Homogeneous Equations and the y = vx Substitution
Now for the method that students find most mysterious at first meeting, and most mechanical once it clicks. Let us build it slowly.
A first-order equation is called homogeneous when every term has the same total degree — where “degree” here means adding the powers of x and y together. In (x² + y²) every term is degree 2. In 2xy every term is degree 2. So (x² + y²)/(2xy) is a ratio of two degree-2 expressions, and the practical consequence is lovely: the x’s cancel and the whole thing depends only on the ratio y/x.
That observation is the entire method. If the right-hand side only cares about y/x, then give that ratio a name — call it v — and the equation should simplify dramatically.
That derivative line is the one people forget. y = vx is a product of two things that both depend on x (v is a function of x, and so is x itself), so dy/dx is not just v. It is v + x dv/dx. Write it out every single time.
Solve (x² + y²) dx − 2xy dy = 0.
Rearrange into derivative form: dy/dx = (x² + y²) / (2xy). Every term is degree 2 above and below, so it is homogeneous.
Put y = vx, so dy/dx = v + x dv/dx. The right-hand side becomes (x² + v²x²)/(2x · vx) = (1 + v²)/(2v).
So v + x dv/dx = (1 + v²)/(2v).
Isolate the derivative term: x dv/dx = (1 + v²)/(2v) − v = (1 + v² − 2v²)/(2v) = (1 − v²)/(2v).
Now it is separable: 2v dv/(1 − v²) = dx/x.
The left side is −(derivative of the bottom)/(the bottom), so it integrates to −log|1 − v²|:
−log|1 − v²| = log|x| − log|C|
Rearranging, log|x(1 − v²)| = log|C|, so x(1 − v²) = C.
Finally put v = y/x: x(1 − y²/x²) = C, which tidies to x² − y² = Cx.
Solve dy/dx = (x + y)/(x − y).
Top and bottom are both degree 1, so it is homogeneous. Put y = vx:
v + x dv/dx = (x + vx)/(x − vx) = (1 + v)/(1 − v)
x dv/dx = (1 + v)/(1 − v) − v = (1 + v − v + v²)/(1 − v) = (1 + v²)/(1 − v)
Separate: (1 − v)/(1 + v²) dv = dx/x.
Split the left side into two standard integrals: 1/(1 + v²) gives tan−1v, and −v/(1 + v²) gives −½ log(1 + v²).
tan−1v − ½ log(1 + v²) = log|x| + C
Put v = y/x. Since ½ log(1 + y²/x²) + log|x| = ½ log(x² + y²), the answer collapses neatly to
tan−1(y/x) − ½ log(x² + y²) = C
Solve xdy/dx = y + x ey/x.
Divide through by x: dy/dx = y/x + ey/x. The right-hand side depends only on y/x, which is the clearest possible signal to substitute.
Put y = vx: v + x dv/dx = v + ev, so x dv/dx = ev.
The v cancels completely — that is the reward for spotting the structure. Separate: e−v dv = dx/x.
Integrate: −e−v = log|x| + C′.
Putting v = y/x and renaming the constant, e−y/x + log|x| = C.
When x = vy Is the Easier Substitution
The substitution y = vx is the default, but it is not compulsory. Sometimes an equation is built around the ratio x/y instead, and forcing y = vx onto it produces a horrible mess. In those cases, put x = vy and everything falls out cleanly.
Practical rule of thumb: look at where the ratio appears. If you see y/x, use y = vx. If you see x/y, use x = vy. If neither ratio is written explicitly, use y = vx by default.
Solve (1 + ex/y) dx + ex/y(1 − x/y) dy = 0.
The ratio x/y is written all over this equation, so put x = vy, which gives dx = v dy + y dv.
Substituting, and writing ev for ex/y:
(1 + ev)(v dy + y dv) + ev(1 − v) dy = 0
Collect the dy terms: v + v ev + ev − v ev = v + ev. So:
(v + ev) dy + y(1 + ev) dv = 0
Separate: (1 + ev)/(v + ev) dv = − dy/y.
