Which statements about initial value problems y′ = f(x, y), y(x₀) = y₀ are true?

y′ = y^{2/3}, y(0) = 0 has more than one solution

if f and ∂f/∂y are continuous on a rectangle about (x₀, y₀), the problem has a unique solution on some interval about x₀

continuity of f near (x₀, y₀) alone guarantees that a solution exists near x₀

y′ = y^{2/3}, y(0) = 0 has more than one solution

y′ = y², y(0) = 1 has a solution defined on all of ℝ

if f and ∂f/∂y are continuous on a rectangle about (x₀, y₀), the problem has a unique solution on some interval about x₀

continuity of f near (x₀, y₀) alone guarantees that a solution exists near x₀

Which properties of the Laplace transform L (with L{f} = F, L{g} = G) are correct?

L{f ∗ g} = F·G, where f ∗ g is the convolution ∫₀ᵗ f(τ)g(t − τ) dτ

First-Order Theory, Linear Equations and the Laplace Transform

Ordinary Differential Equations is Section 5 of the GATE Mathematics (MA) paper. It asks the theory an engineering paper leaves out — when an initial value problem has a solution, when that solution is unique, and how far it extends — alongside the solution methods. This first chapter covers first-order equations with the Peano and Picard–Lindelöf theorems, the standard first-order forms (exact, linear, Bernoulli), linear equations of higher order with constant coefficients, second-order linear equations with variable coefficients through the Wronskian, Abel’s formula, reduction of order and variation of parameters, the Cauchy–Euler equation, and the Laplace transform with its use for initial value problems. The second chapter takes series solutions and the Legendre and Bessel functions, systems, Sturm theory and Sturm–Liouville problems, and stability. The two counterexamples to carry into the paper are y′ = y^2/3, which has two solutions through the origin, and y′ = y², whose solution through (0, 1) exists only on (−∞, 1).

1. Existence and uniqueness for initial value problems

Consider y′ = f(x, y), y(x₀) = y₀ on a rectangle R = {|x − x₀| ≤ a, |y − y₀| ≤ b} with |f| ≤ M on R. Peano: if f is continuous on R, a solution exists on |x − x₀| ≤ h = min(a, b/M). Picard–Lindelöf: if moreover f is Lipschitz in y on R, |f(x, y₁) − f(x, y₂)| ≤ L|y₁ − y₂| (which holds if ∂f/∂y is continuous on R), the solution on that interval is unique, and Picard’s iterates yₙ₊₁(x) = y₀ + ∫_x₀^x f(t, yₙ(t)) dt converge to it uniformly.

What each hypothesis buys, and what happens without it
Equation	Hypothesis status	Outcome
y′ = y^2/3, y(0) = 0	continuous; not Lipschitz at y = 0	y ≡ 0 and y = (x/3)³ are both solutions (and infinitely many others)
y′ = y², y(0) = 1	smooth, locally Lipschitz, not globally	unique solution y = 1/(1 − x), which blows up at x = 1
y′ = p(x)y + q(x), p, q continuous on I	globally Lipschitz on I	unique solution on the whole of I: linear equations never blow up

⚠️ Existence is local

The theorems promise a solution near x₀, not on the whole line. The solution of y′ = y² with y(0) = 1/2 is y = 1/(2 − x), so the maximal interval to the right ends at x = 2; for y(0) = y₀ > 0 it ends at 1/y₀. A solution can be continued as long as it stays in a region where the hypotheses hold, and it fails to continue only by leaving every compact subset.

2. First-order equations that can be solved

Exact: M dx + N dy = 0 with M_y = Nₓ has a potential φ with φₓ = M, φ_y = N, and the solution is φ = c. If (M_y − Nₓ)/N depends on x alone, e^∫(M_{y − Nₓ)/N dx} is an integrating factor; if (Nₓ − M_y)/M depends on y alone, e^{∫(Nₓ − M_{y})/M dy} is one.
Linear: y′ + p(x)y = q(x). Multiply by μ = e^∫p: (μy)′ = μq. For y′ + (2/x)y = x, μ = x², (x²y)′ = x³ and y = x²/4 + C/x²; with y(1) = 1, C = 3/4 and y(2) = 1 + 3/16 = 1.1875.
Bernoulli: y′ + p(x)y = q(x)yⁿ, n ≠ 0, 1: v = y¹⁻ⁿ makes it linear, v′ + (1 − n)pv = (1 − n)q. For y′ + y = y², v = 1/y gives v′ − v = −1, v = 1 + Ceˣ.

