Non Exact Differential Equations MCQ Quiz - Objective Question with Answer for Non Exact Differential Equations - Download Free PDF

Explanation:

$t\frac{dx}{dt}+t-x=0$

$\frac{dx}{dt}-\frac{x}{t}+1=0$

$\frac{dx}{dt}-\frac{1}{t}x=-1$

This is a linear differential equation of the form,

$\frac{dx}{dt}+Px=Q$

where, $P=-\frac{1}{t}\text{ and }Q=-1$

Integrating factor, $IF=e^{\int -\frac{1}{t}dx}=e^{-\ln t}=e^{\ln \left(\frac{1}{t}\right)}=\frac{1}{t}$

Thus, the solutions is

$x\times IF=\int Q\cdot IFdt+C$

$x\cdot \frac{1}{t}=\int -1\cdot \frac{1}{t}dt+c$

$\frac{x}{t}=-\ln t+c\Rightarrow \frac{x}{t}=\ln \left(\frac{1}{t}\right)+c$

$\frac{x}{t}=\ln \frac{1}{t}+c\Rightarrow x=t\cdot \ln \frac{1}{t}+ct$

Given, at t = 1, x = 0

⇒ 0 = 0 + c ⇒ c = 0

Now, for t = 2,

$x=2\ln \frac{1}{2};x=-2\ln 2$

x = -2 × 0.693 = -1.386 ≃ -1.39

Concept Used:

Separation of variables and integration for solving first-order differential equations.

Calculation

Given:

Differential equation: $\frac{dy}{dx}=xy-2x-2y+4$

⇒ $\frac{dy}{dx}=x(y-2)-2(y-2)$

⇒ $\frac{dy}{dx}=(x-2)(y-2)$

⇒ $\frac{dy}{y-2}=(x-2)dx$

Integrate both sides:

⇒ $\int \frac{dy}{y-2}=\int (x-2)dx$

⇒ ${\log }_e|y-2|=\frac{(x-2{)}^2}{2}+C$

∴ The general solution is ${\log }_e|y-2|=\frac{(x-2{)}^2}{2}+C$

Hence option 2 is correct

Calculation:

$\frac{dy}{cosy}=dx$

sec y dy = dx

Integrating both sides:

$\int secydy=\int dx$

log |sec y + tan y| = x + C

where C is the constant of integration.

∴ The solution is log |sec y + tan y| = x + C

Hence option 1 is correct

Explanation:

We are given the differential equation:

$3xy+y^2+(x^2+xy)\frac{dy}{dx}=0$  

$(3xy+y^2)+(x^2+xy)\frac{dy}{dx}=0$

$(x^2+xy)\frac{dy}{dx}+(3xy+y^2)=0$  

$(x^2+xy)dy+(3xy+y^2)dx=0$   

$(3xy+y^2)dx+(x^2+xy)dy=0$  

which is of the form:

M(x, y)  dx+ N(x, y) dy = 0

where:  $M=3xy+y^2$  and  $N=x^2+xy$

A differential equation is exact if:

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

$\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}(3xy+y^2)=3x+2y$    

$\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}(x^2+xy)=2x+y$   

Since  $3x+2y\ne 2x+y$  , the equation is not exact, and we need an integrating factor (IF)

An integrating factor  $\mu (x,y)$  makes the equation exact by multiplying both M and N by $\mu (x,y)$ 

(A) x as an Integrating Factor: 

Multiplying by x :

$x(3xy+y^2)dx+x(x^2+xy)dy=0$    

New  $M^{\prime }=x(3xy+y^2)=3x^{2}y+x{y}^2$

New  $N^{\prime }=x(x^2+xy)=x^3+x^{2}y$

Check exactness:

$\frac{\partial M^{\prime }}{\partial y}=\frac{\partial }{\partial y}(3x^{2}y+x{y}^2)=3x^2+2xy$   

$\frac{\partial N^{\prime }}{\partial x}=\frac{\partial }{\partial x}(x^3+x^{2}y)=3x^2+2xy$  

