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

Ans : (4)

Solution :

Let y = \(\rm \sum_{j=0}^{\infty} a_{j} X^{j}, \frac{d y}{d x}=\sum_{j=0}^{\infty} j a_{j} X^{j-1} \text { and } \frac{d^{2} y}{d x^{2}}=\sum_{j=0}^{\infty} j(j-1) a_{j} X^{j-2}\)

∵ \(\rm \frac{d^{2} y}{d x^{2}}+\frac{1}{x^{2}} \frac{d y}{d x}-\frac{4}{x^{2}} y=0 \Rightarrow x^{2} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-4 y=0\)

⇒ \(\rm \sum_{j=0}^{\infty} j(j-1) a_{j} x^{j}+\sum_{j=0}^{\infty} j a_{j} x^{j-1}-4 \sum_{j=0}^{\infty} a_{j} x^{j}=0\)

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 = \(\rm \frac{4-j(j-1)}{j+1} a_{j}\)

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, \(\rm \displaystyle \lim _{t \rightarrow \infty} y(t)=\frac{\gamma}{\infty}+\frac{c}{\infty}=0\)

Hence, the correct answer is Option 1. 

Explanation:

\(\frac{d y}{d x}\) + 2y sec²x = 2sec²x + 3 tan x sec²x 

I.F. = \(e^{\int 2 \sec ^{2} x d x}\)

I.F. = e^2tanx

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

Put tan x = u

sec²xdx = du 

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

\(y \cdot e^{2 u} \Rightarrow \frac{2 e^{2 u}}{2}+3 \int e^{2 u} \cdot u d u\)

\(y \cdot e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\int \frac{e^{2 u}}{2}\right]\)

\(y e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\frac{e^{2 u}}{4}\right]+C\)

\(y e^{2 \tan x}=e^{2 \tan x}+3\left[\frac{\tan x e^{2 \tan x}}{2}-\frac{e^{2 \tan x}}{4}\right]+C\)

F(0) = \(\frac{5}{4}\)

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

\(\frac{5}{4}-\frac{1}{4}=C\)

1 = C

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

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

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

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

Calculation: 

\(\rm \frac{d y}{d x}-\frac{y}{x}=y^{3}\left(1+\log _{e} x\right)\)

⇒ \(\rm \frac{1}{y^{3}} \frac{d y}{d x}-\frac{1}{x y^{2}}=1+\log _{e} x\)

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

∴ \(\rm \frac{d t}{d x}+\frac{2 t}{x}=2\left(1+\log _{e} x\right)\)

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

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

y(1) = 3

\(\rm \frac{y^{2}}{9}=\frac{x^{2}}{5-2 x^{3}\left(2+\log _{e} x^{3}\right)}\)

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:

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

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

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

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: \(\rm \frac{d^3y}{dx^3} + \cos\left(\frac{d^2y}{dx^2}\right) = 0\)

The highest order derivative presents in the differential equation is \(\rm \frac{d^3y}{dx^3}\)

Hence, its order is three.

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

Concept:

\(\rm \displaystyle \int \frac{dx}{1+x^2} = \tan^{-1} x + c\)

Calculation:

Given: dy = (1 + y²) dx

\(\rm \Rightarrow \frac{dy}{1+y^2}=dx\)

Integrating both sides, we get

\(\rm \Rightarrow \displaystyle \int \frac{dy}{1+y^2}=\displaystyle \int dx\\\rm \Rightarrow \tan^{-1} y = x + c \)

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

= {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.

Calculation:

Given: \(\ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) - {\rm{a}} = 0\)

\( \Rightarrow \ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{a}}\)

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {{\rm{e}}^{\rm{a}}}\)

\(\Rightarrow {\rm{\;}}\smallint \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \smallint {{\rm{e}}^{\rm{a}}}\)

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:

\(\rm y = x \frac{dy}{dx}+\left(\frac{dy}{dx}\right)^{-2} \\ \rm y = x\frac{dy}{dx}+\frac{1}{(\frac{dy}{dx})^2} \\ y(\frac{dy}{dx} )^2= x(\frac{dy}{dx})^3 + 1\)

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.

Concept:

\(\rm \int \frac{1}{x}dx = \log x + c\)

\(\rm \int x^ndx = \frac{x^{n+1}}{n+1} + c\)

Calculation:

Given: \(\rm\left( xy \frac {dy}{dx} -1 \right)= 0\)

\(\Rightarrow \rm xy \frac {dy}{dx} =1 \)

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

Integrating both sides, we get

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

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:

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

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: 

\(\rm \int \frac{1}{a^{2}+x^{2}}dx = \frac{1}{a}\tan ^{-1}\frac{x}{a}+ C\) 

Calculation: 

Given : \(\rm dy = \left ( 4 + y^{2} \right )dx\) 

⇒ \(\rm \frac{dy}{4+y^{2}}= dx\) 

Integrating both sides, we get 

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

⇒ \(\rm \frac{1}{2}\tan^{-1}\frac{y}{2}= x+c\) 

⇒ \(\rm \tan^{-1}\frac{y}{2}= 2x+ 2c\)

⇒ \(\rm \tan^{-1}\frac{y}{2}= 2x+ C\)  [∵ 2c = C]

⇒ \(\rm \frac{y}{2}= \tan(2x+ C)\)

 \(\rm y = 2\tan \left ( 2x+C \right )\) 

The correct option is 2 .