3. Linear equations with constant coefficients

For y⁽ⁿ⁾ + a₁y⁽ⁿ⁻¹⁾ + … + aₙy = 0, try y = e^mx: m solves the characteristic polynomial. A real root m of multiplicity k contributes e^mx, xe^mx, …, x^k−1e^mx; a complex pair α ± iβ contributes e^αxcos βx and e^αxsin βx (times powers of x if repeated). The n functions obtained form a basis of the solution space, which has dimension exactly n.

⚠️ Resonance in the method of undetermined coefficients

For y″ + y = sin x the natural guess A cos x + B sin x solves the homogeneous equation and cannot produce sin x. Multiply by x: y_p = x(A cos x + B sin x), which gives −2A sin x + 2B cos x = sin x, so y_p = −(x/2)cos x. In general, if the forcing term’s exponent is a root of multiplicity s of the characteristic polynomial, multiply the guess by xˢ.

4. Second-order equations with variable coefficients; the Cauchy–Euler equation

For y″ + p(x)y′ + q(x)y = 0 with p, q continuous on I, the solutions form a 2-dimensional space. The Wronskian W(y₁, y₂) = y₁y₂′ − y₁′y₂ satisfies Abel’s formula W(x) = W(x₀) exp(−∫_x₀^x p): it is either never zero on I (y₁, y₂ independent) or identically zero (dependent). For n functions W is the determinant of the matrix of derivatives; for e^x, e^2x, e^3x at 0 it is the Vandermonde determinant (2 − 1)(3 − 1)(3 − 2) = 2.

Reduction of order: if y₁ is a solution, y₂ = y₁∫ e^−∫p/y₁² dx is a second, independent one.
Variation of parameters: for y″ + py′ + qy = g, y_p = −y₁∫ y₂g/W dx + y₂∫ y₁g/W dx.
Cauchy–Euler: x²y″ + axy′ + by = 0 on x > 0. Put y = xᵐ: m(m − 1) + am + b = 0. Distinct real roots give x^m₁, x^m₂; a double root m gives xᵐ, xᵐ ln x; roots α ± iβ give x^αcos(β ln x), x^αsin(β ln x). The substitution x = eᵗ turns it into a constant-coefficient equation.

⚠️ The Wronskian test is one-directional for arbitrary functions

W ≢ 0 always implies independence. But W ≡ 0 implies dependence only for solutions of one linear equation with continuous coefficients: x² and x|x| on ℝ have W ≡ 0 and are independent. They cannot both solve such an equation, because at x = 0 both vanish together with their derivatives.

5. The Laplace transform

For f piecewise continuous of exponential order, F(s) = L{f}(s) = ∫₀^∞ e^−stf(t) dt exists for s large. The properties used to solve initial value problems: linearity; L{f′} = sF − f(0) and L{f″} = s²F − sf(0) − f′(0); first shifting L{e^atf} = F(s − a); second shifting L{u(t − a)f(t − a)} = e^−asF(s); L{tf} = −F′(s); and the convolution theorem L{f ∗ g} = F·G with (f ∗ g)(t) = ∫₀ᵗ f(τ)g(t − τ) dτ. The transform of a product fg is not F·G.

Transforms to know
f(t)	F(s)
1, tⁿ	1/s, n!/sⁿ⁺¹
e^at, t e^at	1/(s − a), 1/(s − a)²
sin bt, cos bt	b/(s² + b²), s/(s² + b²)
u(t − a), δ(t − a)	e^−as/s, e^−as

Worked IVP. y″ + 4y = sin t, y(0) = y′(0) = 0. Transforming, (s² + 4)Y = 1/(s² + 1), so Y = 1/((s² + 1)(s² + 4)) = (1/3)[1/(s² + 1) − 1/(s² + 4)] and y = (1/3)(sin t − ½ sin 2t). At t = π/2, y = 1/3.

