Initial and Boundary Value Problems MCQ Quiz - Objective Question with Answer for Initial and Boundary Value Problems - Download Free PDF

Concept:

Linear Dependence of Solutions: The differential equation and boundary conditions must be such that multiple

solutions satisfy both the equation and the boundary conditions. This can happen when there is a free parameter in

the solution, leading to multiple possible values for the constants.

Parameter Tuning: A specific parameter (often represented as k k" id="MathJax-Element-41-Frame" role="presentation" style="position: relative;" tabindex="0">kk k" id="MathJax-Element-46-Frame" role="presentation" style="position: relative;" tabindex="0">k$k$ $k$ $k$ in boundary conditions) may allow or disallow

multiple solutions. For a certain value of this parameter, the boundary conditions may no longer restrict the solution to a single

unique answer, allowing infinitely many distinct solutions.

Explanation:

The differential equation:

$x^2{y}^{″}-2x{y}^{\prime }+2y=0,1\le x\le 2$

 Boundary conditions:

y(1) - y'(1) = 1, y(2) - k y'(2) = 4

 The question asks for the value of k that allows this problem to have infinitely many distinct solutions.

The given differential equation is $x^2{y}^{″}-2x{y}^{\prime }+2y=0$

This is a Cauchy-Euler differential equation, which has the form:

$x^2{y}^{″}+ax{y}^{\prime }+by=0$

where a = -2 and b = 2 .

To solve Cauchy-Euler equations, we assume a solution of the form:

$y(x)=x^r$

Substituting $y(x)=x^r$ into the differential equation, we get:

$y^{\prime }=r{x}^{r-1}$

$y^{″}=r(r-1)x^{r-2}$

Substitute these into the original equation: 

$x^2\cdot r(r-1)x^{r-2}-2x\cdot r{x}^{r-1}+2x^r=0$

Simplify each term, $r(r-1)x^r-2r{x}^r+2x^r=0$ 

⇒ r(r-1) - 2r + 2 = 0

⇒ (r - 1)(r - 2) = 0

Thus, r = 1 and r = 2 are the roots.

Since we have two distinct roots, the general solution to the differential equation is:

$y(x)=C_{1}x+C_2{x}^2$

where $C_1$ and $C_2$ are constants.

First boundary condition: y(1) - y'(1) = 1

$y(1)=C_1\cdot 1+C_2\cdot 1^2=C_1+C_2-y^{\prime }(x)=C_1+2C_{2}x-y^{\prime }(1)=C_1+2C_2$

⇒ ($C_1+C_2)-(C_1+2C_2)=1$
   
⇒ $C_1+C_2-C_1-2C_2=1\Rightarrow -C_2=1\Rightarrow C_2=-1$
 

 y(2) - k y'(2) = 4:

$y(x)=C_{1}x-x^2$

Now, calculate y(2) and y'(2) :

$y(2)=C_1\cdot 2-(2{)}^2=2C_1-4-y^{\prime }(x)=C_1+2(-1)\cdot x=C_1-2x-y^{\prime }(2)=C_1-2\cdot 2=C_1-4$

Substitute these into the boundary condition:
   
($2C_1-4)-k(C_1-4)=4$
  
$2C_1-4-k{C}_1+4k=4$
   
$(2-k)C_1+4k-4=4$

⇒ ($2-k)C_1=8-4k$

For infinitely many solutions to exist, the coefficient of $C_1$ must be zero (as $C_1$ can take any value). Thus, we set 2 - k = 0 :

 k = 2
   

Hence 2 is the correct answer.

Concept:

Suppose we have the following Sturm-Liouville problem: 

$\frac{d^{2}y}{d{x}^2}+\lambda y=0$ with boundary conditions y(0) = 0 and y(L) = 0 where λ is the eigen value and L is the length of interval.

Here, the value of λ  for which non-trivial solutions exist are called eigenvalues. 