The numerator on the left is precisely the derivative of the denominator, so the integral is a logarithm:
log|v + ev| = − log|y| + log|C|
So y(v + ev) = C. Putting v = x/y gives y(x/y + ex/y) = C, that is
x + y ex/y = C
Solve y² dx + (x² − xy) dy = 0.
Every term is degree 2, so it is homogeneous and either substitution is legal. Because the dy coefficient factorises as x(x − y), the substitution x = vy is the tidier route.
Put x = vy, dx = v dy + y dv, and note x² − xy = v²y² − vy² = y²(v² − v):
y²(v dy + y dv) + y²(v² − v) dy = 0
Divide throughout by y² and collect the dy terms: (v + v² − v) dy + y dv = 0, that is v² dy + y dv = 0.
Separate: dv/v² = − dy/y.
Integrate: −1/v = −log|y| + C′, so 1/v = log|y| + C.
Putting v = x/y: y/x = log|y| + C.
Method 3: Linear Equations of the Type dy/dx + Py = Q
The third and final method, and the one with the most reliable procedure. Once you can spot the form, the rest is a recipe you can follow half asleep.
IF = e∫P dx and y × IF = ∫(Q × IF) dx + C.
What makes an equation linear? Two conditions, both about the dependent variable y. First, y and dy/dx each appear only to the first power. Second, they are never multiplied by each other, and y never appears inside a function like sin y or ey. Note that x can be as wild as it likes — x³, sin x, log x are all fine as coefficients.
Why does the integrating factor work? Briefly, and it is worth reading once: multiplying through by IF makes the left-hand side become exactly the derivative of the product y × IF. So the equation turns into d/dx(y × IF) = Q × IF, and then you just integrate both sides. Understanding this means you will never misremember which side gets multiplied.
Before computing the integrating factor, the equation must be in standard form — the coefficient of dy/dx has to be exactly 1. If your equation starts as xdy/dx + 2y = x³, divide through by x first. Skipping this step is the most common way this method goes wrong.
Solve dy/dx + 2y = e3x.
It is already in standard form, with P = 2 and Q = e3x.
IF = e∫2 dx = e2x.
Apply the formula: y × e2x = ∫ e3x × e2x dx = ∫ e5x dx = e5x/5 + C.
Divide through by e2x:
y = e3x/5 + Ce−2x
Solve xdy/dx − y = x².
The coefficient of dy/dx is x, not 1, so divide everything by x:
dy/dx − y/x = x, so P = −1/x and Q = x.
IF = e∫(−1/x) dx = e−log x = 1/x.
Apply the formula: y × (1/x) = ∫ x × (1/x) dx = ∫ 1 dx = x + C.
Multiply through by x: y = x² + Cx.
Worth a quick check: differentiating gives dy/dx = 2x + C, and then x(2x + C) − (x² + Cx) = 2x² + Cx − x² − Cx = x². Correct.
Solve (1 + x²)dy/dx + 2xy = 1/(1 + x²).
Divide by (1 + x²) to reach standard form:
dy/dx + [2x/(1 + x²)] y = 1/(1 + x²)²
Here P = 2x/(1 + x²). Notice that the numerator is the derivative of the denominator, so ∫P dx = log(1 + x²).
IF = elog(1 + x²) = 1 + x².
Apply the formula: y(1 + x²) = ∫ [1/(1 + x²)²] × (1 + x²) dx = ∫ dx/(1 + x²) = tan−1x + C.
y(1 + x²) = tan−1x + C
A pleasant shortcut worth noticing: the original left-hand side (1 + x²)dy/dx + 2xy was already the derivative of y(1 + x²) by the product rule. When you spot that, you can skip the integrating factor entirely.
Linear Equations of the Type dx/dy + Px = Q
Here is a situation that catches people out. You look at an equation, y appears squared or inside a logarithm, and you conclude it cannot possibly be linear. But sometimes if you flip your point of view — treat x as the dependent variable and y as the independent one — the very same equation becomes beautifully linear.
This is explicitly part of the CBSE 2026-27 syllabus, so it is worth being comfortable with rather than hoping it does not appear.
IF = e∫P dy and x × IF = ∫(Q × IF) dy + C.
