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

Ans : (4)

Solution :

Let y = $\sum _{\mathrm{j}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{j}}{\mathrm{X}}^{\mathrm{j}},\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\sum _{\mathrm{j}=0}^{\mathrm{\infty }}\mathrm{j}{\mathrm{a}}_{\mathrm{j}}{\mathrm{X}}^{\mathrm{j}-1}\text{ and }\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\sum _{\mathrm{j}=0}^{\mathrm{\infty }}\mathrm{j}(\mathrm{j}-1){\mathrm{a}}_{\mathrm{j}}{\mathrm{X}}^{\mathrm{j}-2}$

∵ $\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^2}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-\frac{4}{{\mathrm{x}}^2}\mathrm{y}=0\Rightarrow {\mathrm{x}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}+\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-4\mathrm{y}=0$

⇒ $\sum _{\mathrm{j}=0}^{\mathrm{\infty }}\mathrm{j}(\mathrm{j}-1){\mathrm{a}}_{\mathrm{j}}{\mathrm{x}}^{\mathrm{j}}+\sum _{\mathrm{j}=0}^{\mathrm{\infty }}\mathrm{j}{\mathrm{a}}_{\mathrm{j}}{\mathrm{x}}^{\mathrm{j}-1}-4\sum _{\mathrm{j}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{j}}{\mathrm{x}}^{\mathrm{j}}=0$

Equating coefficient of x^j to zero; j(j - 1)a_j + (j + 1)a_j+1 - 4a_j = 0

⇒ (j + 1)aj+1 = 4aj - j(j - 1)aj ⇒ aj+1 = $\frac{4-\mathrm{j}(\mathrm{j}-1)}{\mathrm{j}+1}{\mathrm{a}}_{\mathrm{j}}$

Calculation: 

$\frac{\mathrm{dy}}{\mathrm{dt}}+\alpha \mathrm{y}=\gamma {\mathrm{e}}^{-\beta \mathrm{t}}$

$\text{ I.F. }={\mathrm{e}}^{\int \alpha \mathrm{dt}}={\mathrm{e}}^{\alpha \mathrm{t}}$

$\text{ Solution }\Rightarrow \mathrm{y}\cdot {\mathrm{e}}^{\alpha \mathrm{t}}=\int \gamma {\mathrm{e}}^{-\beta \mathrm{T}}\cdot {\mathrm{e}}^{\alpha \mathrm{t}}\mathrm{dt}$

$\Rightarrow {\mathrm{ye}}^{\alpha \mathrm{t}}=\gamma \frac{{\mathrm{e}}^{(\alpha -\beta )\mathrm{t}}}{(\alpha -\beta )}+\mathrm{c}$

$\Rightarrow \mathrm{y}=\frac{\gamma }{{\mathrm{e}}^{\beta \mathrm{t}}(\alpha -\beta )}+\frac{\mathrm{c}}{{\mathrm{e}}^{\alpha \mathrm{t}}}$

So, ${\displaystyle \underset{\mathrm{t}\to \mathrm{\infty }}{lim}\mathrm{y}(\mathrm{t})=\frac{\gamma }{\mathrm{\infty }}+\frac{\mathrm{c}}{\mathrm{\infty }}=0}$

Hence, the correct answer is Option 1. 

Explanation:

$\frac{dy}{dx}$ + 2y sec²x = 2sec²x + 3 tan x sec²x 

I.F. = $e^{\int 2{\sec }^{2}xdx}$

I.F. = e^2tanx

^{$y\cdot e^{2\tan x}=\int e^{2\tan x}(2+3\tan x){\sec }^{2}xdx$}

Put tan x = u

sec²xdx = du 

$y\cdot e^{2u}=\int e^{2u}(2+3u)du$

$y\cdot e^{2u}\Rightarrow \frac{2e^{2u}}{2}+3\int e^{2u}\cdot udu$

$y\cdot e^{2u}=e^{2u}+3[\frac{u{e}^{2u}}{2}-\int \frac{e^{2u}}{2}]$

$y{e}^{2u}=e^{2u}+3[\frac{u{e}^{2u}}{2}-\frac{e^{2u}}{4}]+C$

$y{e}^{2\tan x}=e^{2\tan x}+3[\frac{\tan x{e}^{2\tan x}}{2}-\frac{e^{2\tan x}}{4}]+C$

