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

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:

\(\rm \frac{d^2y}{dx^2}+a\sin x = 0\)

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

Now, the power of the highest order derivative is 1.

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

Hence, the degree of the differential equation is 1.

Calculation:

Given: \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}}\)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right)}}{{\phi \left( {\rm{x}} \right)}} - \frac{{{\rm{y}}.{\rm{y}}}}{{\phi \left( {\rm{x}} \right)}}\)

\({\rm{Let}},{\rm{\;}}\frac{{\rm{y}}}{{\phi \left( {\rm{x}} \right)}} = {\rm{z}}\)     ...1)

\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right)}}{{\phi \left( {\rm{x}} \right)}} - \frac{{{\rm{y}}.{\rm{y}}}}{{\phi \left( {\rm{x}} \right)}} = \phi '\left( {\rm{x}} \right){\rm{z}} - \phi \left( {\rm{x}} \right){{\rm{z}}^2}\)

Now, \({\rm{y}} = \phi \left( {\rm{x}} \right){\rm{z}}\) 

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \phi \left( {\rm{x}} \right)\frac{{{\rm{dz}}}}{{{\rm{dx}}}} + \phi '\left( {\rm{x}} \right){\rm{z}}\)

\(\therefore \phi '\left( {\rm{x}} \right){\rm{z}} - \phi \left( {\rm{x}} \right){{\rm{z}}^2} = \phi \left( {\rm{x}} \right)\frac{{{\rm{dz}}}}{{{\rm{dx}}}} + \phi '\left( {\rm{x}} \right){\rm{z}}\)

\(\Rightarrow - {{\rm{z}}^2} = \frac{{{\rm{dz}}}}{{{\rm{dx}}}}\)

\(\Rightarrow - {\rm{dx}} = \frac{{{\rm{dz}}}}{{{{\rm{z}}^2}}}\)

Integrating both sides, we get

\(\Rightarrow {\rm{x}} + {\rm{c}} = \frac{1}{{\rm{z}}}\)      

\({\rm{But\;z}} = \frac{{\rm{y}}}{{\phi \left( {\rm{x}} \right)}}\)     ...(From (1)

\(\therefore {\rm{x}} + {\rm{c}} = \frac{{\phi \left( {\rm{x}} \right)}}{{\rm{y}}}\)

\( \Rightarrow {\rm{y}} = \frac{{\phi \left( {\rm{x}} \right)}}{{{\rm{x}} + {\rm{c}}}}\)

Calculation

Given equation: \(x\frac{dy}{dx} = y + x \tan(\frac{y}{x})\)

Divide by x: \(\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})\)

Let y = vx, then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

Substitute in the equation:

\(v + x\frac{dv}{dx} = v + \tan(v)\)

⇒ \(x\frac{dv}{dx} = \tan(v)\)

⇒ \(\frac{dv}{\tan(v)} = \frac{dx}{x}\)

⇒ \(\cot(v) dv = \frac{dx}{x}\)

Integrate both sides:

\(\int \cot(v) dv = \int \frac{dx}{x}\)

⇒ \(\ln|\sin(v)| = \ln|x| + \ln|C|\)

⇒ \(\ln|\sin(v)| = \ln|Cx|\)

Remove the logarithms:

⇒ \(\sin(v) = Cx\)

Substitute v = y/x:

⇒ \(\sin(\frac{y}{x}) = Cx\)

∴ The general solution is \(\sin(\frac{y}{x}) = Cx\).

Hence option 2 is correct

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 \( x = sin \theta\))

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

Concept:

The solution of first order differential equation \(\frac{dy}{dx}+Py=Q\) is given by y × IF = \(\int Q\times IF \ dx\), where IF = Integrating Factor = \(e^{\int P \ dx}\)

Calculation:

Given, \(\rm (1+x^2)\frac{dy}{dx}+y=e^{\tan^{-1}x}\)

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

∴ I.F. = \(e^{\int \frac{1}{1+x^2} d x}=e^{\tan ^{-1} x}\)

∴ \(\rm y \cdot e^{\tan ^{-1} x}=\int\left(\frac{e^{\tan ^{-1} x}}{1+x^2}\right) e^{\tan ^{-1} x} \cdot \rm d x\)

Let tan^–1 x = z ⇒ \(\frac{\mathrm{dx}}{1+\mathrm{x}^2}=\mathrm{dz}\)

∴ ye^z = \(\int \mathrm{e}^{2 \mathrm{z}} \mathrm{dz}=\frac{\mathrm{e}^{2 \mathrm{z}}}{2}+\mathrm{C}\)

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

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

Now, y(1) = 0 

⇒ 0 = \(\frac{\mathrm{e}^{\pi / 4}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\pi / 4}}\)

⇒ C = \(\frac{-\mathrm{e}^{\pi / 2}}{2}\)

∴ \(\rm y=\frac{e^{\tan ^{-1} x}}{2}-\frac{e^{\pi / 2}}{2 e^{\tan ^{-1} x}}\)

⇒ y(0) = \(\frac{1-\mathrm{e}^{\pi / 2}}{2}\)

∴ The value of y(0) is \(\frac{1-\mathrm{e}^{\pi / 2}}{2}\).

The correct answer is Option 2.

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

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

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.

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: 

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

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 .