Explanation:

$\frac{d^{2}y}{d{x}^2}+\lambda y=0$ , x ∈(0, π), λ > 0

Let λ = k² 

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

Auxiliary equation may be written as m² + k² = 0 ⇒ m = ± k

Solution of equation is y(x) = A coskx + B sinkx

given that y(0) = 0 ⇒ A = 0 ⇒ y(x) = B sinkx 

Now, $y(\pi )-\frac{dy}{dx}(\pi )=0$ ⇒ y(π) - y'(π) = 0 

⇒ B sinkπ - B k coskπ = 0 ⇒ sinkπ = k coskπ 

⇒ k-tankπ = 0 and λ = $k_n^2$, where kn, n = 1, 2, 3, ...

Hence, Option (3) is true

Explanation -

The given boundary value problem is a specific case of the Sturm-Liouville problem. The given differential equation is y'' + λy = 0,

where y'' denotes the second derivative of y with respect to x, and λ is a parameter. The boundary conditions are y(0) = 0 and y(π) = 0.

This is a homogeneous second order differential equation. The general solution of this equation depends on the value of λ:

If λ > 0, then the general solution is

$y(x)=Acos(\sqrt{(}\lambda )x)+Bsin(\sqrt{(}\lambda )x),$ where A and B are constants determined by the initial conditions.

If λ = 0, then the general solution is y(x) = Ax + B.

If λ < 0, then the general solution is$y(x)=Acosh(\sqrt{(-\lambda )}x)+Bsinh(\sqrt{(-\lambda )}x),$

where cosh is the hyperbolic cosine and sinh is the hyperbolic sine.

Since y(0) = 0, we have A = 0 for all three cases. Now, using the second boundary condition y(π) = 0 gives different constraints for the parameter λ. We will solve for the case λ > 0, as that is the most commonly considered scenario.

Setting y(π) = 0, we obtain

$Bsin(\sqrt{(}\lambda )\pi )=0.$

This equation is satisfied for all integer values of sqrt(λ)π, i.e., for λ = (n)², where n is an integer. These are the eigenvalues that satisfy the given boundary conditions, and the corresponding eigenfunctions are y(x) = sin(n x).

So the solutions of the Sturm-Liouville problem are the series of functions y_n(x) = sin(nx) with eigenvalues λ_n = (n)² for n ∈ Z. As per the boundary conditions given, we only consider n ≥ 1; therefore, the system exhibits an infinite number of eigenvalues and associated eigenfunctions, which form a set of orthogonal functions in the space of square integrable functions over the interval (0, π).

Hence the option (i) is correct.

Concept:

Differential equation: It is an equation that involves differential coefficients or differentials.

The solution of differential equations consists of two parts Complementary function (C.F) and Particular Integral (P.I)

To solve the linear second-order differential equation which is in a general form as shown:

  $\frac{d^{n}y}{d{x}^n}+k_1\frac{d^{n-1}y}{d{x}^{n-1}}+k_2\frac{d^{n-2}y}{d{x}^{n-2}}+\cdots +k_{n}y=0$

 this equation in symbolic form is

$\left(D^n+k_1{D}^{n-1}+\cdots +k_n\right)y=0$

We assume operator ‘D’ as d/dx and form an auxiliary Equation.

To find the roots of the differential equation, make the symbolic coefficient to zero.

  $D^n+k_1{D}^{n-1}+\cdots +k_n=0$ is called auxiliary equation and (A.E.)

Let m1, m2, ⋯ , mn be its roots.

There are four different types of roots.

1. If roots are different m1 ≠ m2 then the solution is of the form

$c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$

2. If the roots are equal, m1 = m2 = m

$\left(c_1+c_2\right)e^{mx}$

3. If the roots are complex and not equal m1 = α + ι β and m2 = α - ι β

$e^{\alpha x}\left(C_{1}cos\beta x+C_{2}sin\beta x\right)$

4. If roots are complex and equal, i.e, m1 = m2 = α + ι β and m3 = m4 = α - ι β

$e^{\alpha x}\left(\left(c_{1}x+c_2\right)cos\beta x+\left(c_{3}x+c_4\right)\mathbf{c}\mathbf{o}\mathbf{s}\beta \mathbf{x}\right)$

P.I part is also found in different methods based on the X function which is R.H.S of differential function.