The signal to look for: the equation is awkward in y but tame in x. If x and dx/dy would appear only to the first power, you are in business.
Solve y dx − (x + 2y²) dy = 0.
Try it as dy/dx first and you get dy/dx = y/(x + 2y²), which is neither separable nor linear in y. So flip it.
Rearranged for dx/dy: y dx/dy = x + 2y², so dx/dy − x/y = 2y.
That is dx/dy + Px = Q with P = −1/y and Q = 2y — both functions of y alone. Linear.
IF = e∫(−1/y) dy = e−log y = 1/y.
Apply the formula: x × (1/y) = ∫ 2y × (1/y) dy = ∫ 2 dy = 2y + C.
Multiply by y: x = 2y² + Cy.
Solve (1 + y²) dx = (tan−1y − x) dy.
Write it as dx/dy: (1 + y²) dx/dy = tan−1y − x. Divide by (1 + y²) and bring x across:
dx/dy + x/(1 + y²) = tan−1y / (1 + y²)
So P = 1/(1 + y²), and ∫P dy = tan−1y. Therefore IF = etan−1y.
x × etan−1y = ∫ [tan−1y / (1 + y²)] etan−1y dy
For that integral, substitute t = tan−1y, so dt = dy/(1 + y²). The whole thing becomes ∫ t et dt, which by parts is (t − 1)et + C.
Putting t back: x etan−1y = (tan−1y − 1) etan−1y + C, so
x = tan−1y − 1 + C e−tan−1y
Pinning Down the Constant: Particular Solutions
You have now met all three methods. This short section is about the final step that appears in almost every long-answer question: turning a general solution into a particular one. The method never changes, but doing it under exam pressure is where the arithmetic slips happen, so let us make the routine automatic.
Solve dy/dx = y tan x given that y = 1 when x = 0.
Separate: dy/y = tan x dx.
Integrate: log|y| = log|sec x| + log|A|, using log|A| for the constant so the logs combine.
So y = A sec x.
Apply the condition x = 0, y = 1: 1 = A sec 0 = A × 1, so A = 1.
The particular solution is y = sec x.
Check: differentiating sec x gives sec x tan x, and y tan x = sec x tan x. They agree, and y(0) = sec 0 = 1. Both boxes ticked.
Solve dy/dx + y cot x = 2x + x² cot x, given that y = 0 when x = π/2.
This is linear with P = cot x, so IF = e∫cot x dx = elog sin x = sin x.
Apply the formula: y sin x = ∫ (2x + x² cot x) sin x dx = ∫ (2x sin x + x² cos x) dx.
Look carefully at that integrand — it is exactly the product-rule derivative of x² sin x. So the integral is x² sin x + C, with no integration by parts needed at all.
y sin x = x² sin x + C
Apply x = π/2, y = 0: 0 × 1 = (π/2)² × 1 + C, so C = − π²/4.
y sin x = x² sin x − π²/4, that is y = x² − π²/(4 sin x).
Solve (x² + xy) dy = (x² + y²) dx given that y = 0 when x = 1.
Every term is degree 2, so it is homogeneous. Put y = vx, dy/dx = v + x dv/dx:
v + x dv/dx = (x² + v²x²)/(x² + vx²) = (1 + v²)/(1 + v)
x dv/dx = (1 + v² − v − v²)/(1 + v) = (1 − v)/(1 + v)
Separate: (1 + v)/(1 − v) dv = dx/x. Write the numerator as 2 − (1 − v) to split the fraction:
∫ [2/(1 − v) − 1] dv = −2 log|1 − v| − v
So −2 log|1 − v| − v = log|x| + C. Putting v = y/x and using 1 − y/x = (x − y)/x, this rearranges to
(x − y)² = A x e−y/x , for a constant A.
Apply x = 1, y = 0: (1 − 0)² = A × 1 × e0, so A = 1.
(x − y)² = x e−y/x
Good to Know: Forming a Differential Equation From a Family of Curves
With that caveat clearly stated, here is the idea in brief, because it genuinely deepens your understanding of everything above.
Solving a differential equation takes you from an equation about slopes to a family of curves. Forming one runs the film backwards: you start with a family of curves containing arbitrary constants, and you produce the differential equation that every member of the family satisfies.
