Q: Construct Green's function for the differential equation xy''+ y'=0, 0<x<l under the conditions that y(0) is bounded and y(l)=0.

The differential equation xy''+ y'=0 is an exact equation, expressible as (xy')'=0. Integrating this twice yields the general solution y(x) = C1 ln(x) + C2. Applying the boundary conditions: y(0) bounded implies we choose y1(x)=1. For y(l)=0, we select y2(x)=ln(x/l), as y2(l)=ln(l/l)=ln(1)=0. The Wronskian for these solutions is W(y1, y2) = y1 y2' - y1' y2 = (1)(1/x) - (0)(ln(x/l)) = 1/x. The operator L = d/dx(x d/dx) implies p(x)=x. Green's function G(x, ξ) is defined as (1/(p(ξ)W(ξ))) times

Q: Show that between every successive pair of zeros of J₀(x) there exists a zero of J₁(x).

This property is demonstrated using Rolle's Theorem and a fundamental recurrence relation for Bessel functions, as typically covered in the MMT-007 course. Let x₁ and x₂ be any two successive zeros of J₀(x). Since J₀(x) is continuous and differentiable on the interval [x₁, x₂], Rolle's Theorem applies. It states that there must exist at least one point 'c' such that x₁ < c < x₂ where J₀'(c) = 0. From the properties of Bessel functions, we know the recurrence relation J₀'(x) = -J₁(x). Substitut

Q: Using the transformation y = x¹/2u, 2x3/2 = 3z find the solution of the equation y''+x y=0 in terms of Bessel's functions.

The given differential equation is y'' + xy = 0. We begin by applying the transformation y = x¹/²u. After calculating y' and y'' and substituting them into the original equation, the equation simplifies to x²u'' + xu' + (x³ - 1/4)u = 0. Next, we introduce the second transformation, 2x³/² = 3z. This implies x = ((3/2)z)²/³ and consequently, x³ = (9/4)z². Through the chain rule, all derivatives with respect to x are converted into derivatives with respect to z. Substituting these transformed ter

Question 1

Check whether f(x, y) = x²y² satisfies Lipschitz condition

Accepted Answer

The Lipschitz condition is crucial in the theory of differential equations, particularly for guaranteeing the existence and uniqueness of solutions to initial value problems. For a function f(x, y) to satisfy a Lipschitz condition with respect to y on a domain D, there must exist a constant L > 0 such that |f(x, y₁) - f(x, y₂)| ≤ L |y₁ - y₂| for all (x, y₁), (x, y₂) in D. This constant L is known as the Lipschitz constant. A common way to check this condition is to verify if the partial derivati

Question 2

Find the series solution about x = 1 of the equation x(1-x) d²y/dx² -(1+3x)dy/dx - y = 0.

Accepted Answer

To find the series solution about x = 1 for the given differential equation, we first perform a substitution. Let t = x - 1, which implies x = t + 1. Then, dy/dx = dy/dt and d²y/dx² = d²y/dt². Substituting these into the equation x(1-x) d²y/dx² -(1+3x)dy/dx - y = 0, we get the transformed equation: -t(t+1) d²y/dt² -(3t+4)dy/dt - y = 0, or t(t+1) d²y/dt² +(3t+4)dy/dt + y = 0.

We then analyze the nature of the point t=0. Dividing by t(t+1) to obtain the standard form, P(t) = (3t+4)/(t(t+1)) and Q

Question 3

For the following differential equation locate and classify its singular points on the x-axis

Accepted Answer

Answer available — download the full PDF for complete details.

Question 4

Show that ∫ from -1 to 1 x²Pn-1(x) Pn+1(x) dx = (2n(n+1)) / ((2n-1) (2n+1) (2n+3))

Accepted Answer

The proof employs the Legendre Polynomials' three-term recurrence relation, a fundamental property from MMT-007. This relation states: (2k+1)xPk(x) = (k+1)Pk+1(x) + kPk-1(x).

Applying this recurrence for k=n-1 and k=n+1, we derive two crucial expressions:
1. (2n-1)xPn-1(x) = nPn(x) + (n-1)Pn-2(x)
2. (2n+3)xPn+1(x) = (n+2)Pn+2(x) + (n+1)Pn(x)

Multiplying these two derived expressions together yields: (2n-1)(2n+3)x²Pn-1(x)Pn+1(x) = [nPn(x) + (n-1)Pn-2(x)][(n+2)Pn+2(x) + (n+1)Pn(x)].