$e^{\alpha x}\left(\left(c_{1}x+c_2\right)cos\beta x+\left(c_{3}x+c_4\right)\mathrm{c}\mathrm{o}\mathrm{s}\beta \mathrm{x}\right)$

Calculation:

$\frac{d^{2}y}{d{x}^2}+12\frac{dy}{dx}+36y=0$

Let $\frac{d}{dx}=D$

The auxilliary equation can be written as:

D² + 12D + 36 = 0

(D + 6)² = 0

D = -6 and -6

Complementary solution y = (C₁ + C₂x) e^-6x

Given y(0) = 3 = (C₁ + 0) e^-0

c₁ = 3

$\frac{dy}{dx}=(C_1+C_{2}x)(-6e^{-6x})+e^{-6x}{C}_2$

$\frac{dy}{dx}{|}_{x=0}=-36=C_1(-6)+C_2$

C₂ = -36 + 18 = -18

∴ y = (C₁ + C₂x) e^-6x

y = (3 – 18x) e−6x

Given: 

$\begin{array}{l}\frac{dy}{dx}=f(y),x\in \mathbb{R}\\ y(0)=y(1)=0\end{array}\}$ where f satisfies Lipschitz condition.

Concept used: Integrating the function and putting both initial values yield our solution.

Solution: 

Integrating given D.E. with respect to x from 0 to 1

$y(1)-y(0)={\int }_0^{1}f(y)dx$.; now using the condition $y(0)=y(1)=0$ we get

${\int }_0^{1}f(y)dx=0$ which is only true when $f(y)=0$.

Hence $dy/dx=0$, which shows $y(x)$ is a constant function.

Using initial condition,

$y(x)=0$ for all x ∈ $\mathbb{R}$

(2) is correct 

Concept:

Laplacian function:

div(gradf) = ∇2f = $\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}$

Calculation:

Given:

∇2T = 0 

Square domain, boundary condition:

T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y

if we will solve it by forming equations it will take a lot of time, hence we will solve it by satisfying boundary conditions to the options.

substituing T(x, 0)= x in all options:

1. T(x, y) = x - xy + y = x - x.0 + 0 = x
2. T(x, y) = x + y = x + 0 = x
3. T(x, y) = -x + y = -x + 0 = -x ⇒ does not satisfy hence eliminated
4. T(x, y) = x + xy + y = x + x.0 + 0 = x

​substituting T(x, 1) = 1 + x in options

1. T(x, y) = x - xy + y = x - x.1 + 1 = 1 ⇒ does not satisfy , thus eliminated
2. T(x, y) = x + y = x + 1 ⇒ It satisfy hence correct option
3. T(x, y) = x + xy + y  = x + x.1 + 1 = 2x + 1 ⇒ does not satisfy , thus eliminated

Hence, option 2 satisfy all the boundary condition.

Important Points Laplace equation in a square domain (0 < x < 1 and 0 < y < 1) means:

∇2T = $\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}$ = 0,  0 < x < 1 and 0 < y < 1

Boundary condition:

f₁(x) = T(x, 0) = x 

f₂(y) = T(0, y) = y

f₃(x) = T(x, 1) = 1 + x

f₄(y) = T(1, y) = 1 + y

y’’ + 9y = 0

y’’ + 9y = 0

Differential Eq. is of 2^nd order and Homogeneous.

Differential Eq. → D² + 9 = 0

Auxiliary Eq. →

∴ m² + 9 = 0 ⇒ m² = -9 = 9 i²

m = ± 3i

On comparing α ± iβ on root of above eqn.

i.e. Root of above eq. is imaginary α = 0, β = 3

So, Solution above eq.

y = C.F = e^αx [C₁ cos β_x + C₂ sin β_x]

∴ y = C₁ cos 3x + C₂ sin 3x

From given boundary condition: y (0) = 0

At x = 0, y = 0

⇒ C₁ = 0

y(π/2) = √2

$\sqrt{2}=C_1\cos 3\left(\frac{\pi }{2}\right)+C_2\sin \frac{3\pi }{2}=C_2\sin \frac{3\pi }{2}=-C_2$

⇒ C₂ = -√2

∴ y = -√2 sin 3x

y(π/4) = ?