The procedure is short. Count the arbitrary constants — say there are n. Differentiate the given equation n times. You now have n + 1 equations in total, and you use them to eliminate all n constants. Whatever equation survives is the answer, and it will have order n.
For a quick feel: take the family y = Ce−x, the one drawn in the figure earlier on this page. There is one constant, so differentiate once: dy/dx = −Ce−x. But Ce−x is just y, so dy/dx = −y, that is dy/dx + y = 0. Order 1, as expected, and the constant has vanished. That is why every curve in that purple family had the same slope rule.
Notice how this closes the loop. In the general-solutions section we started with dy/dx + y = 0 and produced y = Ce−x. Here we started with y = Ce−x and recovered dy/dx + y = 0. Solving and forming are the same doorway walked through in opposite directions — and understanding that is useful whether or not it is on your paper.
Common Traps and How to Dodge Them
Let us finish the notes with the errors that cost the most marks. None of these are conceptual gaps — they are all slips of habit, which means they are all fixable in an afternoon.
If a section on this page still feels shaky, go back to it now rather than pushing on. Differential Equations is unusually cumulative — the linear method assumes you are fluent with separation, and every method assumes you can integrate calmly. There is genuinely no rush.
Practice Worksheet
Ten questions, arranged roughly from gentlest to most demanding. Work each one on paper in full before you open the answer — reading a solution feels like learning, but writing one is learning. If a question defeats you, go back to the relevant section above rather than straight to the answer.
Question 1.
Find the order and the degree of (d³y/dx³)⁴ + sin x × d²y/dx² = 0.
Show Answer
The highest-order derivative present is d³y/dx³, so the order is 3.
The equation is already a polynomial in the derivatives — note that sin x contains the independent variable, not a derivative, so it is only a coefficient and causes no trouble.
The highest-order derivative d³y/dx³ is raised to the power 4, so the degree is 4. (The first power on d²y/dx² is irrelevant — it is not the highest-order derivative.)
Question 2.
Find the order and the degree of d²y/dx² + cos(dy/dx) = 0.
Show Answer
The highest-order derivative is d²y/dx², so the order is 2.
However, dy/dx is trapped inside a cosine. The equation can therefore never be written as a polynomial in its derivatives, so the degree is not defined.
Be precise in your wording here: “not defined” earns the mark; “zero” or “one” does not.
Question 3.
Verify that y = e2x(A + Bx) is a solution of d²y/dx² − 4dy/dx + 4y = 0, where A and B are arbitrary constants.
Show Answer
Differentiate using the product rule:
dy/dx = 2e2x(A + Bx) + e2x(B) = e2x(2A + 2Bx + B)
Differentiate again:
d²y/dx² = 2e2x(2A + 2Bx + B) + e2x(2B) = e2x(4A + 4Bx + 4B)
Substitute into the left-hand side and take out the common factor e2x:
LHS = e2x[(4A + 4Bx + 4B) − 4(2A + 2Bx + B) + 4(A + Bx)]
Collect the A terms: 4A − 8A + 4A = 0. The Bx terms: 4Bx − 8Bx + 4Bx = 0. The B terms: 4B − 4B = 0.
So LHS = 0 = RHS, and y = e2x(A + Bx) is a solution. Since the equation has order 2 and the answer carries two arbitrary constants, it is in fact the general solution.
Question 4.
Solve dy/dx = (1 + x)(1 + y²).
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The right-hand side is already a product of a function of x and a function of y, so the equation is separable.
Separate: dy/(1 + y²) = (1 + x) dx.
Integrate both sides:
tan−1y = x + x²/2 + C
Check by differentiating: the left gives (1/(1 + y²))dy/dx, the right gives 1 + x. Rearranging returns the original equation.
Question 5.
Solve x√(1 + y²) dx + y√(1 + x²) dy = 0.
Show Answer
Divide throughout by √(1 + x²) √(1 + y²) to separate the variables:
[x / √(1 + x²)] dx + [y / √(1 + y²)] dy = 0
Each integral is of the form ∫ f′(u)/(2√f(u)) du. Taking the first, substitute t = 1 + x², so dt = 2x dx:
∫ x dx/√(1 + x²) = √(1 + x²)
and identically for the y integral. So the general solution is
√(1 + x²) + √(1 + y²) = C
Question 6.
