Solving Linear Differential Equation MCQ Quiz - Objective Question with Answer for Solving Linear Differential Equation - Download Free PDF

This is a linear differential equation

I.F. \(= e^{\int \dfrac{x}{x^2 -1}dx} = e^{\dfrac{1}{2} ln |x^2 -1|} = \sqrt{1 - x^2}\)

\(\Rightarrow \) solution is

\(y\sqrt{1 - x^2}=\int \dfrac{x(x^3 +2)}{\sqrt{1 - x^2}} \cdot \sqrt{1 -x^2}dx\)

or \( y \sqrt{1 -x^2} = \int (x^4 +2x)dx = \sqrt{x^5}{5} + x^2 +c\)

\(f(0) = 0 \Rightarrow c = 0\)

\(\Rightarrow f(x) \sqrt{1 - x^2} = \dfrac{x^5}{5} + x^2\)

Now,

\(\int_{- \sqrt{3}/ 2}^{\sqrt{3} / 2} f(x) dx = \int_{\sqrt{3}/2}^{\sqrt{3}/2} d\dfrac{x^2}{\sqrt{1- x^2}}dx\) (Using property)

\(= 2 \int_{0}^{\sqrt{3}/2} \dfrac{x^2}{\sqrt{1 -x^2}} dx = 2 \int_0^{\pi/ 3} \dfrac{sin^2 \theta}{cos \theta} cos \theta d \theta\) (Taking \( x = sin \theta\))

\(= 2 \int_{0}^{\pi / 3} sin^2 \theta d \theta = 2 \left [ \frac{\theta}{2} - \frac{sin 2 \theta}{4} \right]_0^{\frac{\pi}{3}} = 2 \left ( \frac{\pi}{6} \right ) - 2 \left ( \frac{\sqrt{3}}{8} \right ) = \frac{\pi}{3} - \frac{\sqrt{3}}{4}\)

If \( \mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\operatorname{tn}(\sec \mathrm{x})}=\sec \mathrm{x} \)

\( \therefore \mathrm{y} \cdot \sec \mathrm{x}=\int\left\{\frac{2+\sec \mathrm{x}}{(1+2 \sec \mathrm{x})^{2}}\right\} \sec \mathrm{xdx} \\ \)

\(=\int \frac{2 \cos \mathrm{x}+1}{(\cos \mathrm{x}+2)^{2}} \mathrm{dx} \) Let \( \cos \mathrm{x}=\frac{1-\mathrm{t}^{2}}{1+\mathrm{t}^{2}} \)

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

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

\(=2 \int \frac{3-\mathrm{t}^{2}}{\left(\mathrm{t}^{2}+3\right)^{2}} \mathrm{dt}\)

Let  \(\mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}\)

\( \left(1-\frac{3}{t^{2}}\right) d t=d u \)

\(=-2 \int \frac{d u}{u^{2}} \)

\(y \cdot(\sec x)=\frac{2}{u}+c \)

\(y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c\) .........(I)

At \(x=\frac{\pi}{3}, t=\tan \frac{x}{2}=\frac{1}{\sqrt{3}} \)

\(2 \cdot \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+c \)

\(2 \cdot \frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+c \Rightarrow C=0 \)

At \(x=\frac{\pi}{4}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \)

\(\therefore \mathrm{y} \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \)

\(\text { y. } \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}} \)

\(y=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7}=\frac{4-\sqrt{2}}{14}\)

Explanation:

A.  \(\frac{x}{\log_e x} \)

This function is defined for x > 0, as the logarithm function is only defined for positive values of x.

To find the interval in which the function is increasing, take the derivative of  \(\frac{x}{\log_e x} \)  with respect to x.

After solving, it can be found that the function is increasing when x > e.

So, the correct interval is (IV).

B. \(\frac{x}{2} + \frac{2}{x}, \, x \neq 0 \)  

This function is defined for \(x \neq 0. \) 

By finding the derivative and setting it to zero,

it can be shown that the function is increasing for x > 2 and x < -2.

So, the correct interval is (I).

C.  \( x^x, \, x > 0 \)

This function is increasing for x > 0.

So, the correct interval is (III).

D. \( \sin x - \cos x \)

The derivative of   \( \sin x - \cos x \) is   \( \cos x + \sin x. \)

Setting this derivative greater than zero gives the interval \(\left( -\frac{\pi}{4}, \frac{\pi}{4} \right) \)  , so the correct interval is (II).