Key takeaways

Continuity of f gives local existence (Peano); a Lipschitz condition in y gives uniqueness (Picard–Lindelöf); neither gives a global solution.
Exact, linear and Bernoulli equations reduce by a potential, an integrating factor and v = y¹⁻ⁿ respectively.
Constant coefficients: roots of the characteristic polynomial, with x-multiplied terms for repeated roots and for resonant forcing.
Abel: W = W(x₀)e^−∫p, so the Wronskian of two solutions is never zero or always zero; Cauchy–Euler uses y = xᵐ.
Laplace turns an IVP into algebra: L{y′} = sY − y(0), shifting theorems, and convolution for products of transforms.

Practice questions (14)

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.

Which statements about initial value problems y′ = f(x, y), y(x₀) = y₀ are true?
1. y′ = y^2/3, y(0) = 0 has more than one solution
2. y′ = y², y(0) = 1 has a solution defined on all of ℝ
3. if f and ∂f/∂y are continuous on a rectangle about (x₀, y₀), the problem has a unique solution on some interval about x₀
4. continuity of f near (x₀, y₀) alone guarantees that a solution exists near x₀
Show answer
Answer: A — y′ = y^{2/3}, y(0) = 0 has more than one solution; C — if f and ∂f/∂y are continuous on a rectangle about (x₀, y₀), the problem has a unique solution on some interval about x₀; D — continuity of f near (x₀, y₀) alone guarantees that a solution exists near x₀
(1) y ≡ 0 and y = (x/3)³ both work: y′ = x²/9 = ((x/3)³)^2/3. (2) The unique solution is 1/(1 − x), which blows up at x = 1. (3) A continuous ∂f/∂y makes f Lipschitz in y on the rectangle: Picard–Lindelöf. (4) Peano’s theorem.
The solution of y′ = y², y(0) = 1/2 exists on an interval [0, b) with b as large as possible. The value of b is ____.
Numerical answer — type the value.
Show answer
Answer: 2
Separating, −1/y = x + C with C = −2, so y = 1/(2 − x). It is defined and smooth for x < 2 and tends to ∞ as x → 2⁻, so it cannot be continued past b = 2 although f(x, y) = y² is smooth everywhere.
The solution of y′ + (2/x)y = x, y(1) = 1 on x > 0 has y(2), correct to two decimal places, equal to ____.
Numerical answer — type the value.
Show answer
Answer: 1.19
μ = e^∫2/x = x², so (x²y)′ = x³ and x²y = x⁴/4 + C. y(1) = 1 gives C = 3/4, so y = x²/4 + 3/(4x²) and y(2) = 1 + 3/16 = 1.1875, i.e. 1.19.
The solution of the Bernoulli equation y′ + y = y², y(0) = 1/2, has y(ln 3) = ____.
Numerical answer — type the value.
Show answer
Answer: 0.25
With v = 1/y, v′ = −y′/y² = −(y² − y)/y² = −1 + v, so v′ − v = −1 and v = 1 + Ceˣ. v(0) = 2 gives C = 1, so y = 1/(1 + eˣ) and y(ln 3) = 1/(1 + 3) = 0.25.
The general solution of y″ − 4y′ + 4y = 0 is
1. (c₁ + c₂x)e^2x
2. c₁e^2x + c₂e^−2x
3. c₁e^2x
4. c₁cos 2x + c₂sin 2x
Show answer
Answer: A — (c₁ + c₂x)e^{2x}
The characteristic polynomial m² − 4m + 4 = (m − 2)² has the double root 2, which contributes e^2x and xe^2x. A second-order equation needs two independent solutions, so c₁e^2x alone is incomplete.
For y″ + y = sin x, the correct form of a trial particular solution in the method of undetermined coefficients is
1. x(A cos x + B sin x)
2. A cos x + B sin x
3. A x sin x
4. A x² sin x
Show answer
Answer: A — x(A cos x + B sin x)
