The partial differential equation 4u_xx + 4u_xy + u_yy = 0 is:

Not classifiable, because it has no time variable

Differential Equations — Higher Order, Euler–Cauchy, Laplace and PDEs

A linear differential equation with constant coefficients is the friendliest object in Engineering Mathematics, because solving it is not really calculus at all — it is finding the roots of a polynomial. Replace every derivative d/dt by the symbol m, solve the resulting auxiliary equation, and read the solution off the roots: distinct real roots give exponentials, a repeated root multiplies one of them by t, and a complex pair gives e^at(A cos bt + B sin bt). That is the whole method for the homogeneous case, and it is mechanical once you accept that the polynomial is doing the work. Two structural facts save more marks than any technique. First, the general solution of an nth-order equation carries exactly n arbitrary constants — so an answer to a second-order problem with one constant in it is wrong before you check anything else, and this alone eliminates options. Second, an initial-value problem and a boundary-value problem are not the same species: an IVP fixes the function and its derivatives at one point and always has exactly one solution, while a BVP fixes values at two different points and may have one solution, none at all, or infinitely many. Examiners exploit that difference deliberately. Beyond the constant-coefficient case sit two special forms worth recognising on sight: the Euler–Cauchy equation, where each derivative is multiplied by a matching power of x and the substitution y = x^m turns it into an algebraic equation, and the Laplace transform, which converts an initial-value problem into algebra in the variable s, solves it there, and transforms back. The three partial differential equations GATE sets — Laplace, heat and wave — are classified by the discriminant B² − 4AC exactly as conic sections are, and the classification is usually the whole question.

The Auxiliary Equation Does All the Work

Roots of the auxiliary equation → form of the solution
Roots	General solution	Worked example
Distinct real m₁, m₂	c₁e^m₁t + c₂e^m₂t	y″ − 5y′ + 6y = 0 gives m² − 5m + 6 = 0, roots 2 and 3, so y = c₁e^2t + c₂e^3t.
Repeated real m	(c₁ + c₂t)e^mt	y″ − 4y′ + 4y = 0 gives (m − 2)² = 0, so y = (c₁ + c₂t)e^2t. The factor t is what supplies the second constant.
Complex a ± ib	e^at(A cos bt + B sin bt)	y″ + 4y = 0 gives m = ±2i, so y = A cos 2t + B sin 2t — an undamped oscillation, because the real part is zero.

🧠 Count the constants before reading the options

An nth-order linear equation has exactly n arbitrary constants in its general solution, and an option with the wrong number of them is wrong without any further analysis. In the repeated-root case the second solution is t·e^mt rather than a second exponential, precisely because two copies of e^mt would supply only one independent function — the t is there to keep the count honest, not as a piece of notation to memorise.

With a non-zero right-hand side the solution is the homogeneous solution plus one particular integral, and the particular integral is guessed from the shape of the right-hand side: a polynomial for a polynomial, Ae^kt for an exponential, and A cos ωt + B sin ωt for either a sine or a cosine — always both terms, even when only one appears on the right. The one case that needs care is resonance, where the guessed form is already a solution of the homogeneous equation; then multiply the guess by t before substituting, exactly as in the repeated-root case.

Euler–Cauchy, and Why a BVP Is Not an IVP

An Euler–Cauchy equation is recognised by its pattern rather than its content: the coefficient of y″ is x², of y′ is x, of y is a constant — each derivative paired with a power of x equal to its order. Substituting y = x^m turns it into an algebraic equation in m. For x²y″ − 3xy′ + 4y = 0 that is m(m − 1) − 3m + 4 = m² − 4m + 4 = 0, a repeated root m = 2, and the repeated-root rule takes the same form as before with ln x in place of t: y = (c₁ + c₂ ln x)x².