Solve dy/dx = (x − y)/(x + y).
Show Answer
Top and bottom are both degree 1, so the equation is homogeneous. Put y = vx, giving dy/dx = v + x dv/dx:
v + x dv/dx = (x − vx)/(x + vx) = (1 − v)/(1 + v)
x dv/dx = (1 − v)/(1 + v) − v = (1 − v − v − v²)/(1 + v) = (1 − 2v − v²)/(1 + v)
Separate: (1 + v) dv/(1 − 2v − v²) = dx/x.
The derivative of the denominator 1 − 2v − v² is −2 − 2v = −2(1 + v), which is −2 times the numerator. So the left integral is −½ log|1 − 2v − v²|:
−½ log|1 − 2v − v²| = log|x| − ½ log|C|
Multiply by −2 and combine: log|1 − 2v − v²| + log|x²| = log|C|, so x²(1 − 2v − v²) = C.
Put v = y/x: x²(1 − 2y/x − y²/x²) = C, that is
x² − 2xy − y² = C
Question 7.
Solve (x² + y²) dx = 2xy dy given that y = 0 when x = 1.
Show Answer
This is the homogeneous equation from the study notes, so the general solution is x² − y² = Cx. (If you need the working: put y = vx, reduce to 2v dv/(1 − v²) = dx/x, integrate to −log|1 − v²| = log|x| − log|C|, then substitute v = y/x.)
Apply the condition x = 1, y = 0:
1² − 0² = C × 1, so C = 1.
The particular solution is x² − y² = x.
Check: at (1, 0) the left side is 1 and the right side is 1. Correct.
Question 8.
Solve dy/dx + y tan x = sec x given that y = 1 when x = 0.
Show Answer
This is linear with P = tan x and Q = sec x.
∫tan x dx = −log|cos x| = log|sec x|, so IF = elog|sec x| = sec x.
Apply the formula: y sec x = ∫ sec x × sec x dx = ∫ sec²x dx = tan x + C.
Multiply through by cos x: y = sin x + C cos x. That is the general solution.
Apply x = 0, y = 1: 1 = sin 0 + C cos 0 = C, so C = 1.
The particular solution is y = sin x + cos x.
Check: dy/dx = cos x − sin x, and dy/dx + y tan x = cos x − sin x + (sin x + cos x)tan x = cos x + sin²x/cos x = (cos²x + sin²x)/cos x = sec x. Correct.
Question 9.
Solve (x log x) dy/dx + y = 2 log x, for x > 1.
Show Answer
Divide by x log x to reach standard form:
dy/dx + y/(x log x) = 2/x
So P = 1/(x log x). For the integral, substitute u = log x, du = dx/x:
∫ dx/(x log x) = ∫ du/u = log|u| = log|log x|
IF = elog|log x| = log x.
Apply the formula: y log x = ∫ (2/x) log x dx. Substituting u = log x again gives ∫ 2u du = u².
y log x = (log x)² + C
Question 10.
Solve y dx/dy − x = 2y³.
Show Answer
Here y appears to the third power, so this will not behave as a linear equation in y. But x and dx/dy are both first-power and unmultiplied, so treat x as the dependent variable.
Divide by y to reach standard form:
dx/dy − x/y = 2y²
So P = −1/y and Q = 2y², both functions of y alone.
IF = e∫(−1/y) dy = e−log y = 1/y.
Apply the formula: x × (1/y) = ∫ 2y² × (1/y) dy = ∫ 2y dy = y² + C.
Multiply through by y: x = y³ + Cy.
Check: dx/dy = 3y² + C, so y(3y² + C) − (y³ + Cy) = 3y³ + Cy − y³ − Cy = 2y³. Correct.
Kaizen for today: do not try to master the whole chapter this evening. Aim for one more correct question than yesterday — just one. Ten days of that quietly turns into a chapter you own, and it is far kinder to your nerves than a single heroic all-nighter.