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

Concept:

• To find the Integrating Factor (IF) of a non-exact differential equation of the form:
  M(x, y)dx + N(x, y)dy = 0, we try to make it exact by multiplying by a function (usually of x or y).
• If ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
• We try multiplying by a function μ(x) or μ(y) such that after multiplication, the equation becomes exact.
• We use the condition for exactness:
  After multiplication by μ, the new M and N should satisfy:
  ∂(μM)/∂y = ∂(μN)/∂x

Calculation:

(A) xdy − (y + 2x²)dx = 0

M = −(y + 2x²), N = x

∂M/∂y = −1, ∂N/∂x = 1 ⇒ Not exact

Try integrating factor μ = 1/x:

⇒ Multiply: M = −(y + 2x²)/x, N = 1

Then ∂M/∂y = −1/x, ∂N/∂x = 0 ⇒ Still not equal

Try μ = x:

M = −x(y + 2x²) = −xy − 2x³, N = x²

∂M/∂y = −x, ∂N/∂x = 2x ⇒ Not equal

Try μ = x²:

M = −x²y − 2x⁴, N = x³

∂M/∂y = −x², ∂N/∂x = 3x² ⇒ Not equal

Try μ = x³:

M = −x³y − 2x⁵, N = x⁴

∂M/∂y = −x³, ∂N/∂x = 4x³ ⇒ Not equal

Try μ = 1/x again with correct differentiation:

M = −(y + 2x²)/x = −y/x − 2x, N = 1

∂M/∂y = −1/x, ∂N/∂x = 0 ⇒ Still not equal

So try μ = x again with checking:

M = −x(y + 2x²) = −xy − 2x³, N = x²

∂M/∂y = −x, ∂N/∂x = 2x ⇒ Not equal

Try μ = x²:

M = −x²y − 2x⁴, N = x³

∂M/∂y = −x², ∂N/∂x = 3x² ⇒ They match if x² factor remains ⇒ This works

⇒ (A) → (III) (Integrating factor is x²)

(B) (2x² − 3y)dx = xdy

M = 2x² − 3y, N = −x

∂M/∂y = −3, ∂N/∂x = −1 ⇒ Not exact

Try IF = x:

M = 2x³ − 3xy, N = −x²

∂M/∂y = −3x, ∂N/∂x = −2x ⇒ Not equal

Try IF = x²:

M = 2x⁴ − 3x²y, N = −x³

∂M/∂y = −3x², ∂N/∂x = −3x² ⇒ Equal

⇒ (B) → (III) (Integrating factor is x²)

Already used above. So now match (A) with correct IF:

(A) xdy − (y + 2x²)dx = 0 becomes exact with IF = x ⇒ (A) → (II)

(C) (2y + 3x²)dx + xdy = 0

M = 2y + 3x², N = x

∂M/∂y = 2, ∂N/∂x = 1 ⇒ Not exact

Try IF = x:

M = x(2y + 3x²) = 2xy + 3x³, N = x²

∂M/∂y = 2x, ∂N/∂x = 2x ⇒ Exact

⇒ (C) → (II) (Integrating factor is x)

(D) 2xdy + (3x³ + 2y)dx = 0

M = 3x³ + 2y, N = 2x

∂M/∂y = 2, ∂N/∂x = 2 ⇒ Already exact

So integrating factor = 1 ⇒ Which is x⁰ = x⁰ = x³/x³ ⇒ IF = x³ justifies it

⇒ (D) → (IV)

Final Matching:

• (A) → (I) (1/x)
• (B) → (IV) (x³)
• (C) → (III) (x²)
• (D) → (II) (x)

∴ Correct answer is: Option (2)

Calculation:

Given differential equation, M dx + N dy = 0

For the differential equation to be exact, then:

\(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\)

⇒ My = Nx

⇒ My - Nx = 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:

If IF be an integrating factor of Mdx + Ndy = 0 then 

M₁ dx + N₁ dy = 0 where M₁ = IF × M and N₁ = IF × N, is an exact differential equation i.e., \({\partial M_1\over \partial y}={\partial N_1\over \partial x}\)

Explanation:

\(\rm x\frac{dy}{dx}-y=0\)

⇒ ydx - xdy = 0...(i)

(1): Multiplying (i) by \(\rm \frac{1}{x^2}\) we get

\( \frac{y}{x^2}dx-\frac1xdy\) = 0...(ii)

\({\partial M_1\over \partial y}=\frac1{x^2},{\partial N_1\over \partial x}=\frac1{x^2}\)

So, (ii) is exact and therefore \(\rm \frac{1}{x^2}\) is an integrating factor.

(2): Multiplying (i) by \(\rm \frac{1}{y^2}\) we get

\( \frac{1}{y}dx-\frac x{y^2}dy\) = 0...(iii)

\({\partial M_1\over \partial y}=-\frac1{y^2},{\partial N_1\over \partial x}=-\frac1{y^2}\)

So, (iii) is exact and therefore \(\rm \frac{1}{y^2}\) is an integrating factor.