Final Answer:

The correct matching is:

(A) → (IV)
(B) → (I)
(C) → (III)
(D) → (II)

The correct option is (3).

Concept:

Integrating Factor:

• The integrating factor is used to solve linear first-order differential equations.
• For a linear differential equation of the form: dy/dx + P(x)y = Q(x), the integrating factor is given by:
• Integrating Factor = e^∫P(x)dx

Calculation:

Given the differential equation:

y log_e(y) dx/dy + x = 2 log_e(y)

Rewriting the equation:

y log_e(y) dx/dy = 2 log_e(y) - x

Dividing both sides by y log_e(y):

dx/dy = 2/y - x/(y log_e(y))

This is in the standard form of a linear first-order differential equation:

dx/dy + P(y) x = Q(y)

Identifying P(y) = 1/(y log_e(y)) and Q(y) = 2/y, we find the integrating factor:

μ(y) = e^∫P(y)dy = e^{∫1/(y log_e(y))dy}

The integral of 1/(y log_e(y)) is log_e(log_e(y)), so:

μ(y) = log_e(y)

Hence, the integrating factor is: log_e(y)

Concept:

Integration and Constant of Integration:

• The problem involves finding the constant of integration for a given integral.
• The integral is of the form:
  • \(f(x) = ∫ (e^x) / √(4 - e^{(2x)}) dx \)
• To solve this, we use the substitution method to simplify the integral.
• The substitution used is:
  • \(u = e^x \), and hence \(du = e^x dx \)
• The integral then reduces to a standard form:
  • ∫ 1 / √(4 - u²) du = \(sin^{-1}(u / 2) + C \)  
• We substitute back \(u = e^x \) to get the final result:
  • \(f(x) = sin^{-1} (e^x / 2) + C \)  
• To find the constant C, we use the initial condition that f(0) = π/2.

Calculation:

Given,

\(f(x) = ∫ (e^x) / √(4 - e^{(2x)} ) dx \)  

Substitute  \( u = e^x \), so that   \(du = e^x dx \).

The integral becomes  \( f(x) = sin^{-1} (e^x / 2) + C. \)

Use the initial condition:

f(0) = π/2

Substituting x = 0 into the equation:

\(f(0) = sin^{-1} (e^0 / 2) + C = sin^{-1} (1/2) + C = π/6 + C \)   

Set this equal to π/2:

π/6 + C = π/2

Solve for C:

C = π/2 - π/6 = 3π/6 - π/6 = 2π/6 = π/3

Hence, the constant of integration is: C = π/3

Given:

x = 1

x2 + y2 + z2 = xy + yz + zx

Calculations:

x² + y2 + z2 - xy - yz - zx = 0

⇒(1/2)[(x - y)² + (y - z)² + (z - x)²] = 0

⇒x = y , y = z and z = x

But x = y = z = 1

so, \(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\)

= {10(1)⁴ + 5(1)⁴ + 7(1)⁴}/{13(1)²(1)²+ 6(1)²(1)² + 3(1)²(1)²}

= 22/22

= 1

Hence, the required value is 1.

Given:

x + \(\frac{1}{2x}\) = 3

Concept Used:

Simple calculations is used

Calculations:

⇒ x + \(\frac{1}{2x}\) = 3

On multiplying 2 on both sides, we get

⇒ 2x + \(\frac{1}{x}\) = 6  .................(1)

Now, On cubing both sides,

⇒ \((2x + \frac{1}{x})^3 = 6^3\)

⇒ \(8x^3 + \frac{1}{x^3} + 3(4x^2)(\frac{1}{x})+3(2x)(\frac{1}{x^2})=216\)

⇒ \(8x^3 + \frac{1}{x^3} + 12x+\frac{6}{x}=216\)

⇒ \(8x^3 + \frac{1}{x^3}= 216 - 6(2x+\frac{1}{x})\)

⇒ \(8x^3 + \frac{1}{x^3}= 216- 6(6)\)  ..............from (1)

⇒ \(8x^3 + \frac{1}{x^3}= 216- 36\)

⇒ \(8x^3 + \frac{1}{x^3}= 180\)

⇒ Hence, The value of the above equation is 180

Given:

9-digit number = 83P93678Q

Divisor = 72

Concept Used:

Divisibility of 8 = Last three digits should be divisible by 8.

Divisibility of 9 = Sum of digits is divisible by 9.

Calculation:

As the divisor 72, is divisible by 8 and 9, so the divisibility will be checked.

For divisible by 8,

78Q should be divisible by 8, so, Q should be 4 as 784 is divisible by 8.

