Ordinary Differential Equations MCQ Quiz in मल्याळम - Objective Question with Answer for Ordinary Differential Equations - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Solution - The Green's Function for ODE 

P(x)y" + Q(x)y'+ r(x)y = f(x) is given by 

G(x,t) = $\frac{-1}{P(t)}\frac{y_1(x)y_2(t)-y_1(t)y_2(x)}{y_1(t)y_2^{\prime }(t)-y_2(t)y_1^{\prime }(t)}$ 

y1 and y2 are L.I solution of Homogenous Differential equation 

here the homogenous differential equation is 

y"+5y'+6y=0 

$m^2+5m+6=0;(m+2)(m+3)=0$

$y(x)=c_1{e}^{-2x}+c_2{e}^{-3x}$

Now G(x,t) = $-1\cdot \frac{(e^{-2x}{e}^{-3t}-e^{-2t}{e}^{-3x})}{e^{-2t}(-3e^{-3t})-e^{-3t}(-2e^{-2t})}$

 G(x,t) = e2(t-x) - e3(t-x)

Therefore, Correct Option is Option 2.

Solution:

 Given Ordinary Differential equation is 

$(x-1)y^{″}+(cot\pi x)y^{\prime }+(cose{c}^2\pi x)y=0$

Dividing (x-1) in L.H.S and R.H.S 

$y^{″}+\frac{(cot\pi x)y^{\prime }}{x-1}+\frac{(cose{c}^2\pi x)y}{x-1}$ = 0

Now for x = 0 

$\underset{x\to 0}{lim}\frac{(x-0)(cot\pi x)}{x-1}$ (0/0 form)

Using L' Hospital we get 

$\underset{x\to 0}{lim}\frac{xcos\pi x}{(x-1)(sin\pi x)}$ (0/0 form)

Again using L' Hospital form we get

L= $\frac{-1}{\pi }$ 

now, $li{m}_{x\to 0}\frac{x^{2}cose{c}^2\pi x}{x-1}=\frac{2}{-2{\pi }^2}$

for x=1 

$li{m}_{x\to 1}\frac{x-1cot\pi x}{x-1}=\mathrm{\infty }$

so, 0 is regular point where 1 is irregular 

Therefore Option 1 is correct . 

Explanation:

y(x) = v sec(x) ⇒ y' = v' sec x + v sec x tan x

⇒ y'' = v'' sec x + v' sec x tan x + v' sec x tan x + v sec x tan² x + v sec³ x

Substituting these values in the given differential equation 

y'' - (2tan x)y' + 5y = 0

⇒ v'' sec x + v' sec x tan x + v' sec x tan x + v sec x tan2 x + v sec3 x - 2 v' sec x tan x - 2 v sec x tan² x + 5v sec(x) = 0

⇒ v'' sec x + v sec3 x - v sec x tan2 x + 5v sec(x) = 0  

⇒ v'' sec x + v sec x(sec2 x - tan2 x + 5) = 0

⇒ v'' sec x + 6v sec x = 0 (∵ sec2 x - tan2 x = 1)

⇒ v'' + 6v = 0

⇒ v = c₁ cos(√6 x) + c₂ sin(√6 x) ...(i)

Given y(0) = 0 and y'(0) = √6 ⇒ v(0) = 0 and v'(0) = √6

Substituting initial conditions 

 v(0) = 0  ⇒ c1 = 0

So v = c2 sin(√6 x)

v' = c2 √6 cos(√6 x)

v'(0) = √6 ⇒ c2 √6 = √6 ⇒ c2 = 1

Hence v = sin(√6 x)

∴ v($\frac{\pi }{6\sqrt{6}}$) =  sin(√6 $\frac{\pi }{6\sqrt{6}}$) = sin($\frac{\pi }{6}$) = 0.5

∴ Option (3) is correct

Concept:

Bendixon's Criterion: If f_x and g_y are continuous in a simply connected region ℝ² and fx + gx ≠ 0 then the system of differential equations

$\frac{dx}{dt}$ = f(x,y)

​$\frac{dy}{dt}$ = g(x,y)​

has no closed trajectories inside ℝ

Explanation:

Here f(x,y) = 4x3y2 - x5y4  g(x,y) = x4y5 + 2x2y3

fx = 12x2y2 - 5x4y4, g_y = 5x4y4 + 6x2y2

Both fx and gy are continuous and 

 fx + gx = 12x2y2 - 5x4y4 + 5x4y4 + 6x2y2 = 18x2y2  ≠ 0 in whole ℝ2 as it is zero on (0,0) only.

Hence by Bendixsion Criterion, there is no closed path in ℝ2

Option (4) is correct.

Concept:

If the eigenvalues of the Jacobian matrix at the critical point are negative then that critical point is an asymptotically stable node

Explanation:

Given system of ordinary differential equations

$\frac{dx}{dt}=x(3-2x-2y),$

$\frac{dy}{dt}=y(2-2x-y),$

So F(x,y) = x(3 - 2x - 2y) = 3x - 2x² - 2xy and G(x,y) = y(2 - 2x - y) = 2y - 2xy - y². 

