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. $=e^{\int {\displaystyle \frac{x}{x^2-1}}dx}=e^{{\displaystyle \frac{1}{2}}ln|x^2-1|}=\sqrt{1-x^2}$

$\Rightarrow$ solution is

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

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

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

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

Now,

${\int }_{-\sqrt{3}/2}^{\sqrt{3}/2}f(x)dx={\int }_{\sqrt{3}/2}^{\sqrt{3}/2}d{\displaystyle \frac{x^2}{\sqrt{1-x^2}}}dx$ (Using property)

$=2{\int }_0^{\sqrt{3}/2}{\displaystyle \frac{x^2}{\sqrt{1-x^2}}}dx=2{\int }_0^{\pi /3}{\displaystyle \frac{si{n}^2\theta }{cos\theta }}cos\theta d\theta$ (Taking $x=sin\theta$)

$=2{\int }_0^{\pi /3}si{n}^2\theta d\theta =2{[\frac{\theta }{2}-\frac{sin2\theta }{4}]}_0^{\frac{\pi }{3}}=2\left(\frac{\pi }{6}\right)-2\left(\frac{\sqrt{3}}{8}\right)=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$

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

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

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

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

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

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

Let  $\mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}$

$(1-\frac{3}{t^2})dt=du$

$=-2\int \frac{du}{u^2}$

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

$y\cdot \sec x=\frac{2}{t+\frac{3}{t}}+c$ .........(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. $\frac{x}{2}+\frac{2}{x},x\ne 0$  

This function is defined for $x\ne 0.$ 

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 $(-\frac{\pi }{4},\frac{\pi }{4})$  , 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:
  • $f(x)=\int (e^x)/\surd (4-e^{(2x)})dx$
• 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 = $si{n}^{-1}(u/2)+C$  
• We substitute back $u=e^x$ to get the final result:
  • $f(x)=si{n}^{-1}(e^x/2)+C$  
• To find the constant C, we use the initial condition that f(0) = π/2.

Calculation:

Given,

$f(x)=\int (e^x)/\surd (4-e^{(2x)})dx$  

Substitute  $u=e^x$, so that   $du=e^{x}dx$.

The integral becomes  $f(x)=si{n}^{-1}(e^x/2)+C.$

Use the initial condition:

f(0) = π/2

Substituting x = 0 into the equation:

$f(0)=si{n}^{-1}(e^0/2)+C=si{n}^{-1}(1/2)+C=\pi /6+C$   

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, $\frac{10{\mathrm{x}}^4+5{\mathrm{y}}^4+7{\mathrm{z}}^4}{13{\mathrm{x}}^2{\mathrm{y}}^2+6{\mathrm{y}}^2{\mathrm{z}}^2+3{\mathrm{z}}^2{\mathrm{x}}^2}$

= {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, $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}+\mathrm{P}\mathrm{x}=\mathrm{Q}$, where P and  Q are given continuous function of y. 

IF = ${\mathrm{e}}^{\int \mathrm{P}\mathrm{d}\mathrm{y}}$ 

Calculation:

Given differential equation

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{x}\mathrm{y}=\mathrm{x}$

Now, this differential equation is in the form

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

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

Integrating Factor (I.F.) = ${\mathrm{e}}^{\int \mathrm{P}(\mathrm{x})}\mathrm{d}\mathrm{x}$

I.F. = ${\mathrm{e}}^{\int \mathrm{x}}\mathrm{d}\mathrm{x}$ = ${\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}$ 

Concept:

First Order Linear Differential Equation:
A differential equation of the from ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$ + 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 ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$ + P × y = Q, where P and Q are constants or functions of x only.
• Find the Integrating Factor (F) by using the formula: F = ${\mathrm{e}}^{\int \mathrm{P}\mathrm{d}\mathrm{x}}$.
• Write the solution using the formula: $\mathrm{y}\times \mathrm{F}={\displaystyle \int (\mathrm{Q}\times \mathrm{F})\mathrm{d}\mathrm{x}+\mathrm{C}}$ where C is the constant of integration.

Calculation:

$\mathrm{x}{\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}-\mathrm{y}=3$

$\Rightarrow {\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}+\left({\displaystyle \frac{-1}{\mathrm{x}}}\right)\mathrm{y}={\displaystyle \frac{3}{\mathrm{x}}}$

⇒ P = ${\displaystyle \frac{-1}{\mathrm{x}}}$ and Q = ${\displaystyle \frac{3}{\mathrm{x}}}$.

