Exact Differential Equations MCQ Quiz in தமிழ் - Objective Question with Answer for Exact Differential Equations - இலவச PDF ஐப் பதிவிறக்கவும்

$\frac{d^{2}y}{d{x}^2}=-12x^2+24x-20$

Integrating both sides w.r.t x

$\frac{dy}{dx}=-4x^3+12x^2-20x+c_1$

Again integrating both sides w.r.t x

y = -x⁴ + 4x³ – 10x² + c₁x + c₂                                    ……….(i)

At x = 0 and y = 5

5 = c₂

At x = 2, y =21

y = -x4 + 4x3 – 10x2 + c1x + c2   

21 = – 16 + 32 – 40 + 2c₁ + c₂

2c₁ = 21 + 16 – 32 + 40 – 5

2c₁ = 40

c₁ = 20

y = – x⁴ + 4x³ – 10 x² + 20x + 5  

Put x = 1

y = – 1 + 4 – 10 + 20 + 5

y = 18

Concept:

Let Mdx + Ndy = 0 be a differential equation and if $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}$  = f(x) or constant then the integration factor is given as 

I. F = $e^{\int f(x)dx}$

Calculation:

Given:

Consider the first differential equation:

$x{y}^{{}^{\prime }}+2y=x\Rightarrow y^{{}^{\prime }}+\left(\frac{2}{x}\right)\times y=1$

I. F = $e^{\int f(x)dx}$

${\mathrm{I}}_1\left(\mathrm{I}.\mathrm{F}\right)={\mathrm{e}}^{\int \frac{2}{\mathrm{x}}\mathrm{d}\mathrm{x}}={\mathrm{e}}^{2.\mathrm{l}\mathrm{n}\mathrm{x}}={\mathrm{e}}^{\mathrm{l}\mathrm{n}{\mathrm{x}}^2}={\mathrm{x}}^2$

For the second differential equation:

$x{y}^{{}^{\prime }}-2y=x\Rightarrow y^{{}^{\prime }}+(-\frac{2}{x})\times y=1$

I. F = $e^{\int f(x)dx}$

 ${\mathrm{I}}_2\left(\mathrm{I}.\mathrm{F}\right)={\mathrm{e}}^{-\int \frac{2}{\mathrm{x}}\mathrm{d}\mathrm{x}}={\mathrm{e}}^{-2\mathrm{l}\mathrm{n}\mathrm{x}}={\mathrm{x}}^{-2}$

$I_1\times I_2=x^2\times x^{-2}=1$

∴  $I_1\times I_2=1$

1) If the function is in x terms then the integration factor is I.F = $e^{\int f(x)dx}$

2) If the function is in y terms then the integration factor is I.F = $e^{\int f(y)dy}$

Leibnitz’s linear equation

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

Solution is

$y\left(I.f\right)=\int Q\left(I.f\right)dx+c$

$I.f.=e^{\int pdx}$

$\frac{dy}{dx}+\frac{1}{x\log x}y=\frac{\log x^2}{x\log x}=\frac{2\log x}{x\log x}=\frac{2}{x}$

$\int pdx=\int \frac{1}{x\log x}dx=\ln \left(\log x\right)$

$I.f.=e^{\int pdx}=e^{\ln \left(\log x\right)}=\log x$

$y\left(\log x\right)=\int \frac{2}{x}\log xdx+c=2\int \log x.\frac{dx}{x}+c$

y(log x) = (log x)² + c

Concept:

The general form of a differential equation is,

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ or Mdx + Ndy = 0

Where M and N are functions of x and y.

By variable separable form,

In an equation, if it is possible to collect all terms of x and dx on one side and all the terms of y and dy on the other side, then the variables are said to be separable.