F(0) = $\frac{5}{4}$

$\frac{5}{4}=1-\frac{3}{4}+C$

$\frac{5}{4}-\frac{1}{4}=C$

1 = C

$y=1+3(\frac{\tan x}{2}-\frac{1}{4})+1\cdot e^{-2\tan x}$

$y\left(\frac{\pi }{4}\right)=1+3(\frac{1}{2}-\frac{1}{4})+\frac{1}{e^2}$

$y\left(\frac{\pi }{4}\right)=\frac{7}{4}+\frac{1}{e^2}$

$12(y\left(\frac{x}{4}\right)-\frac{1}{e^2})=12(\frac{7}{4}+\frac{1}{e^2}-\frac{1}{e^2})$ = 21

Calculation: 

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{y}}^3(1+{\log }_{\mathrm{e}}\mathrm{x})$

⇒ $\frac{1}{{\mathrm{y}}^3}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-\frac{1}{\mathrm{x}{\mathrm{y}}^2}=1+{\log }_{\mathrm{e}}\mathrm{x}$

Let $-\frac{1}{{\mathrm{y}}^2}=\mathrm{t}\Rightarrow \frac{2}{{\mathrm{y}}^3}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$

∴ $\frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{x}}+\frac{2\mathrm{t}}{\mathrm{x}}=2(1+{\log }_{\mathrm{e}}\mathrm{x})$

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

⇒ $\frac{-{\mathrm{x}}^2}{{\mathrm{y}}^2}=\frac{2}{3}((1+{\log }_{\mathrm{e}}\mathrm{x}){\mathrm{x}}^3-\frac{{\mathrm{x}}^3}{3})+\mathrm{C}$

y(1) = 3

$\frac{{\mathrm{y}}^2}{9}=\frac{{\mathrm{x}}^2}{5-2{\mathrm{x}}^3(2+{\log }_{\mathrm{e}}{\mathrm{x}}^3)}$

Hence, the correct answer is Option 1. 

Concept:

Transformation of Variable in Second Order Differential Equations:

• A second-order linear differential equation with variable coefficients can sometimes be transformed into one with constant coefficients by a suitable change of variables.
• Given equation: d²y/dx² + P(x) dy/dx + e^2x y = 0
• Let the new variable be s = s(x) such that the new equation in s has constant coefficients.
• Using chain rule for derivatives:
  • dy/dx = dy/ds × ds/dx
  • d²y/dx² = d²y/ds² × (ds/dx)² + dy/ds × d²s/dx²
• Substituting into the equation yields transformed coefficients in terms of ds/dx and d²s/dx².
• To achieve constant coefficients, expressions involving P(x) and e^2x must cancel appropriately.

Calculation:

Given,

d²y/dx² + P(x) dy/dx + e^2x y = 0

Let s = s(x), such that ds/dx = e^x

⇒ dy/dx = dy/ds × e^x

⇒ d²y/dx² = d²y/ds² × e^2x + dy/ds × e^x

Substitute in original equation:

⇒ e^2x d²y/ds² + e^x dy/ds + P(x) e^x dy/ds + e^2x y = 0

⇒ e^2x d²y/ds² + e^x (1 + P(x)) dy/ds + e^2x y = 0

To make coefficients constant:

⇒ e^x(P(x) + 1) must be constant

⇒ e^−x(P(x) + 1) = constant

∴ e^−x(P(x) + 1) is a constant function on ℝ.

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given:

$\mathrm{y}=\mathrm{x}{\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)}^2+\left(\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}\right)$

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

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

For the given differential equation the highest order derivative is 1.