Integrating factor F = ${\mathrm{e}}^{\int \mathrm{P}\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\int \frac{-1}{\mathrm{x}}\mathrm{d}\mathrm{x}}={\mathrm{e}}^{-\ln \mathrm{x}}={\displaystyle \frac{1}{\mathrm{x}}}$.

The solution of the given differential equation is:

$\mathrm{y}\times {\displaystyle \frac{1}{\mathrm{x}}}={\displaystyle \int ({\displaystyle \frac{3}{\mathrm{x}}}\times {\displaystyle \frac{1}{\mathrm{x}}})\mathrm{d}\mathrm{x}+\mathrm{C}}$

$\Rightarrow {\displaystyle \frac{\mathrm{y}}{\mathrm{x}}}=3\left({\displaystyle \frac{-1}{\mathrm{x}}}\right)+\mathrm{C}$

⇒ 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 $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}+\mathrm{P}\mathrm{x}=\mathrm{Q}$ then, where P and Q are functions of y.

The solution is given as, $\mathrm{x}\times \mathrm{I}.\mathrm{F}.=\int \mathrm{I}.\mathrm{F}.\times \mathrm{Q}\mathrm{d}\mathrm{y}+\mathrm{c}$

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

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

Calculation:

Given: ydx – (x + 2y²) dy = 0

$\mathrm{y}\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}=\mathrm{x}+2{\mathrm{y}}^2$

$\Rightarrow \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}=\frac{\mathrm{x}}{\mathrm{y}}+2\mathrm{y}$

$\Rightarrow \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}-\frac{\mathrm{x}}{\mathrm{y}}=2\mathrm{y}$

Differential equation is in form of, $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}+\mathrm{P}\mathrm{x}=\mathrm{Q}$

Integrating factor, $\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{P}\mathrm{d}\mathrm{y}}$

$\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int -\frac{1}{\mathrm{y}}\mathrm{d}\mathrm{y}}$

$\Rightarrow \mathrm{I}.\mathrm{F}.={\mathrm{e}}^{-\ln \mathrm{y}}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=\frac{1}{\mathrm{y}}$

Differential equation is given as,

$\mathrm{x}\times \frac{1}{\mathrm{y}}=\int \frac{1}{\mathrm{y}}\times \left(2\mathrm{y}\right)\mathrm{d}\mathrm{y}+\mathrm{c}$

$\Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=2\mathrm{y}+\mathrm{c}$

⇒ x = 2y² + cy

Concept:

The solution of the linear differential equation $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}(\mathrm{x})\mathrm{y}=\mathrm{Q}(\mathrm{x})$ 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 $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=\frac{1+\mathrm{y}}{\mathrm{x}}$ ⇒  $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ + y - $\frac{y}{x}$ = $\frac{1}{x}$

⇒  $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ + $(1-\frac{1}{x})$y = $\frac{1}{x}$

This is a differential equation of the form $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}(\mathrm{x})\mathrm{y}=\mathrm{Q}(\mathrm{x})$

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:

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

1. find I.F = $e^{\int Pdx}$
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×$\frac{1}{cosx}$(sin x - y) dx

⇒ dy = (y $\frac{sinx}{cosx}$ - y²$\frac{1}{cosx}$) dx ⇒ $\frac{dy}{dx}$ = y tan x - y² sec x 

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

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

Putting these values we get 

$-\frac{dt}{dx}-ttanx=-secx$  ⇒ $\frac{dt}{dx}+ttanx=secx$

Now,  I.F = $e^{\int tanxdx}=e^{logsecx}=secx$

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 first order linear differential equation;

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}\mathrm{y}=\mathrm{Q}$, where P and Q are function of x

Integrating factor (IF) = e∫ P dx

y × (IF) = ∫ Q(IF) dx

Calculation:

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\frac{\mathrm{y}}{\mathrm{x}}=3\sin \mathrm{x}$

IF = e∫  $\frac{1}{x}$dx

⇒ IF = e^ln x

⇒ IF = x

Concept:

In the first-order linear differential equation;

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

Where P and Q are functions of x

Integrating factor (IF) = e∫ P dx

y × (IF) = ∫ Q(IF) dx

Calculation:

x$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ + y = 4x3 + x

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\frac{\mathrm{y}}{\mathrm{x}}=4{\mathrm{x}}^2+1$

IF = e∫ $\frac{1}{\mathrm{x}}$ 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 + $\frac{{\mathrm{x}}^2}{2}$+ c (where c is integration constant)

⇒ y = $\text{boldsymbol}{\mathrm{x}}^3+\frac{\mathrm{x}}{2}+\frac{\mathrm{c}}{\mathrm{x}}$