±i are simple roots of m² + 1, so sin x and cos x solve the homogeneous equation and must be multiplied by x. Keeping both the sine and cosine terms is needed: substitution gives B = 0, A = −1/2, y_p = −(x/2)cos x, which the form A x sin x cannot produce.
y₁ and y₂ are solutions of x²y″ + xy′ + (x² − 1)y = 0 on x > 0 with Wronskian W(y₁, y₂)(1) = 2. The value of W(y₁, y₂)(4) is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
In standard form y″ + (1/x)y′ + (1 − 1/x²)y = 0, so p = 1/x and Abel’s formula gives W(x) = W(1)exp(−∫₁ˣ dt/t) = 2/x. Hence W(4) = 0.5. The coefficient of y plays no part; forgetting to divide by x² would give p = x and a wrong exponential.
The Wronskian of eˣ, e^2x, e^3x at x = 0 is ____.
Numerical answer — type the value.
Show answer
Answer: 2
At x = 0 the rows are the functions, first and second derivatives: (1, 1, 1), (1, 2, 3), (1, 4, 9). This Vandermonde determinant is (2 − 1)(3 − 1)(3 − 2) = 2. In general W(x) = 2e^6x, never zero, so the three functions are independent.
The general solution of the Cauchy–Euler equation x²y″ − 3xy′ + 4y = 0 on x > 0 is
1. x²(c₁ + c₂ ln x)
2. c₁x² + c₂x⁻²
3. c₁x + c₂x⁴
4. (c₁ + c₂x)e^2x
Show answer
Answer: A — x²(c₁ + c₂ ln x)
y = xᵐ gives m(m − 1) − 3m + 4 = m² − 4m + 4 = (m − 2)², a double root. The second solution is x² ln x, just as a double root of a constant-coefficient equation gives xe^mx (put x = eᵗ). Writing (c₁ + c₂x)e^2x confuses the two equations.
The solution of x²y″ − 2xy′ + 2y = 0, y(1) = 1, y′(1) = 3, has y(2) = ____.
Numerical answer — type the value.
Show answer
Answer: 6
m(m − 1) − 2m + 2 = m² − 3m + 2 = (m − 1)(m − 2), so y = c₁x + c₂x². Then c₁ + c₂ = 1 and c₁ + 2c₂ = 3 give c₂ = 2, c₁ = −1, and y(2) = −2 + 8 = 6.
The Laplace transform of t e^2t is
1. 1/(s − 2)²
2. 1/(s − 2)
3. 2/(s − 2)³
4. 1/(s + 2)²
Show answer
Answer: A — 1/(s − 2)²
L{t} = 1/s², and the first shifting theorem replaces s by s − 2: 1/(s − 2)². 2/(s − 2)³ is the transform of t²e^2t, and 1/(s + 2)² shifts the wrong way.
Which properties of the Laplace transform L (with L{f} = F, L{g} = G) are correct?
1. L{f′}(s) = sF(s) − f(0)
2. L{e^atf(t)}(s) = F(s − a)
3. L{f ∗ g} = F·G, where f ∗ g is the convolution ∫₀ᵗ f(τ)g(t − τ) dτ
4. L{f·g} = F·G
Show answer
Answer: A — L{f′}(s) = sF(s) − f(0); B — L{e^{at}f(t)}(s) = F(s − a); C — L{f ∗ g} = F·G, where f ∗ g is the convolution ∫₀ᵗ f(τ)g(t − τ) dτ
(1) follows by integrating by parts. (2) is the first shifting theorem. (3) is the convolution theorem. (4) is false: with f = g = 1, L{1} = 1/s but F·G = 1/s².
The solution of y″ + 4y = sin t, y(0) = 0, y′(0) = 0 has y(π/2), correct to two decimal places, equal to ____.
Numerical answer — type the value.
Show answer
Answer: 0.33
(s² + 4)Y = 1/(s² + 1), so Y = (1/3)[1/(s² + 1) − 1/(s² + 4)] and y = (1/3)sin t − (1/6)sin 2t. At t = π/2: (1/3)(1) − (1/6)(0) = 1/3 = 0.33. Check: y″ + 4y = (1/3)(−sin t + 4 sin t) = sin t.
Let f(t) be the inverse Laplace transform of 1/(s(s² + 1)). The value of f(π) is ____.
Numerical answer — type the value.
Show answer
Answer: 2
1/(s(s² + 1)) = 1/s − s/(s² + 1), so f(t) = 1 − cos t; equivalently f = 1 ∗ sin t = ∫₀ᵗ sin τ dτ. Then f(π) = 1 − (−1) = 2.