⇒ x is an integrating factor

(B) $x^2$  as an Integrating Factor:

Multiplying by $x^2$  :

$x^2(3xy+y^2)dx+x^2(x^2+xy)dy=0$   

New  $M^{\prime }=x^2(3xy+y^2)=3x^{3}y+x^2{y}^2$

New  $N^{\prime }=x^2(x^2+xy)=x^4+x^{3}y$

Check exactness:

$\frac{\partial M^{\prime }}{\partial y}=\frac{\partial }{\partial y}(3x^{3}y+x^2{y}^2)=3x^3+2x^{2}y$     

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

⇒  $x^2$  is not an integrating factor

(C) 3x as an Integrating Factor:

Multiplying by 3x :

$3x(3xy+y^2)dx+3x(x^2+xy)dy=0$    

New  $M^{\prime }=3x(3xy+y^2)=9x^{2}y+3x{y}^2$

New  $N^{\prime }=3x(x^2+xy)=3x^3+3x^{2}y$

Check exactness:

$\frac{\partial M^{\prime }}{\partial y}=\frac{\partial }{\partial y}(9x^{2}y+3x{y}^2)=9x^2+6xy$    

$\frac{\partial N^{\prime }}{\partial x}=\frac{\partial }{\partial x}(3x^3+3x^{2}y)=9x^2+6xy$      

⇒  3x is not an integrating factor

(D)  $\frac{1}{xy(2x+y)}$  as an Integrating Factor:

Multiplying by  $\frac{1}{xy(2x+y)}$ :

$\frac{1}{xy(2x+y)}(3xy+y^2)dx+\frac{1}{xy(2x+y)}(x^2+xy)dy=0$

$M^{\prime }=\frac{(3x+y)}{x(2x+y)}$

$N^{\prime }=\frac{(x+y)}{y(2x+y)}$  

We now check if the equation becomes exact, meaning: $\frac{\partial M^{\prime }}{\partial y}=\frac{\partial N^{\prime }}{\partial x}$  

$\frac{\partial }{\partial y}\left(\frac{3x+y}{x(2x+y)}\right)=\frac{\partial }{\partial x}\left(\frac{x+y}{y(2x+y)}\right)$  

which confirms exactness 

Thus, $\frac{1}{xy(2x+y)}$  is a valid integrating factor

⇒ (A), (C), and (D) are  correct only 

Hence Option(2) is the correct answer.

Given:

$(y-x{y}^2)dx-(x+x^{2}y)dy=0$

Concept used:

If an equation of the form Mdx + Ndy = 0 is not exact, it can always be made exact by multiplying by some function of x and y. Such a multiplier is called an integrating factor. 

Solution:

We have,

(y - xy2) dx - (x + x2y)dy = 0

Comparing with Mdx + Ndy = 0 Here M = (y - xy²) and N = - (x + x²y)

Since Mx - Ny $\ne$ 0 then Integrating factor is $\frac{1}{Mx-Ny}$

Mx - Ny = (y - xy²)x - {- (x + x²y)y} ⇒ Mx - Ny = xy - x²y² + xy + x²y²

⇒ Mx - Ny = 2xy Hence, Mx - Ny $\ne$ 0

 So, I.F = 1/2xy

$\therefore$ Option 3 is correct.

 $\frac{dy}{dx}=y^2$and y(0) = 1

$\Rightarrow \frac{dy}{y^2}=dx$(separation of variable)

⇒ from the Integrating the equation, we get

$\Rightarrow \int \frac{dy}{y^2}=\int dx$

$\Rightarrow \frac{-1}{y}=x+c$

⇒ at x = 0, y = 1, hence -1/1 = c ⇒ c = -1

$\Rightarrow \frac{-1}{y}=x-1$

$\Rightarrow \frac{1}{y}=1-x$

$\Rightarrow y=\frac{1}{1-x}$

x ≠ 1, at (x = 1) y will be undefined

Concept: 

The standard form of a linear equation of the first order, commonly known as Leibnitz's linear equation, is