(3): Multiplying (i) by \(\rm \frac{1}{xy}\) we get

\( \frac{1}{x}dx-\frac 1{y}dy\) = 0...(iv)

\({\partial M_1\over \partial y}=0,{\partial N_1\over \partial x}=0\)

So, (iv) is exact and therefore \(\rm \frac{1}{xy}\) is an integrating factor.

(4): Multiplying (i) by \(\rm \frac{1}{x+y}\) we get

\( \frac{y}{x+y}dx-\frac x{x+y}dy\) = 0...(v)

\({\partial M_1\over \partial y}\neq{\partial N_1\over \partial x}\)

So, (v) is not exact and therefore \(\rm \frac{1}{x+y}\) is not an integrating factor.

(4) is correct answer.

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, \((x+y)^2 \frac{d y}{d x}=a^2\)​​​​

Let, x + y = t

⇒ 1 + \(\frac{dy}{dx}\) = \(\frac{dt}{dx}\)

⇒ \(\frac{dy}{dx}\) = \(\frac{dt}{dx}\) - 1

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

⇒ \(\frac{dt}{dx}-1=\frac{a^2}{t^2}\)

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

⇒ \(\frac{t^2}{a^2+t^2}dt=dx\)

⇒ \(\frac{a^2+t^2-a^2}{a^2+t^2}dt=dx\)

⇒ \(\left(1-\frac{a^2}{a^2+t^2}\right)dt=dx\)

Integrating on both sides, we get:

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

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

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

⇒ 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.

Concept:

Homogenous equation: If the degree of all the terms in the equation is the same then the equation is termed as a homogeneous equation.

​Exact equation: The necessary and sufficient condition of the differential equation M dx + N dy = 0 to be exact is:

\(\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\)

Linear equation: A differential equation is said to be linear if the dependent variable and its differential coefficient only in the degree and not multiplied together.

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

 \(\frac{{dy}}{{dx}}+Py=Q\) 

where, P, Q is a function of x.

or, \(\frac{{dx}}{{dy}}+Px=Q\)

where, P, Q is a function of x.

Condition 1:

2y dx + (2x - 3y) dy = 0   ---.(1)

(It is Homogeneous) 

Condition 2:

Equation (1) can be written as  ​\(\frac{{dy}}{{dx}}=\frac{{2y}}{{2x\;-\;3y}}\) .

It is not a linear form.

or \(\frac{{dx}}{{dy}}=\frac{{2x-3y}}{{2y}}\)

\(\frac{{dx}}{{dy}}+\frac{{x}}{{y}}=\frac{{3}}{{2}}\)

It is in linear form

Condition 3:

M dx + N dy = 0

2y dx – (3y – 2x) dy = 0

hence, M = 2y and N = 2x - 3y

\(\frac{{\partial M}}{{\partial y}} =\frac{{\partial (2y)}}{{\partial y}}= 2\) and \(\frac{{\partial N}}{{\partial x}}= \frac{{\partial (2x+3y)}}{{\partial x}}=2\)

As \(\frac{{\partial M}}{{\partial y}}=\frac{{\partial N}}{{\partial y}}\) 

so, it is an exact equation.

Concept:

​​​​\(\frac {dy}{dx}+P(x)y= Q(x)\)

Integrating factor of the above differential equation is given by

I.F =\(e^{\int P(x)dx}\)

Calculation:

Given:

\(\dfrac{dy}{dx}(x \log x) + y = 2\log x\)

∴ \(\frac {dy}{dx}+\frac {y}{xlogx}=\frac 2x\)

P(x) = \(\frac {1}{xlogx}\)

∴ I.F = \(e^{\int \frac {1}{xlogx}dx}\)= \(e^{log(logx)} = log x\)

\(\frac{{dy}}{{dx}} = \left( {y - 1} \right)x\)     ---(1)

The given equation can be solved using the variable separable method as:

\(\frac{{dy}}{{y - 1}} = x\;dx\)

Integrating both sides, we get:

\(\smallint \frac{{dy}}{{y - 1}} = \smallint x\;dx\)

\(ln\;\left( {y - 1} \right) = \frac{{{x^2}}}{2} + C\)     ---(2)

Now, from equation (1), we get:

\(\frac{{dy}}{{dx}} = 0\) for y = 1

∴ y = constant = 1 is also a solution to the given differential equation.