Integrating

Question 5

Construct Green's function for the differential equation xy''+ y'=0, 0<x<l under the conditions that y(0) is bounded and y(l)=0.

Accepted Answer

The differential equation xy''+ y'=0 is an exact equation, expressible as (xy')'=0. Integrating this twice yields the general solution y(x) = C1 ln(x) + C2.

Applying the boundary conditions: y(0) bounded implies we choose y1(x)=1. For y(l)=0, we select y2(x)=ln(x/l), as y2(l)=ln(l/l)=ln(1)=0.

The Wronskian for these solutions is W(y1, y2) = y1 y2' - y1' y2 = (1)(1/x) - (0)(ln(x/l)) = 1/x. The operator L = d/dx(x d/dx) implies p(x)=x.

Green's function G(x, ξ) is defined as (1/(p(ξ)W(ξ))) times

Question 6

Show that between every successive pair of zeros of J₀(x) there exists a zero of J₁(x).

Accepted Answer

This property is demonstrated using Rolle's Theorem and a fundamental recurrence relation for Bessel functions, as typically covered in the MMT-007 course.

Let x₁ and x₂ be any two successive zeros of J₀(x). Since J₀(x) is continuous and differentiable on the interval [x₁, x₂], Rolle's Theorem applies. It states that there must exist at least one point 'c' such that x₁ < c < x₂ where J₀'(c) = 0.

From the properties of Bessel functions, we know the recurrence relation J₀'(x) = -J₁(x). Substitut

Question 7

Using the transformation y = x¹/2u, 2x3/2 = 3z find the solution of the equation y''+x y=0 in terms of Bessel's functions.

Accepted Answer

The given differential equation is y'' + xy = 0. We begin by applying the transformation y = x¹/²u. After calculating y' and y'' and substituting them into the original equation, the equation simplifies to x²u'' + xu' + (x³ - 1/4)u = 0.

Next, we introduce the second transformation, 2x³/² = 3z. This implies x = ((3/2)z)²/³ and consequently, x³ = (9/4)z². Through the chain rule, all derivatives with respect to x are converted into derivatives with respect to z.

Substituting these transformed ter

Question 8

Show that ∫ from 0 to ∞ e^(-ax) J₀(bx) dx = 1 / sqrt(a² + b²), a>0 b>0.

Accepted Answer

To demonstrate the given integral, we employ the series expansion of the Bessel function of the first kind, J₀(x), and the properties of Laplace transforms. The integral ∫ from 0 to ∞ e^(-ax) J₀(bx) dx can be viewed as the Laplace transform of J₀(bx) with respect to the variable 'a'.

We begin by recalling the power series expansion for J₀(x): J₀(x) = Σ from n=0 to ∞ (-1)ⁿ (x/2)²ⁿ / (n!)². Substituting bx for x, we get J₀(bx) = Σ from n=0 to ∞ (-1)ⁿ (bx/2)²ⁿ / (n!)². Placing this into the integr

MMT-007 Solved Assignment — January 2026 / July 2026

Differential Equations and Numerical Solutions | IGNOU Master of Science (Mathematics with Applications in Computer Science) (MSCMACS)

MMT-007—January 2026 / July 2026

Differential Equations and Numerical Solutions

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MMT-007 Assignment Questions — January 2026

Q1. Check whether f(x, y) = x²y² satisfies Lipschitz condition

Q2. Find the series solution about x = 1 of the equation x(1-x) d²y/dx² -(1+3x)dy/dx - y = 0.

Q3. For the following differential equation locate and classify its singular points on the x-axis

Q4. Show that ∫ from -1 to 1 x²Pn-1(x) Pn+1(x) dx = (2n(n+1)) / ((2n-1) (2n+1) (2n+3))

Q5. Construct Green's function for the differential equation xy''+ y'=0, 0<x<l under the conditions that y(0) is bounded and y(l)=0.

Q6. Show that between every successive pair of zeros of J₀(x) there exists a zero of J₁(x).

Q7. Using the transformation y = x¹/2u, 2x3/2 = 3z find the solution of the equation y''+x y=0 in terms of Bessel's functions.

Q8. Show that ∫ from 0 to ∞ e^(-ax) J₀(bx) dx = 1 / sqrt(a² + b²), a>0 b>0.

Q9. Find the Laplace transform of (sin√t)/√t

Q10. If km and kn are distinct roots of Bessel function Jp(kb) = 0 with p≥0, b>0 then show that ∫ from 0 to b x Jp(kmx) Jp(knx)dx = { 0 if m≠n ; b²/2 [Jp+1(kb)]² if m=n }.