F_x = 3 - 4x - 2y, F_y = - 2x, G_x = - 2y, G_y = 2 - 2x - 2y

At (0, 2), Fx = - 3 - 4 = - 1, Fy = 0, Gx = - 4, Gy = 2 - 0 - 4 = - 2

Hence at (0,2) Jacobian is

J(0,2) = $\left[\begin{array}{cc}-1& 0\\ -4& -2\end{array}\right]$

This is an upper triangular matrix so eigenvalues are -1, -2.

Both eigenvalues are negative so (0,2) is a stable node.

Option (3) is correct.

Concept:

• Ordinary Point: A point x = x₀ is called an ordinary point of differential equation y'' + P(x)y' + Q(x) = 0, if P(x) and Q(x) are both analytical at x = x₀.

• A singular point x = x0 is called regular singular point if both (x - x₀)P(x) and (x - x₀)²Q(x) are analytic at x = x₀. Otherwise it is called irregular singular point.

• The indicial equation in variable m for regular singular point x₀ is represented by m(m - 1) + pm + q = 0, where p =$\underset{x\to x_0}{lim}(x-x_0)P(x)$  and q = $\underset{x\to x_0}{lim}(x-x_0{)}^{2}Q(x)$.

Calculation:

We have, 4xy" + 2y' + y = 0

⇒ $y^{″}+\frac{1}{2x}\frac{dy}{dx}+\frac{1}{4x}y=0$ 

⇒ P(x) = $\frac{1}{2x}$ and Q(x) = $\frac{1}{4x}$

⇒ x = 0 is a singular point.

Also, $\underset{x\to 0}{lim}xP(x)$ = $\underset{x\to 0}{lim}x(\frac{1}{2x})$ = $\frac{1}{2}$ = p

$\underset{x\to 0}{lim}{x}^{2}Q(x)$ = $\underset{x\to 0}{lim}{x}^2(\frac{1}{4x})$ = 0 = q

⇒ x = 0 is a regular singular point.

Now, indicial equation for the given differential equation is given by m(m - 1) + pm + q = 0

⇒ $m^2-m+\frac{m}{2}=0$

⇒ $m^2-\frac{m}{2}=0$

⇒ $m=0,\frac{1}{2}$ [Distinct roots]

Therefore, we get two independent solutions corresponding to two different value of m.

Since, x = x₀ is regular singular point, we have to use Forbenious method to get the required solution.

Let, $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{m+n}$

⇒ $y^{\prime }=\sum _{n=0}^{\mathrm{\infty }}(m+n)a_n{x}^{m+n-1}$

⇒ $y^{″}=\sum _{n=0}^{\mathrm{\infty }}(m+n)(m+n-1)a_n{x}^{m+n-2}$

Substituting the values of y, y' and y" in the given equation, we have,

$4x\sum _{n=0}^{\mathrm{\infty }}(m+n)(m+n-1)a_n{x}^{m+n-2}$

$+2\sum _{n=0}^{\mathrm{\infty }}(m+n)a_n{x}^{m+n-1}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{m+n}$ = 0

⇒ $\sum _{n=0}^{\mathrm{\infty }}4(m+n)(m+n-1)a_n{x}^{m+n-1}$

$+\sum _{n=0}^{\mathrm{\infty }}2(m+n)a_n{x}^{m+n-1}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{m+n}$ = 0

Shifting the index of first two terms to m+n, we have

⇒ $\sum _{n=0}^{\mathrm{\infty }}4(m+n+1)(m+n)a_{n+1}{x}^{m+n}$

$+\sum _{n=0}^{\mathrm{\infty }}2(m+n+1)a_{n+1}{x}^{m+n}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{m+n}$ = 0

In general, equating co-efficient of $x^{m+n}$ to zero, we have

⇒ $[4(m+n+1)(m+n)+2(m+n+1)]a_{n+1}+a_n=0$

⇒ $a_{n+1}=\frac{a_n}{[4(m+n+1)(m+n)+2(m+n+1)]},n\ge 0$

When m = 0:

$a_{n+1}=\frac{a_n}{[4(n+1)(n)+2(n+1)]},n\ge 0$

$a_1=\frac{a_0}{2}$

$a_2=\frac{a_1}{12}=\frac{a_0}{24}$, and so on.

Therefore, when m=0, one of the solution of y(x) is

$y(x)=x^0(a_0+a_{1}x+a_2{x}^2+\cdots )$

⇒ $y(x)=a_0+\frac{a_0}{2}+\frac{a_0}{24}{x}^2+\cdots .$

Substituting the initial condition y(0) = 1, we get a₀ = 1

∴ $y(x)=1+\frac{x}{2}+\frac{x^2}{24}+\cdots$

⇒ $y^{\prime }(x)=\frac{1}{2}+\frac{x}{12}+\cdots$

⇒ $y^{″}(x)=\frac{1}{12}+\underset{\text{higher power of }x}{\underbrace{\cdots \cdots \cdots }}$

∴ $y^{″}(0)=\frac{1}{12}$

The correct answer is Option 2.