$y=-\sqrt{2}\sin \frac{3\pi }{4}=-\sqrt{2}\times \frac{1}{\sqrt{2}}=-1$

Concept:

Differential equation: It is an equation which involves differential coefficients or differentials.

The solution of differential equations consists of two parts Complementary function (C.F) and Particular Integral (P.I)

To solve the linear second-order differential equation which is in a general form as shown:

  $\frac{d^{n}y}{d{x}^n}+k_1\frac{d^{n-1}y}{d{x}^{n-1}}+k_2\frac{d^{n-2}y}{d{x}^{n-2}}+\cdots +k_{n}y=0$

 this equation in symbolic form is

$\left(D^n+k_1{D}^{n-1}+\cdots +k_n\right)y=0$

We assume operator ‘D’ as d/dx and form auxiliary Equation.

To find the roots of the differential equation, make symbolic coefficient to zero.

  $D^n+k_1{D}^{n-1}+\cdots +k_n=0$ is called auxiliary equation and (A.E.)

Let m1, m2, ⋯ , mn be its roots.

There are four different types of roots.

1. If roots are different m1 ≠ m2 then the solution is of the form

$c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$

2. If the roots are equal, m1 = m2 = m

$\left(c_1+c_2\right)e^{mx}$

3. If the roots are complex and not equal m1 = α + ι β and m2 = α - ι β

$e^{\alpha x}\left(C_{1}cos\beta x+C_{2}sin\beta x\right)$

4. If roots are complex and equal, i.e, m1 = m2 = α + ι β and m3 = m4 = α - ι β

$e^{\alpha x}\left(\left(c_{1}x+c_2\right)cos\beta x+\left(c_{3}x+c_4\right)\mathbf{c}\mathbf{o}\mathbf{s}\beta \mathbf{x}\right)$

P.I part is also found in different methods based on the X function which is R.H.S of differential function.

$e^{\alpha x}\left(\left(c_{1}x+c_2\right)cos\beta x+\left(c_{3}x+c_4\right)\mathrm{c}\mathrm{o}\mathrm{s}\beta \mathrm{x}\right)$

Calculation:

Given DE is  $\frac{d^{2}y}{d{x}^2}+4\frac{dy}{dx}+4y=0$ and

Initial conditions are y(0) = 𝑦′(0) = 1,

A.E is this is (D+2)² = 0. And roots are again equal.

Solution for the above equation is:

$y=\left(A+Bx\right)e^{-2x}$  ⋯ (i)

from the initial condition substituting x = 0

$\left(A+0\right)e^0=1$

A = 1

Differentiating eq.(i) we get:

$y^{\prime }\left(x\right)=B{e}^{-2x}-2\left(A+Bx\right)e^{-2x}$ ⋯ (ii)

In equation(ii) using initial value we get

1 = B - 2 

B = 3

Substitute A and B values in solution

$y\left(x\right)=\left(1+3x\right)e^{-2x}$

∴ $y\left(1\right)=\left(1+3\right)e^{-2}$

$=\frac{4}{e^2}$

= 0.541

Given:

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

Comparing with

$\frac{dy}{dx}+py=q$

We get,

$p=\frac{1}{x};q=x$

$TF=e^{\int pdx}=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$

Now,

General solution can be written as,

Y (IF) = ∫ q (IF) dxt C

Y (x) = ∫ x (x) dxt C

xy = ∫ x² dx + C

$xy=\frac{x^3}{3}+C$

At, x = 1 & y  =1

$\left(1\right)\left(1\right)=\frac{\left(1\right)}{3}+C$

$\therefore C=\frac{2}{3}$

$\therefore xy=\frac{x^3}{3}+\frac{2}{3}$

$\therefore y=\frac{x^2}{3}+\frac{2}{3x}$

3y”(x) + 27 y(x) = 0

y”(x) + 9y(x) = 0

(D² + 9) y = 0

So characteristic equation is given by:

m² + 9 = 0

m = ± 3i = 0 ± 3i

y = (C₁ cos3x + C₂ sinx)e^0x

y = C₁ cos 3x + C₂ sin 3x

y’ = -3C₁ sin 3x + 3C₂ cos 3x

y’(0) = 2000

$\Rightarrow 3C_2=2000\Rightarrow C_2=\frac{2000}{3}$

y(0) = 0 ⇒ C₁(1) + C₂(0) = 0

⇒ C₁ = 0

$y=\frac{2000}{3}\sin 3x$

$y\left(1\right)=\frac{2000}{3}\sin (3\times 1\times \frac{180}{\pi })=94.08$

Concept:

Differential equation: It is an equation that involves differential coefficients or differentials.

The solution of differential equations consists of two parts Complementary function (C.F) and Particular Integral (P.I)

To solve the linear second-order differential equation which is in a general form as shown:

  $\frac{d^{n}y}{d{x}^n}+k_1\frac{d^{n-1}y}{d{x}^{n-1}}+k_2\frac{d^{n-2}y}{d{x}^{n-2}}+\cdots +k_{n}y=0$

 this equation in symbolic form is

$\left(D^n+k_1{D}^{n-1}+\cdots +k_n\right)y=0$

We assume operator ‘D’ as d/dx and form an auxiliary Equation.

To find the roots of the differential equation, make the symbolic coefficient to zero.

  $D^n+k_1{D}^{n-1}+\cdots +k_n=0$ is called auxiliary equation and (A.E.)

Let m1, m2, ⋯ , mn be its roots.

There are four different types of roots.

1. If roots are different m1 ≠ m2 then the solution is of the form

$c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$

2. If the roots are equal, m1 = m2 = m

$\left(c_1+c_2\right)e^{mx}$

3. If the roots are complex and not equal m1 = α + ι β and m2 = α - ι β

$e^{\alpha x}\left(C_{1}cos\beta x+C_{2}sin\beta x\right)$

4. If roots are complex and equal, i.e, m1 = m2 = α + ι β and m3 = m4 = α - ι β

$e^{\alpha x}\left(\left(c_{1}x+c_2\right)cos\beta x+\left(c_{3}x+c_4\right)\mathbf{c}\mathbf{o}\mathbf{s}\beta \mathbf{x}\right)$

P.I part is also found in different methods based on the X function which is R.H.S of differential function.

$e^{\alpha x}\left(\left(c_{1}x+c_2\right)cos\beta x+\left(c_{3}x+c_4\right)\mathrm{c}\mathrm{o}\mathrm{s}\beta \mathrm{x}\right)$

Calculation:

$\frac{d^{2}y}{d{x}^2}+12\frac{dy}{dx}+36y=0$

Let $\frac{d}{dx}=D$

The auxilliary equation can be written as:

D² + 12D + 36 = 0

(D + 6)² = 0

D = -6 and -6

Complementary solution y = (C₁ + C₂x) e^-6x

Given y(0) = 3 = (C₁ + 0) e^-0

c₁ = 3

$\frac{dy}{dx}=(C_1+C_{2}x)(-6e^{-6x})+e^{-6x}{C}_2$

$\frac{dy}{dx}{|}_{x=0}=-36=C_1(-6)+C_2$

C₂ = -36 + 18 = -18

∴ y = (C₁ + C₂x) e^-6x

y = (3 – 18x) e−6x

Concept:

The solution of a linear differential equation of n^th order comprises of two parts. They are:

• Complementary function
• Particular integral

i.e. y = C.F. + P.I.

Calculation:

Given:

$\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=e^x,y=3and\frac{dy}{dx}=3whenx=0$

Complimentary function (CF):

Put R.H.S = 0, and calculate the roots.

$\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0$

For finding the roots put d/dx = m.

∴ m² - 3m + 2 = 0

∴ m = 1 and 2.

∴ y = C₁e^x + C₂e^2x

Particular Integral (PI):

$\frac{1}{D^2-3D+2}{e}^x$

$\frac{1}{2D-3}x{e}^x$    [∵ f(D) = 0 when D = 1]

∴ P.I = -xe^x

General solution is y = C1ex + C2e2x - xe^x

When x = 0, y = 3.