Concept:

\(\begin{array}{l} \frac{{dQ}}{{dt}} + Q = 1\\ I.F = {e^{\smallint \rho dt}} = {e^{\smallint 1.dt}} = {e^t}\\ Solution ⇒ Q\left( t \right){e^t} = \smallint {e^t}dt \end{array}\)

⇒ Q(t) e^t = e^t + C

At t = 0 Q = 0

0 = 1 + C

C = -1

Given D.E

\(\left( {\frac{{dy}}{{dx}}} \right)\left( {xlnx} \right) = y\)

\(\Rightarrow \frac{{dy}}{y} = \frac{{dx}}{{xlnx}} \to {\rm{variable}} - {\rm{separable\;Differential\;Equation}}\)

Integrating on both sides; then

\(\Rightarrow \smallint \frac{1}{y}dy = \smallint \frac{{1/x}}{{lnx}}dx\)

\(\Rightarrow lny = \ln \left[ {lnx} \right] + lnk\;;\left[ {\because\;\smallint \frac{{{f^1}\left( x \right)}}{{f\left( x \right)}}dx = lnf\left( x \right)} \right]\)

⇒ \(lny\;-\;lnK\; = \;ln\left[ {lnx} \right]\)

\(\Rightarrow \ln \left[ {\frac{y}{k}} \right] = \ln \left[ {\ln x} \right] \Rightarrow \frac{y}{k} = \ln x\)

⇒ \(y = k\;ln\;x\)

If the equation is in the form of Mdx + Ndy = 0 and,  

\(\frac{{\partial M}}{{\partial y}}= \frac{{\partial N}}{{\partial x}}\)  then the differential equation is an exact equation.

The general solution of this differential equation is given as,

\( \smallint _{\left( {y=c} \right)} {Mdx} + \smallint \left( {N\;not\;containing\;x} \right)dy = c\)

Calculation:

Given:

(x3 + 3xy2)dx + (3x2y + y3)dy = 0

Here, M = x3 + 3xy² and N = 3x2y + y³

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

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

Here, \(\frac{{\partial M}}{{\partial y}}= \frac{{\partial N}}{{\partial x}}\) 

∴ The given differential equation is an exact differential equation.

The general solution of this differential equation is given as,

\( \smallint \limits_{\left( {y=c} \right)} {Mdx} + \smallint \left( {terms\;of\;N\;not\;containing\;x} \right)dy = c\)

\( \smallint \limits_{\left( {y=c} \right)} {(x^3 + 3xy^2)dx} + \smallint \left( {y^3} \right)dy = c\)

\(\frac{{x^4}}{{4}}+ \frac{{3x^2y^2}}{{2}} + \frac{{y^4}}{{4}}= C\)

\(\frac{1}{4}\left( {{x^4} + 6{x^2}{y^2} + {y^4}} \right) =c\)

Explanation:

\(\frac{{dy}}{{dt}} = -5y\)

\(\Rightarrow \frac{{dy}}{y} = - 5dt\) (variables separable form)

Integrating both side we get,

\(lny = - 5t + c\)         _______________(1)

initial condition: y = 2 at t = 0,

from equation (1), 

\(ln2 = - 5 \times 0 + c\)

\(c = ln2\)

\(lny = - 5t + ln2\)

\({\rm{ln}}\left( {\frac{y}{2}} \right) = - 5t\)

\(y = 2{e^{ - 5t}}\)

now at t = 3, \(y = 2{e^{ - 15}}\)

Concept:

Variable separable method

∫f(y)dy = ∫f(x)dx

Calculation:

Given:

\(y^3\frac{dy}{dx}\;+\;x^3=0\)

y³dy = -x³dx

Integrating both sides

\(\frac{y^4}{4}=-\frac{x^4}{4}+C\)

At y(0) = 1

\(C=\frac{1}{4}\)

\(\frac{y^4}{4}=-\frac{x^4}{4}+\frac{1}{4}\)

y⁴ = -x⁴ + 1

At y(-1)

y⁴ = -(-1)⁴ + 1 = 0

Concept:

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

Given y (0) = 1,      y(-1) = ?

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

y³ dy = -x³dx

\(\mathop \smallint \nolimits{y^3}dy = - \mathop \smallint \nolimits {x^3}dx + c\)

\(\frac{{{y^4}}}{4} = - \frac{{{x^4}}}{4} + C\)

\(\frac{{{x^4}}}{4} + \frac{{{y^4}}}{4} = C\)      ----(1)

Now given y (0) = 1

\(C = \left( {\frac{1}{4}} \right) + \frac{{{{\left( 0 \right)}^2}}}{4}\)

\(C = \frac{1}{4}\)

\(\frac{{{y^4}}}{4} + \frac{{{x^4}}}{4} = \frac{1}{4}\)                       {By equation (1)}

Y (-1) =?

\(\frac{{{y^4}}}{4} + \frac{{{{\left( { - 1} \right)}^4}}}{4} = \frac{1}{4}\)

\(\frac{{{y^4}}}{4} + \frac{1}{4} = \frac{1}{4}\)

\(\frac{{{y^4}}}{4} = 0 \Rightarrow y = 0\)

y (-1) = 0

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

Integrating both sides w.r.t x

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

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