⚠️ A boundary-value problem can have no solution at all

For y″ + y = 0 the general solution is A cos x + B sin x. Impose the initial conditions y(0) = 0, y′(0) = 1 and there is exactly one solution, sin x — an IVP with continuous coefficients always has one and only one. Now impose the boundary conditions y(0) = 0 and y(π) = 1 instead: the first forces A = 0, so y = B sin x, and then y(π) = B sin π = 0 ≠ 1 for every B. There is no solution. Change the second condition to y(π) = 0 and every B works — infinitely many solutions. Same equation, three different outcomes, decided entirely by where the conditions are imposed. If a question mentions two points, do not assume a unique solution exists.

Laplace Transforms — Turning Calculus into Algebra

The transforms worth knowing without a table
f(t)	F(s)	Note
1	1/s	And tⁿ gives n!/sⁿ⁺¹, so t → 1/s².
e^at	1/(s − a)	The shifting rule: multiplying f(t) by e^at replaces s by s − a everywhere. So t·e^−2t has transform 1/(s + 2)².
sin ωt, cos ωt	ω/(s² + ω²), s/(s² + ω²)	Numerator ω for sine, s for cosine — the commonest pair to swap under pressure.
f′(t)	sF(s) − f(0)	This is the whole point: the initial condition enters the algebra automatically instead of being applied at the end.

The method in three steps, on y′ + y = 1 with y(0) = 0. Transform both sides: sY − 0 + Y = 1/s. Solve algebraically: Y = 1/[s(s + 1)]. Invert by partial fractions, 1/s − 1/(s + 1), to get y = 1 − e^−t. No integration was performed anywhere, and the initial condition was used in the first line rather than the last. That is why Laplace is the method of choice for an IVP with a discontinuous or impulsive right-hand side, where the classical route needs a separate solution on each interval.

The Three PDEs, and Classification by Discriminant

For a second-order equation A·u_xx + B·u_xy + C·u_yy + … = 0 the sign of B² − 4AC classifies it, exactly as it classifies a conic. Negative is elliptic, zero is parabolic, positive is hyperbolic. So 4u_xx + 4u_xy + u_yy = 0 has B² − 4AC = 16 − 16 = 0 and is parabolic, and the arithmetic is the entire question.

Laplace, u_xx + u_yy = 0 — elliptic. B² − 4AC = 0 − 4 = −4. It is a steady-state equation: no time appears in it, so it describes an equilibrium, and its solutions are the harmonic functions that also turn up in vector calculus and complex variables.
Heat, u_t = k·u_xx — parabolic. One time derivative against two space derivatives. It smooths its initial data out and cannot be run backwards, which is the mathematical form of irreversibility.
Wave, u_tt = c²·u_xx — hyperbolic. Two time derivatives against two space derivatives, so B² − 4AC > 0. It propagates disturbances at finite speed c and is reversible, which is why it keeps a sharp pulse sharp while the heat equation would blur it.

🎯 Why separation of variables works, and when it does not

Assume u(x, t) = X(x)·T(t), substitute, and divide by XT: every term now involves x alone or t alone. A function of x that equals a function of t for all x and t must be a constant, and that constant is the separation constant. This converts one PDE into two ODEs, and the boundary conditions then quantise the constant — which is where the eigenvalue problems and Fourier series come from. It works because the equation is linear and homogeneous with boundaries that respect the coordinates; a non-homogeneous boundary condition has to be removed by a substitution first, and a nonlinear PDE does not separate at all.

Key takeaways

Replace d/dt by m and solve the auxiliary polynomial. Distinct roots give exponentials, a repeated root brings in a factor t, a complex pair gives e^at(A cos bt + B sin bt).
The general solution of an nth-order equation has exactly n arbitrary constants, so miscounted options can be discarded without solving anything.
An IVP with continuous coefficients always has exactly one solution. A BVP may have one, none or infinitely many — never assume existence when two points are named.
An Euler–Cauchy equation pairs each derivative with a matching power of x. Substituting y = x^m makes it algebraic; a repeated root brings in ln x rather than t.
Laplace transforms feed the initial condition into the algebra through L{f′} = sF(s) − f(0), so an IVP is solved without integrating once.
The shifting rule is the transform table in one line: multiplying by e^at replaces s by s − a, so t·e^−2t transforms to 1/(s + 2)².
Classify a second-order PDE by B² − 4AC: negative elliptic (Laplace), zero parabolic (heat), positive hyperbolic (wave). That arithmetic is often the whole question.