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}.\mathrm{y}=\mathrm{Q}$

Where, P and Q are the functions of 'X'

The integrating factor is given as,

$\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \mathrm{P}\mathrm{d}\mathrm{x}}$

The solution of the above equation is given as,

$\begin{array}{l}\left(\mathrm{I}.\mathrm{F}\right)\mathrm{y}=\int \left(\mathrm{I}.\mathrm{F}\right)\mathrm{Q}\mathrm{d}\mathrm{x}+C\end{array}$

Calculation:

Given equation,

$\frac{\mathrm{d}\mathrm{P}}{\mathrm{d}\mathrm{t}}+{\mathrm{K}}_2\mathrm{P}={\mathrm{K}}_1{\mathrm{L}}_0{\mathrm{e}}^{-{\mathrm{K}}_1\mathrm{t}}$

This equation is a linear equation of first order, therefore comparing it with the general equation

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}.\mathrm{y}=\mathrm{Q}$

Hence P = K₂

∴ $\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \mathrm{P}\mathrm{d}\mathrm{t}}$

∴ $\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int {\mathrm{K}}_2\mathrm{d}\mathrm{t}}$

∴ $\mathrm{I}.\mathrm{F}={\mathrm{e}}^{{\mathrm{K}}_2\mathrm{t}}$

Concept:

Separation of Variables:

f₁(x)g₁(y)dx + f₂(x)g₂(y)dy = 0

$\int \frac{f_1(x)}{f_2(x)}dx+\int \frac{g_2(y)}{g_1(y)}dy=C$

Calculation:

Given:

Differential equation:

$\frac{dy}{dx}=\left(1+y^2\right)x$

By separation of variables:

$\frac{1}{1+y^2}dy=xdx$

Integrating on both sides we will get:

$\Rightarrow {\tan }^{-1}y=\frac{x^2}{2}+C$

$\therefore y=\tan \left(\frac{x^2}{2}+C\right)$

The given differential equation is, (dy/dx) = ky

$\frac{dy}{y}=kdx$

On integrating both the sides, we get

ln y = kx + ln c

Given differential equation is,

$y\sqrt{1-x^2}dy+x\sqrt{1-y^2}dx=0$

$\Rightarrow \frac{y}{\sqrt{1-y^2}}dy=\frac{-x}{\sqrt{1-x^2}}dx$

By integrating both sides,

$\int \frac{y}{\sqrt{1-y^2}}dy=-\int \frac{x}{\sqrt{1-x^2}}dx+C_1$

(1 – y²)^1/2 = - (1 – x²)^1/2 + C

Concept:

One dimensional wave equation:

 $\frac{{\partial }^{2}u}{\partial t^2}=C^2\frac{{\partial }^{2}u}{\partial x^2}$

Two-dimensional wave equation:

 $\frac{{\partial }^{2}u}{\partial t^2}=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}$

Analysis:

The solution is given as:

F = f(x ± ct)

option 1 and 3 satisfies the given conditions. 

Concept:

This is the case of an In-exact differential equation first we need to reduce it to exact differential

(1) If 'M' and 'N' are a homogeneous function of some degree then

• I.F = $\frac{1}{\mathrm{M}\mathrm{x}+\mathrm{N}\mathrm{y}}$

(2) If 'M' and 'N' are not homogeneous but M = yf₁(x,y), N = xf₂(x,y) then

• I.F = $\frac{1}{\mathrm{M}\mathrm{x}-\mathrm{N}\mathrm{y}}$

(3) If $\frac{\frac{\partial \mathrm{M}}{\partial \mathrm{y}}-\frac{\partial \mathrm{N}}{\partial \mathrm{x}}}{\mathrm{N}}=\mathrm{f}\left(\mathrm{x}\right)\left(\mathrm{o}\mathrm{r}\right)\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$         

• $\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}}$

(4) If $\frac{\frac{\partial \mathrm{N}}{\partial \mathrm{x}}-\frac{\partial \mathrm{M}}{\partial \mathrm{x}}}{\mathrm{M}}=\mathrm{f}\left(\mathrm{y}\right)\left(\mathrm{o}\mathrm{r}\right)\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$