Concept: 

Second-Order Differential Equations: Solve the homogeneous equation first, then find the particular solution.

Boundary Value Problems: Use boundary conditions to determine the constants in the general solution.

Explanation:
$(x{y}^{\prime }{)}^{\prime }-2y^{\prime }+\frac{y}{x}=1,1<x<e^4$

with boundary conditions $y(1)=0,y(e^4)=4e^4$

Rewriting the equation:

The equation is simplified to

 $x^2{y}^{″}-x{y}^{\prime }+y=x$

Let x = e^z then it becomes

⇒ {D'(D' - 1) - D' + 1}y = e^z

⇒ (D'² - 2D' + 1)y = e^z

Auxiliary equation is

m² - 2m + 1 = 0

⇒ $(m-1{)}^2=0$

⇒ $m=1,1$  

CF = C₁e^z + C₂ze^z 

 i.e., CF = $C_{1}x+C_{2}x\ln (x)$

PI = $\frac{1}{(D^{\prime 2}-2D^{\prime }+1)}{e}^z$

   = $e^z\frac{1}{((D^{\prime }+1{)}^2-2(D^{\prime }+1)+1)}.1$

  = $e^z\frac{1}{D^{\prime 2}}.1$

  = $\frac{z^2}{2}{e}^z$ = $\frac{x(\ln x{)}^2}{2}$
 
$y(x)=C_{1}x+C_{2}x\ln (x)$+ $\frac{x(lnx{)}^2}{2}$

Applying Boundary Conditions:

y(1) = 0:

 $C_1\cdot 1+C_2\cdot 1\ln (1)+\frac{1^2}{2}.0=0$
 
⇒ $C_1=0$

$y(e^4)=4e^4$: 

$\frac{1}{2}{e}^4+C_2{e}^4\ln (e^4)+e^4\frac{(ln{e}^4{)}^2}{2}=4e^4$
   
⇒  $C_2{e}^4\cdot 4+8e^4=4e^4$

⇒ $c_2=-1$
   
$y(x)=-xlnx+$$\frac{x(lnx{)}^2}{2}$

⇒ $y(e)=-e+\frac{e}{2}$

⇒ $y(e)=-\frac{e}{2}$

Hence option 1) is correct.

Concept:

Picard’s Existence and Uniqueness Theorem: Consider the Initial Value Problem (IVP) $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{f}(\mathrm{x},\mathrm{y})$, y(x₀) = y₀, suppose that f(x, y) and $\frac{\partial f}{\partial y}$ are continuous functions in some open rectangle R = {(x, y): a < x < b, c < y < d} that contains the point (x₀, y₀) . Then the IVP has a unique solution in some closed interval I = [x₀ - h,x0 + h] where h > 0.

Explanation: 

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\cos (\mathrm{x}\mathrm{y}),$ x ∈ ℝ, y(0) = y0,

Here f(x, y) = cos(xy)

$\frac{\partial f}{\partial y}$(x, y) = - x sin(xy)

Both are continuous in a open rectangular region R = {(x, y): a < x < b, c < y < d} containing (0, y₀) 

Now, |$\frac{\partial f}{\partial y}$(x, y)| = |-x sin(xy)| = |x||sin(xy)| ≤ |x| < b (as |sin(xy)| ≤ 1 for all x, y ℝ)

Hence by Picard’s existence and uniqueness theorem, 

the given IVP has a unique solution

Option (1) is true

Explanation:

y0 > 0, z0 > 0 and α > 1. 

(∗) $\{\begin{array}{l}\frac{dy}{dt}=y^{\alpha }\text{ for }t>0,\\ y(0)=y_0\end{array}$

(∗∗) $\{\begin{array}{l}\frac{dz}{dt}=-z^{\alpha }\text{ for }t>0,\\ z(0)=z_0\end{array}$

Let us assume α = 2

then (∗) ⇒

 $\frac{dy}{dt}=y^2,y(0)=y_0$ 

⇒ $\frac{dy}{y^2}$ = dt

⇒ $-\frac{1}{y}$ = t + c (integrating)

Using y(0) = y₀ we get

c = $-\frac{1}{y_0}$

⇒ $-\frac{1}{y}$ = t $-\frac{1}{y_0}$

⇒ y = $-\frac{y_o}{1-t{y}_0}$

y is not defined if

1 - ty₀ = 0 ⇒ t = $\frac{1}{y_0}$ > 0  as  y0 > 0

So (∗) does not have a global solution.

(1), (2) are false

$\underset{t\to \frac{1}{y_0}}{lim}|y(t)|=+\mathrm{\infty }$

(4) is correct

If we check (∗∗) by taking α = 2 we can see that (3) is false

Concept:

(i) If y1 and y2 are linearly independent then W(x) ≠ 0 ∀ x

(ii) If y1 and y2 are linearly dependent then W(x) = 0 ∀ x

Explanation:

By direct result, (4) is correct only