Practice questions (9)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The general solution of y″ − 5y′ + 6y = 0 is:
1. c₁e^2t + c₂e^3t
2. c₁e^−2t + c₂e^−3t
3. (c₁ + c₂t)e^5t
4. c₁ cos 2t + c₂ sin 3t
Show answer
Answer: A — c₁e^{2t} + c₂e^{3t}
The auxiliary equation is m² − 5m + 6 = 0, which factorises as (m − 2)(m − 3) = 0, so the roots are 2 and 3 and the solution is c₁e^2t + c₂e^3t. Option B flips both signs, which is the mistake of reading the roots off the coefficients rather than off the factorisation — the roots of m² − 5m + 6 are positive because both the sum, 5, and the product, 6, are positive. Option C would need a repeated root, and option D a complex pair; both are ruled out the moment the discriminant 25 − 24 = 1 is seen to be positive and non-zero.
The solution of y″ + 4y = 0 with y(0) = 0 and y′(0) = 2 is:
1. sin 2t
2. 2 sin 2t
3. cos 2t
4. 2 sin t
Show answer
Answer: A — sin 2t
The roots are m = ±2i, so y = A cos 2t + B sin 2t. Then y(0) = A = 0, leaving y = B sin 2t, and y′ = 2B cos 2t gives y′(0) = 2B = 2, hence B = 1 and y = sin 2t. Option B is what you get by setting B equal to the given derivative value without dividing by the factor 2 that the chain rule produces — the commonest slip in this family. Option D solves y″ + y = 0 instead: the 4 in the equation becomes the 2 inside the sine, since m² = −4.
The general solution of the Euler–Cauchy equation x²y″ − 3xy′ + 4y = 0 is:
1. c₁x² + c₂x⁴
2. (c₁ + c₂ ln x)x²
3. c₁e^2x + c₂xe^2x
4. (c₁ + c₂x)x²
Show answer
Answer: B — (c₁ + c₂ ln x)x²
Substituting y = x^m gives m(m − 1) − 3m + 4 = m² − 4m + 4 = (m − 2)² = 0, a repeated root m = 2. For an Euler–Cauchy equation the second independent solution multiplies the first by ln x, not by x, so the answer is (c₁ + c₂ ln x)x². Option D is the constant-coefficient repeated-root rule transplanted without the substitution x = e^t that justifies it, and it fails on substitution. Option C is the answer to a completely different equation, y″ − 4y′ + 4y = 0 — recognising which of the two families you are in is the whole skill here, and the tell is that each derivative carries a matching power of x.
The Laplace transform of t·e^−2t is:
1. 1/(s + 2)²
2. 1/(s − 2)²
3. 1/(s² + 2)
4. 2/(s + 2)²
Show answer
Answer: A — 1/(s + 2)²
Start from L{t} = 1/s². The shifting rule says that multiplying by e^at replaces s by s − a, and here a = −2, so s becomes s + 2 and the answer is 1/(s + 2)². Option B applies the shift in the wrong direction, which is the single commonest error with this rule — the sign of the exponent flips as it moves into the denominator, so a decaying exponential gives s plus the decay rate. Option D multiplies by an extra 2, which would be the transform of 2t·e^−2t.
Solving y′ + y = 1 with y(0) = 0 by Laplace transforms gives:
1. y = 1 − e^−t
2. y = e^−t
3. y = 1 + e^−t
4. y = t·e^−t
Show answer
Answer: A — y = 1 − e^{−t}
Transform both sides using L{y′} = sY − y(0) = sY: (s + 1)Y = 1/s, so Y = 1/[s(s + 1)] = 1/s − 1/(s + 1), and inverting term by term gives y = 1 − e^−t. Two checks confirm it and are worth doing in the exam: y(0) = 1 − 1 = 0 satisfies the initial condition, and as t → ∞ the solution tends to 1, which is exactly the steady state you would expect from y′ = 0 in the original equation. Option C satisfies the equation but gives y(0) = 2, and it is the answer produced by getting the sign wrong in the partial fractions.