• $\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \mathrm{f}\left(\mathrm{y}\right)\mathrm{d}\mathrm{y}}$

The solution is given by

$\int {\mathrm{M}}_1\mathrm{d}\mathrm{x}+\int \left(\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}\mathrm{i}\mathrm{n}{\mathrm{N}}_1\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{x}\right)\mathrm{d}\mathrm{y}=0$...(i)

where, M₁ = (I.F) × M, N₁ = (I.F) × N

Calculation:

Given:

2 sin (y²)dx + xy cos (y²)dy = 0, $\mathrm{y}\left(2\right)=\sqrt{\left(\frac{\pi }{2}\right)}$

M = 2 sin (y²), N = xy cos (y²)dy

$\frac{\partial \mathrm{M}}{\partial \mathrm{y}}=2\cos \left(y^2\right)2y$ & $\frac{\partial \mathrm{N}}{\partial \mathrm{x}}=y\cos \left(y^2\right)2y$

$\frac{\partial \mathrm{M}}{\partial \mathrm{y}}\ne \frac{\partial \mathrm{N}}{\partial \mathrm{x}}$

By using 3rd statement

$\frac{\frac{\partial \mathrm{M}}{\partial \mathrm{y}}-\frac{\partial \mathrm{N}}{\partial \mathrm{x}}}{\mathrm{N}}=\frac{3\mathrm{y}\mathrm{c}\mathrm{o}\mathrm{s}\left({\mathrm{y}}^2\right)}{\mathrm{x}\mathrm{y}\mathrm{c}\mathrm{o}\mathrm{s}\left({\mathrm{y}}^2\right)}=\frac{3}{\mathrm{x}}=\mathrm{f}\left(\mathrm{x}\right)$

$\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \frac{3}{\mathrm{x}}dx}=x^3$

M₁ = x³ × 2 sin (y2),

N₁ = x³ × xy cos (y2)

from equation '1' we have

$\int {\mathrm{M}}_1\mathrm{d}\mathrm{x}+\int \left(\mathrm{T}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}\mathrm{i}\mathrm{n}{\mathrm{N}}_1\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{x}\right)\mathrm{d}\mathrm{y}=0$

$\int 2x^3\sin \left(y^2\right)dx+\int x^{4}ycos\left(y^2\right)dy=0$

$\frac{x^4}{2}\sin \left(y^2\right)=c$, $\mathrm{y}\left(2\right)=\sqrt{\left(\frac{\pi }{2}\right)}$

$\frac{x^4}{2}\sin \left(\frac{\pi }{2}\right)=c\Rightarrow c=8$

$\frac{x^4}{2}\sin \left(\frac{\pi }{2}\right)=8$

$x^4\sin \left(y^2\right)=16$

y log y dx + (x – log y) dy = 0

$\frac{dx}{dy}+\frac{x-\log y}{y\log y}=0$

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

It is a Leibnitz’s equation in x.

Concept:

In the equation M dx + N dy = 0

If $\frac{\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)}{N}$ be a function of x only = f(x), then $e^{\int f\left(x\right)dx}$ is an integrating factor.

Calculation:

Given differential equation is

(x + y³)dx + 6xy² dy = 0

M = x + y³, N = 6xy²

$\frac{\partial M}{\partial y}=3y^2$ 

$\frac{\partial N}{\partial x}=6y^2$ 

$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\frac{3y^2-6y^2}{6x{y}^2}=\frac{-1}{2}\left(\frac{1}{x}\right)=f\left(x\right)$ 

Integrating factor $e^{\int \frac{-1}{2}\left(\frac{1}{x}\right)dx}=\frac{1}{\sqrt{x}}=x^{-0.5}$ 

Given that x^r is an integrating factor.

3yy’ + 4x = 0

$3y\frac{dy}{dx}=-4x$

3ydy = - 4x dx

On integration both the sides:

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

Where c is the constant of integration.

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

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