∴ C₁ + C₂ = 3 ...... (1)

When x = 0, dy/dx = 3.

$\frac{dy}{dx}=C_1{e}^x+2C_2{e}^{2x}-x{e}^x-e^x$

∴ 3 = C₁ + 2C₂ - 1

∴ C₁ + 2C₂ = 4  ........ (2)

Solving (1) and (2) gives C₁ = 2 and C₂ = 1.

∴ y = 2ex + e2x - xex

Concept:

Replace $\frac{d^n}{d{z}^n}with{D}^n$ & solving for polynomial in D;

If f(D) = 0 has distinct & real roots, then the solution of the given differential equation will be :

f(x) = c₁ e^+m1x + c₂ e^+m2x + c₃e^+m3x + …. + c_ke^+kx [with k distinct & real roots]

Calculation:

Given Differential Equatin $\frac{d^{2}x\left(t\right)}{d{t}^2}+3\frac{dx\left(t\right)}{dt}+2x\left(t\right)=0withx\left(0\right)=20\mathrm{\&}x\left(1\right)=\frac{10}{e}$

⇒ f(D) = (D²+3D+2)

⇒ D = -2 & -1

So Its Solution will be f(x) = c₁e^-x + c₂e^-2x

Applying the Boundary Conditions:

f(0) = c₁ + c₂ = 20 and f(1) = c₁e^-2 + c₂e^-1.

Solving the above equations for c₁ & c₂, we get;

$c_1=\frac{10e-20}{e-1}and{c}_2={\frac{10e}{e-1}}^{}$ 

$f\left(x\right)=\frac{{(10e-20)}^{e^{-x}}}{(e-1)}+\left(\frac{10e}{e-1}\right).e^{-2x}{}^{}$ 

(D² – D – 2) y = 3 e^2x

A.E. is. D² – D – 2 = 0

⇒ (D – 2) (D + 1) = 0

⇒ D = -1, 2

C.F. = C₁ e^-x + C₂ e^2x

$P.I.=\frac{1}{\left(D-2\right)\left(D+1\right)}.3e^{2x}$

$=\frac{1}{D^2-D-2}3e^{2x}$

$f\left(D\right)=D^2-D-2$

$f\left(2\right)=0$

$f^{\prime }\left(D\right)=2D-1$

$f^{\prime }\left(2\right)=3$

$PI=\frac{1}{D^2-D-2}3e^{2x}=x\frac{1}{2D-1}3e^{2x}$

$P.I.=x.\frac{1}{3}.3e^{2x}=x{e}^{2x}$

Complete solution is

$y=C_1{e}^{-x}+C_2{e}^{2x}+x{e}^{2x}$

Given that y(0) = 0

⇒ 0 = C₁ + C₂ ⇒ C₁ = -C₂

y’(0) = -2

$y^{\prime }=-C_1{e}^{-x}+2C_2{e}^{2x}+e^{2x}+2x{e}^{2x}$

⇒ -2 = -C₁ + 2C₂ + 1

→ -C₁ + 2C₂ = -3

⇒ -C₁ -2C₁ = - 3 ⇒ C₁ = 1, C₂ = -1

Concept:

Initial value problem: A differential equation together with an initial condition is called an initial value problem (IVP).

Calculation:

Given:

The given ordinary differential equation is $\frac{dy}{dx}=-2xy$

by variable separation method,

$\frac{dy}{y}=-2xdx$

integrating both sides,

$\int \frac{dy}{y}=\int -2xdx$

∴ $\ln y=\frac{-2x^2}{2}+c$

∴ $\ln y=-x^2+c$

At, x = 0, y = 2

hence put these values in above equation to find value of constant c,

∴ ln 2 = c + 0

⇒ c = ln 2

the equation becomes

→ ln y = -x² + ln2

$y=e^{-x^2+ln2}$

$y=e^{-x^2}\times e^{ln2}$

$y=e^{-x^2}\times 2$

$\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{-}\mathbf{2}}}$