The partial differential equation 4u_xx + 4u_xy + u_yy = 0 is:
1. Elliptic
2. Parabolic
3. Hyperbolic
4. Not classifiable, because it has no time variable
Show answer
Answer: B — Parabolic
Read off A = 4, B = 4, C = 1 and compute the discriminant: B² − 4AC = 16 − 16 = 0, which is the parabolic case. The arithmetic is the entire question, and the only trap is in B: the coefficient of the mixed derivative is B itself, not half of it, so a candidate who halves it gets 4 − 16 = −12 and answers "elliptic". Option D is a false rule — the classification depends only on the three coefficients and does not care what the variables are called; the heat equation is parabolic whether its second variable is time or not.
For y″ + y = 0 with the boundary conditions y(0) = 0 and y(π) = 1, the number of solutions is:
1. Exactly one
2. None
3. Infinitely many
4. Exactly two
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Answer: B — None
The general solution is A cos x + B sin x. The condition y(0) = 0 forces A = 0, so y = B sin x. But then y(π) = B sin π = 0 for every value of B, and the second condition demands 1. No B works, so there is no solution. This is the difference between a boundary-value problem and an initial-value problem in one example: the same equation with y(0) = 0, y′(0) = 1 has exactly one solution. Change the second boundary condition to y(π) = 0 and every B satisfies it, giving infinitely many — so all three of options A, B and C are reachable from this one equation depending on the conditions, which is why examiners set it.
For y″ + 3y′ + 2y = 0 with y(0) = 1 and y′(0) = 0, the solution is y = c₁e^−t + c₂e^−2t. The value of c₁ is _____.
Numerical answer — type the value.
Show answer
Answer: 2
The auxiliary equation m² + 3m + 2 = 0 has roots −1 and −2, confirming the given form. Now impose the conditions: y(0) = c₁ + c₂ = 1 and y′(0) = −c₁ − 2c₂ = 0. From the second, c₁ = −2c₂; substituting into the first gives −2c₂ + c₂ = 1, so c₂ = −1 and c₁ = 2. The solution is 2e^−t − e^−2t, and both conditions check: y(0) = 2 − 1 = 1 and y′(0) = −2 + 2 = 0. Note the sign pattern worth carrying — with both roots negative the solution decays to zero, and the coefficient of the slower-decaying term is the larger one.
Which of the following are solutions of y″ + 4y = 0? (More than one option may be correct.)
1. e^2x
2. cos 2x
3. sin 2x
4. x·sin 2x
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Answer: B — cos 2x; C — sin 2x
The auxiliary equation m² + 4 = 0 gives m = ±2i, so the general solution is A cos 2x + B sin 2x and both cos 2x and sin 2x are solutions — options B and C. Option A fails on substitution: (e^2x)″ + 4e^2x = 4e^2x + 4e^2x = 8e^2x ≠ 0, and the trap is the shared 2, which comes from m² = −4 and therefore belongs inside a sine or cosine rather than in a real exponent. Option D also fails: (x sin 2x)″ + 4x sin 2x = 4 cos 2x ≠ 0. The factor x appears only when the auxiliary equation has a repeated root, and ±2i is a distinct pair. Substituting each candidate is four lines of work and settles all four judgements, which is the right use of a question that costs nothing to get wrong.