After separating, integrate on both the side of the equation,

$\int \mathrm{f}\left(\mathrm{y}\right)\mathrm{d}\mathrm{y}=\int \varphi \left(\mathrm{x}\right)\mathrm{d}\mathrm{x}+\mathrm{C}$

Calculation:

Given,

9yy’ + 4x = 0

Where,

$\left({\mathrm{y}}^{\prime }=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}},\mathrm{C}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\right)$

$9\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+4\mathrm{x}=0$

$9\mathrm{y}\mathrm{d}\mathrm{y}=-4\mathrm{x}\mathrm{d}\mathrm{x}$

Integrating on both the side,

$\int 9\mathrm{y}\mathrm{d}\mathrm{y}=\int \left(-4\mathrm{x}\right)\mathrm{d}\mathrm{x}+{\mathrm{C}}_1$

$\frac{9y^2}{2}=-\frac{4x^2}{2}+{\mathrm{C}}_1$

$9{\mathrm{y}}^2+4{\mathrm{x}}^2=2{\mathrm{C}}_1$

$4{\mathrm{x}}^2+9{\mathrm{y}}^2=\mathrm{C}$

Hence the general solution of the given differential equation is $4{\mathrm{x}}^2+9{\mathrm{y}}^2=\mathrm{C}$

Explanation:

$\frac{du}{dx}=\frac{-x{u}^2}{2+x^{2}u}$

(2 + x²u) du + (xu²) dx = 0

For exact differential, $\frac{dP}{dx}=\frac{dQ}{du}$

Q = (2 + x2u), P = (xu2)

$\frac{dQ}{dx}=\frac{d}{dx}(2+x^{2}u)$ = 2ux

$\frac{dP}{du}=\frac{d}{du}(x{u}^2)$ = 2ux

Since dP/du = dQ/dx, hence it is an exact differential.

Thus its solution is given by:

∫Pdx + ∫terms in Q(x,y) not containing x du = C

${\int }_{C}x{u}^{2}dx+\int 2du=C$

$\frac{x^2{u}^2}{2}+2u=C$

y = $A{e}^{3x}+B{e}^{2x}$

, $\frac{dy}{dx}=3A{e}^{3x}+2B{e}^{2x}$

, $\frac{d^{2}y}{d{x}^2}=9A{e}^{3x}+4B{e}^{2x}$

Now, according to option (a)

$\frac{d^{2}y}{d{x}^2}+5\frac{dy}{dx}-6y=9A{e}^{3x}+4B{e}^{2x}+15A{e}^{3x}+10B{e}^{2x}-e^{3x}+6B{e}^{2x}\ne 0$

Hence, option (a) is not correct.

Now, according to option (b)

$\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}+6y=9A{e}^{3x}+4B{e}^{2x}-15A{e}^{3x}-10B{e}^{2x}+e^{3x}+6B{e}^{2x}=0$

So, option (b) is the correct option.

Calculation:

Given differential equation, M dx + N dy = 0

For the differential equation to be exact, then:

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

⇒ M_y = N_x

⇒ M_y - N_x = 0

∴ The differential equation M dx + N dy = 0 will be exact if and only if My - Nx = 0.

The correct answer is Option 2.

Concept:

Variable Seperable Method:

Consider the first-order differential equation, 

P(y)$\frac{dy}{dx}$ = Q(x), where Q(x) and P(y) are functions involving x and y only, respectively.

We can solve this by separating variables:

P(y)$\frac{dy}{dx}$ = Q(x) ⇒ $\int P(y)dy=\int Q(x)dx$

Calculation:

Given, $(x+y{)}^2\frac{dy}{dx}=a^2$​​​​

Let, x + y = t

⇒ 1 + $\frac{dy}{dx}$ = $\frac{dt}{dx}$

⇒ $\frac{dy}{dx}$ = $\frac{dt}{dx}$ - 1

∴ The differential equation becomes, $t^2(\frac{dt}{dx}-1)=a^2$

⇒ $\frac{dt}{dx}-1=\frac{a^2}{t^2}$

⇒ $\frac{dt}{dx}=\frac{a^2}{t^2}+1=\frac{a^2+t^2}{t^2}$

⇒ $\frac{t^2}{a^2+t^2}dt=dx$

⇒ $\frac{a^2+t^2-a^2}{a^2+t^2}dt=dx$

⇒ $(1-\frac{a^2}{a^2+t^2})dt=dx$

Integrating on both sides, we get:

$\int (1-\frac{a^2}{a^2+t^2})dt=\int dx$

⇒ $\int 1dt-a^2\int \frac{1}{a^2+t^2}dt=\int dx$

⇒ $t-a^2\times \frac{1}{a}{\tan }^{-1}\left(\frac{t}{a}\right)=x+c$

⇒ x + y - a tan^-1($\frac{x+y}{a}$) = x + c

⇒ y - a tan-1($\frac{x+y}{a}$) = c

⇒ tan-1($\frac{x+y}{a}$) = $\frac{y-c}{a}$

⇒ $\frac{x+y}{a}$ = tan($\frac{y-c}{a}$)

⇒ y + x = a tan($\frac{y-c}{a}$), where c is the constant of integration. 