Now, the power of the highest order derivative is 3.

We know that the degree of a differential equation is the power of the highest derivative

Hence, the degree of the differential equation is 3.

Mistake PointsNote that, there is a term (dx/dy) which needs to convert into the dy/dx form before calculating the degree or order. 

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

The differential equation is given as: $\frac{{\mathrm{d}}^3\mathrm{y}}{\mathrm{d}{\mathrm{x}}^3}+\cos \left(\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}\right)=0$

The highest order derivative presents in the differential equation is $\frac{{\mathrm{d}}^3\mathrm{y}}{\mathrm{d}{\mathrm{x}}^3}$

Hence, its order is three.

Here the given differential equation is not a polynomial equation, Hence its degree is not defined.

Concept:

${\displaystyle \int \frac{\mathrm{d}\mathrm{x}}{1+{\mathrm{x}}^2}={\tan }^{-1}\mathrm{x}+\mathrm{c}}$

Calculation:

Given: dy = (1 + y²) dx

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

Integrating both sides, we get

$$\begin{gathered} \Rightarrow {\displaystyle \int \frac{\mathrm{d}\mathrm{y}}{1+{\mathrm{y}}^2}={\displaystyle \int \mathrm{d}\mathrm{x}}} \\ \Rightarrow {\tan }^{-1}\mathrm{y}=\mathrm{x}+\mathrm{c} \end{gathered}$$

⇒ y = tan (x + c)

∴ The solution of the given differential equation is y = tan (x + c).

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.

Calculation:

Given: $\ln \left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)-\mathrm{a}=0$

$\Rightarrow \ln \left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)=\mathrm{a}$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{a}}$

$\Rightarrow \int \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\int {\mathrm{e}}^{\mathrm{a}}$

On integrating both sides, we get

⇒ y = xe^a + c

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given:

$$\begin{gathered} \mathrm{y}=\mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+{\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)}^{-2} \\ \mathrm{y}=\mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\frac{1}{(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}{)}^2} \\ \mathrm{y}(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}{)}^2=\mathrm{x}(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}{)}^3+1 \end{gathered}$$

For the given differential equation the highest order derivative is 1.

Now, the power of the highest order derivative is 3.

We know that the degree of a differential equation is the power of the highest derivative.

Hence, the degree of the differential equation is 3.

Concept:

$\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\log \mathrm{x}+\mathrm{c}$

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

Given: $(\mathrm{x}\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-1)=0$

$\Rightarrow \mathrm{x}\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=1$

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

Integrating both sides, we get

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

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

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

${\displaystyle \frac{d^{2}y}{d{x}^2}}+3{\left({\displaystyle \frac{dy}{dx}}\right)}^2=x^2\log \left({\displaystyle \frac{d^{2}y}{d{x}^2}}\right)$

For the given differential equation the highest order derivative is 2.

The given differential equation is not a polynomial equation because it involved a logarithmic term in its derivatives hence its degree is not defined.

Concept: 

$\int \frac{1}{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\frac{1}{\mathrm{a}}{\tan }^{-1}\frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}$ 

Calculation: 

Given : $\mathrm{d}\mathrm{y}=(4+{\mathrm{y}}^2)\mathrm{d}\mathrm{x}$ 

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

Integrating both sides, we get 

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

⇒ $\frac{1}{2}{\tan }^{-1}\frac{\mathrm{y}}{2}=\mathrm{x}+\mathrm{c}$ 

⇒ ${\tan }^{-1}\frac{\mathrm{y}}{2}=2\mathrm{x}+2\mathrm{c}$

⇒ ${\tan }^{-1}\frac{\mathrm{y}}{2}=2\mathrm{x}+\mathrm{C}$  [∵ 2c = C]

⇒ $\frac{\mathrm{y}}{2}=\tan (2\mathrm{x}+\mathrm{C})$

 $\mathrm{y}=2\tan (2\mathrm{x}+\mathrm{C})$ 

The correct option is 2 .