For divisible by 9,

⇒ 8 + 3 + P + 9 + 3 + 6 + 7 + 8 + 4 = 48 + P

For being divisible by 9, the nearest number to be added is 6 which gives 54.

Now, \(\sqrt{P^2+Q^2+12}=\sqrt{6^2+4^2+12}\)

⇒ \(\sqrt{36+16+12}=\sqrt{64}=8\)

Therefore, the required value is 8.

Concept: 

Integrating factor, (IF) for a differential equation, \(\rm\frac{dx}{dy}+Px= Q\), where P and  Q are given continuous function of y. 

IF = \(\rm e^{\int Pdy}\) 

Calculation:

Given differential equation

\(\rm \frac{dy}{dx} + xy = x\)

Now, this differential equation is in the form

\(\rm \frac{dy}{dx} + y P(x) = Q(x)\)

where, P(x) = x and Q(x) = x

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

I.F. = \(\rm e^{\int x} dx\) = \(\rm e^{\frac{x^{2}}{2}}\) 

Concept:

First Order Linear Differential Equation:
A differential equation of the from \(\rm \dfrac{dy}{dx}\) + P × y = Q, where P and Q are constants or functions of x only, is known as a first order linear differential equation.

Steps to solve a First Order Linear Differential Equation:

• Convert into the standard form \(\rm \dfrac{dy}{dx}\) + P × y = Q, where P and Q are constants or functions of x only.
• Find the Integrating Factor (F) by using the formula: F = \(\rm e^{\int P\ dx}\).
• Write the solution using the formula: \(\rm y\times F=\displaystyle \int (Q\times F)\ dx+C\) where C is the constant of integration.

Calculation:

\(\rm x \dfrac{dy}{dx}-y=3\)

\(\rm \Rightarrow \dfrac{dy}{dx}+\left(\dfrac{-1}{x}\right)y=\dfrac{3}{x}\)

⇒ P = \(\rm \dfrac{-1}{x}\) and Q = \(\rm \dfrac{3}{x}\).

Integrating factor F = \(\rm e^{\int P\ dx}=e^{\int \tfrac{-1}{x}\ dx}=e^{-\ln x}=\dfrac{1}{x}\).

The solution of the given differential equation is:

\(\rm y\times \dfrac{1}{x}=\displaystyle \int \left(\dfrac{3}{x}\times \dfrac{1}{x}\right)\ dx+C\)

\(\rm \Rightarrow \dfrac{y}{x}=3\left(\dfrac{-1}{x}\right)+C \)

⇒ y = Cx - 3, which is the equation of a straight line.

Concept:

Solution of Linear Differential equation:

If the D.E. has a form of \(\frac{{{\rm{dx}}}}{{{\rm{dy}}}} + {\rm{Px}} = {\rm{Q}}\) then, where P and Q are functions of y.

The solution is given as, \({\rm{x}} \times {\rm{I}}.{\rm{F}}.{\rm{\;}} = \smallint {\rm{I}}.{\rm{F}}.{\rm{\;}} \times {\rm{Qdy}} + {\rm{c}}\)

where, I.F. is integrating factor which is given as,

\({\rm{I}}.{\rm{F}}. = {{\rm{e}}^{\smallint {\rm{Pdy}}}}\)

Calculation:

Given: ydx – (x + 2y²) dy = 0

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

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

\( \Rightarrow \frac{{{\rm{dx}}}}{{{\rm{dy}}}} - \frac{{\rm{x}}}{{\rm{y}}} = 2{\rm{y}}\)

Differential equation is in form of, \(\frac{{{\rm{dx}}}}{{{\rm{dy}}}} + {\rm{Px}} = {\rm{Q}}\)

Integrating factor, \({\rm{I}}.{\rm{F}}. = {{\rm{e}}^{\smallint {\rm{Pdy}}}}\)

\({\rm{I}}.{\rm{F}}. = {{\rm{e}}^{\smallint - \frac{1}{{\rm{y}}}{\rm{dy}}}}\)

\(\Rightarrow {\rm{I}}.{\rm{F}}. = {{\rm{e}}^{ - \ln {\rm{y}}}}\)

\(\Rightarrow {\rm{I}}.{\rm{F}}. = \frac{1}{{\rm{y}}}\)

Differential equation is given as,

\({\rm{x}} \times \frac{1}{{\rm{y}}} = \smallint \frac{1}{{\rm{y}}} \times \left( {2{\rm{y}}} \right){\rm{dy}} + {\rm{c}}\)

