Differential Equations MCQ Quiz - Objective Question with Answer for Differential Equations - Download Free PDF
Last updated on Apr 22, 2025
Latest Differential Equations MCQ Objective Questions
Differential Equations Question 1:
What is the degree of the differential equation \(\rm \frac{d^2y}{dx^2}+a\sin x = 0\)
Answer (Detailed Solution Below)
Differential Equations Question 1 Detailed Solution
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.
Differential Equations Question 2:
The solution of the differential equation \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}}\) is
Answer (Detailed Solution Below)
Differential Equations Question 2 Detailed Solution
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)
\(\therefore {\rm{x}} + {\rm{c}} = \frac{{\phi \left( {\rm{x}} \right)}}{{\rm{y}}}\)
\( \Rightarrow {\rm{y}} = \frac{{\phi \left( {\rm{x}} \right)}}{{{\rm{x}} + {\rm{c}}}}\)
Hence, option (4) is correct.Differential Equations Question 3:
The general solution of the differential equation \(x \frac{d y}{d x}=y+x \tan \left(\frac{y}{x}\right)\) is
Answer (Detailed Solution Below)
Differential Equations Question 3 Detailed Solution
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
Differential Equations Question 4:
The function \(y = f(x)\) is the solution of the differential equation \(\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}\) in \((-1, 1)\) satisfying \(f(0) = 0\). Then \(\int_{- \frac{\sqrt{3}}{2}}^{ \frac{\sqrt{3}}{2}} f(x)dx\) is
Answer (Detailed Solution Below)
Differential Equations Question 4 Detailed Solution
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 \( x = sin \theta\))
\(= 2 \int_{0}^{\pi / 3} sin^2 \theta d \theta = 2 \left [ \frac{\theta}{2} - \frac{sin 2 \theta}{4} \right]_0^{\frac{\pi}{3}} = 2 \left ( \frac{\pi}{6} \right ) - 2 \left ( \frac{\sqrt{3}}{8} \right ) = \frac{\pi}{3} - \frac{\sqrt{3}}{4}\)
Differential Equations Question 5:
Let y = y(x) be the solution of the differential equation \(\rm (1+x^2)\frac{dy}{dx}+y=e^{\tan^{-1}x}, y(1)=0\) Then y(0) is
Answer (Detailed Solution Below)
Differential Equations Question 5 Detailed Solution
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 ⇒ \(\frac{\mathrm{dx}}{1+\mathrm{x}^2}=\mathrm{dz}\)
∴ yez = \(\int \mathrm{e}^{2 \mathrm{z}} \mathrm{dz}=\frac{\mathrm{e}^{2 \mathrm{z}}}{2}+\mathrm{C}\)
⇒ \(\rm y . e^{\tan ^{-1} x}=\frac{e^{2 \tan ^{-1} x}}{2}+C\)
⇒ \(\rm y=\frac{e^{\tan ^{-1} x}}{2}+\frac{C}{e^{\tan ^{-1} x}}\)
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.
Top Differential Equations MCQ Objective Questions
What is the degree of the differential equation \({\rm{y}} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} + {\left( {\frac{{{\rm{dx}}}}{{{\rm{dy}}}}} \right)} \)?
Answer (Detailed Solution Below)
Differential Equations Question 6 Detailed Solution
Download Solution PDFConcept:
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.
The order and degree of the differential equation \(\rm \frac{d^3y}{dx^3} + \cos\left(\frac{d^2y}{dx^2}\right) = 0\) are respectively
Answer (Detailed Solution Below)
Differential Equations Question 7 Detailed Solution
Download Solution PDFConcept:
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.
The solution of the differential equation dy = (1 + y2) dx is
Answer (Detailed Solution Below)
Differential Equations Question 8 Detailed Solution
Download Solution PDFConcept:
\(\rm \displaystyle \int \frac{dx}{1+x^2} = \tan^{-1} x + c\)
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).
If x2 + y2 + z2 = xy + yz + zx and x = 1, then find the value of \(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\)
Answer (Detailed Solution Below)
Differential Equations Question 9 Detailed Solution
Download Solution PDFGiven:
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.
What is the solution of the differential equation \(\ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) - {\rm{a}} = 0?\)
Answer (Detailed Solution Below)
Differential Equations Question 10 Detailed Solution
Download Solution PDFCalculation:
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 = xea + c
What is the degree of the differential equation \(\rm y = x \dfrac{dy}{dx}+\left(\dfrac{dy}{dx}\right)^{-2} \ ?\)
Answer (Detailed Solution Below)
Differential Equations Question 11 Detailed Solution
Download Solution PDFConcept:
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.
Find general solution of \(\rm\left( xy \frac {dy}{dx} -1 \right)= 0\)
Answer (Detailed Solution Below)
Differential Equations Question 12 Detailed Solution
Download Solution PDFConcept:
\(\rm \int \frac{1}{x}dx = \log x + c\)
\(\rm \int x^ndx = \frac{x^{n+1}}{n+1} + c\)
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
\(\rm \Rightarrow \frac{y^2}{2} = \log x + c\)
The degree of the differential equation
\(\dfrac{d^2y}{dx^2}+3\left(\dfrac{dy}{dx}\right)^2 =x^2 \log \left(\dfrac{d^2y}{dx^2}\right)\)
Answer (Detailed Solution Below)
Differential Equations Question 13 Detailed Solution
Download Solution PDFConcept:
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.
If x + \(\frac{1}{2x}\) = 3, then evaluate 8x3 + \(\rm \frac{1}{x^3}\).
Answer (Detailed Solution Below)
Differential Equations Question 14 Detailed Solution
Download Solution PDFGiven:
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
The solution of differential equation \(\rm dy = \left ( 4 + y^{2} \right )dx\) is
Answer (Detailed Solution Below)
Differential Equations Question 15 Detailed Solution
Download Solution PDFConcept:
\(\rm \int \frac{1}{a^{2}+x^{2}}dx = \frac{1}{a}\tan ^{-1}\frac{x}{a}+ C\)
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 .