∴ The solution of the given differential equation is (y + x) = a tan$\left(\frac{y-c}{a}\right)$.

The correct answer is Option 1.

Calculation:

Laplace equation is $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$  ...(i)

Its solution using separation of variables method is

∴ u(x, y) = (C₁cos λ x + C₂ sin λx)(C₃e^λy + C₄^-λy) ...(ii)

Let AB = CD = 1

Using u(x, 0) = 0

⇒ C₃ = –C₄

Using (0, y) = 0

⇒ C₁ = 0

Substituting constants values in equation (ii)

u(x, y) = (0 + C₂ sinλx) (C₃e^λy - C₃e^-λy)

u(x, y) = a_n sin λ x(eλy - e^-λy)

Now, using u(1, y) = 0,

sin λ = 0 ⇒ λ = nπ

Hence u(x, y) = $\sum {\mathrm{a}}_{\mathrm{n}}\sin (\mathrm{n}\pi \mathrm{x})[{\mathrm{e}}^{\mathrm{n}\pi \mathrm{y}}-{\mathrm{e}}^{-\mathrm{n}\pi \mathrm{y}}]$

Using u(x, 1) = 100

⇒ $a_1=\frac{100}{\sin \pi x[e^{\pi }-e^{-\pi }]}$

Using equation (iii),

$\mathrm{u}(\mathrm{x},\mathrm{y})=\frac{100}{\mathrm{s}\mathrm{i}\mathrm{n}\pi \mathrm{x}[{\mathrm{e}}^{\pi }-{\mathrm{e}}^{-\pi }]}({\mathrm{e}}^{\pi \mathrm{y}}-{\mathrm{e}}^{-\pi \mathrm{y}})$

At mid point, ie.

$\mathrm{u}(\frac{1}{2},\frac{1}{2})={\mathrm{T}}_0$

⇒ ${\mathrm{T}}_0=\frac{100}{({\mathrm{e}}^{\pi }-{\mathrm{e}}^{-\pi })}({\mathrm{e}}^{\pi /2}-{\mathrm{e}}^{-\pi 2})$

= $\frac{100}{({\mathrm{e}}^{\pi /2}+{\mathrm{e}}^{-\pi /2})}=\frac{100}{2\cos \mathrm{h}\frac{\pi }{2}}=\frac{50}{2.509}=19.928={19.83}^{\circ }\mathrm{C}$

Concept:

In equation M dx + N dy = 0,

1. if $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial y}}{N}$ be function of x only = f(x) say, then e^∫f(x) dx is an integrating factor.

2. if $\frac{\frac{\partial N}{\partial y}-\frac{\partial M}{\partial y}}{M}$ be function of y only = F(y) say, then e∫F(y) dy is an integrating factor.

The derivative of standard function

$\frac{d}{dx}{x}^n=n{x}^{n-1}$

The integration of standard function

$e^{\int \frac{1}{x}}dx=e^{logx}=x$

Calculation:

Given:

(xy3 + y) dx + 2 (x2y2 + x + y4) dy = 0

where, M = (xy3 + y) and N = 2 (x2y2 + x + y4) 

∴ $\frac{1}{M}(\frac{\partial N}{\partial y}-\frac{\partial M}{\partial y})=\frac{1}{y(x{y}^2+1)}(\frac{2\partial (x^2{y}^2+x+y^4)}{\partial y}-\frac{\partial (x{y}^3+y)}{\partial y})$ 

$\frac{1}{M}(\frac{\partial N}{\partial y}-\frac{\partial M}{\partial y})=\frac{1}{y(x{y}^2+1)}(4x{y}^2+2-3x{y}^2-1)$

$\frac{1}{M}(\frac{\partial N}{\partial y}-\frac{\partial M}{\partial y})=\frac{1}{y}$

$\frac{1}{y}$ is a function of y alone.

∴$Integratingfactor=e^{\int \frac{1}{y}}dy=e^{logy}=y$