\(\Rightarrow \frac{{\rm{x}}}{{\rm{y}}} = 2{\rm{y}} + {\rm{c}}\)

⇒ x = 2y² + cy

Concept:

The solution of the linear differential equation \(\rm\frac{dy}{dx}+P(x)y = Q(x)\) is given by

 y × I.F = \(\int{Q(x)(I.F) }dx+ C\)

Where P and Q are the functions of 'x' and I.F = \(e^{\int{P(x)}dx}\) 

Calculation:

Given \(\rm\frac{dy}{dx} + y = \rm\frac{1+y}{x}\) ⇒  \(\rm\frac{dy}{dx}\) + y - \(\frac{y}{x}\) = \(\frac{1}{x}\)

⇒  \(\rm\frac{dy}{dx}\) + \((1-\frac{1}{x})\)y = \(\frac{1}{x}\)

This is a differential equation of the form \(\rm\frac{dy}{dx}+P(x)y = Q(x)\)

Here P(x) = 1 - \(\frac{1}{x}\)

Integrating factor (I.F) = \(e^{\int{P(x)}dx}\)  = \(e^{\int(1-\frac{1}{x})dx}\) = \(e^{x - \log x}\)

⇒ I.F = \(\frac{e^x}{e^{\log x}}\) = \(\frac{e^x}{x}\)

The integral factor is \(\frac{e^x}{x}\).

The correct answer is option 2.

Concept:

In first order linear differential equation;

\(\rm {dy\over dx}+Py=Q\), where P and Q are function of x

Integrating factor (IF) = e∫ P dx

y × (IF) = ∫ Q(IF) dx

Calculation:

\(\rm {dy\over dx}+ \frac{y}{x} = 3\sin x\)

IF = e∫  \(\frac{1}{x}\)dx

⇒ IF = e^ln x

⇒ IF = x

Concept:

Equation of the form \(\frac{dy}{dx} + Py = Q\) solve by following the steps

1. find I.F = \(e^{∫ Pdx}\)
2. The solution will be y I.F = ∫ Q I.F dx + C

Formula used:

1. sin θ/cos θ = tan θ      2.1/cos θ = sec θ 

3. e^{ln x} = x       4. ∫ sec² x = tan x

Calculation:

cos x dy = y (sin x - y) dx

dy = y×\(\frac{1}{cos x}\)(sin x - y) dx

⇒ dy = (y \(\frac{sin\ x}{cos\ x}\) - y²\(\frac{1}{cos x}\)) dx ⇒ \(dy\over{dx}\) = y tan x - y² sec x 

⇒ \(\frac{1}{y^{2}}\frac{dy}{dx}-\frac{1}{y}tanx=-secx\)

Now, let y = \(1\over{t}\) therefore \(\frac{1}{y^{2}}\frac{dy}{dx}=-\frac{dt}{dx}\)

Putting these values we get 

\(-\frac{dt}{dx}-t \ tanx=-secx\)  ⇒ \(\frac{dt}{dx}+t \ tanx=secx\)

Now,  I.F = \(e^{∫ tan\ x\ dx}=e^{log\ sec\ x}= sec\ x\)

The solution of the equation will be 

⇒ t (I.F) =  ∫ (I.F) sec x dx + c ⇒ t (sec x) =  ∫ (I.F) sec x dx + c

⇒ t sec x = ∫ sec² x + c ⇒ sec x = (tan x + c)y

∴ The solution of an equation is sec x =  (tan x + c)y.

Concept:

In the first-order linear differential equation;

\(\rm {dy\over dx}+Py=Q\),

Where P and Q are functions of x

Integrating factor (IF) = e∫ P dx

y × (IF) = ∫ Q(IF) dx

Calculation:

x\(\rm \frac {dy}{dx}\) + y = 4x3 + x

⇒ \(\rm \frac{dy}{dx}+{y\over x}=4x^2+1\)

IF = e∫ \(\rm 1\over x\) dx

⇒ IF = eln x

⇒ IF = x                 

(∵ e^{ln x} = x)

Now, y × (IF) = ∫ Q (IF) dx

⇒ y × x = ∫ (4x² + 1) × x dx

⇒ yx = ∫ 4x³ + x  dx

Integrating,

⇒ yx = x⁴ + \(\rm x^2\over 2\)+ c (where c is integration constant)

⇒ y = \(\boldsymbol{\rm x^3 + {x\over2}+{c\over x}}\)