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

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.

Calculation

\( \cfrac { dx }{ dy } =\cos { \left( x+y \right) } \)

Let \( x+y=v \implies \dfrac{dx}{dy}+1=\dfrac{dv}{dy} \)

⇒ \( \dfrac{dv}{dy}=1+\cos{v}=2\cos^2{\dfrac{v}{2}} \)

\(\Rightarrow \dfrac{dv}{\cos^2{\dfrac{v}{2}}}=2dy \)

⇒ \( \sec^2{\dfrac{v}{2}}dv=2dy \)

Integrating both sides:-

⇒ \( 2\tan{\dfrac{v}{2}}=2y+k \)

⇒ \( \tan{\left(\dfrac{x+y}{2}\right)}=y+c \)

Hence, option 1 is correct

Calculation

\(\dfrac {dy}{dx} = \tan \left (\dfrac {y}{x}\right ) + \left (\dfrac {y}{x}\right )\) ..... \((i)\)

Take, \(\dfrac {y}{x} = v\)

\(\implies y = vx\)

\(\implies \dfrac {dy}{dx} = v + x\dfrac {dv}{dx}\)

\(\therefore\) The given equation \((i)\) becomes

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

\(\implies \dfrac {1}{\tan v}dv = \dfrac {1}{x}dx\)

\(\implies \displaystyle \int \cot v\ dv = \int \dfrac {1}{x}dx\)

\(\implies \log |\sin v| = \log x + \log c=\log|xc|\)

\(\implies \sin v = xc\)

\(\therefore \sin \left (\dfrac {y}{x}\right ) = xc\)

Hence option 2 is correct

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:

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

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

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: 

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

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 . 

Concept:

Some useful formulas are:

\(\rm \int{dx\over ax}={1\over a}logx+c\)

If log x = z then we can write x = e^z

Calculation:

\(\rm {dx\over dt}= 3x+8\)

Rearranging the equation and integrating we get,

⇒\(\rm \int{dx\over 3x+8}=\int dt\)

⇒\(\rm {1\over 3 }log({3x+8})=t+c\), c = constant of integration

⇒ log(3x + 8) = 3(t + c)

⇒ 3x + 8 = e^3(t+c) 

⇒ 3x = e^3(t+c) - 8

∴ \(\rm x ={ 1\over 3}e^{3(t+c)}-{8\over 3}\)

Concept:

\(\rm \int \frac{dx}{\sqrt{a^{2}-x^{2}}}= sin^{-1}\frac{x}{a}\) 

Calculation:

Given: dy = \(\rm \sqrt{1-y^2}\) dx 

⇒ \(\rm \frac{dy}{\sqrt{1^{2}-y^{2}}} = dx\) 

Integrating both sides, we get

⇒ \(\rm \int \frac{dy}{\sqrt{1^{2}-y^{2}}} =\int dx\) 

⇒ \(\rm sin^{-1}\left ( y \right )\) = x + c 

⇒ y = sin ( x + c ) . 

The correct option is 2.

Concept:

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

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

Calculation:

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

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

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

Integrating both sides, we get

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

\(\rm \Rightarrow y+\frac{y^3}{3} = \tan^{-1}x + c\)

Calculation:

Given: ​\(\rm y \frac {dy}{dx} = x + 1\)

⇒ ydy = (x + 1) dx

Integrating both sides, we get

⇒ ∫ ydy = ∫ (x + 1) dx

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

⇒ y² = x² + 2x + 2c

∴ y2 - x2 - 2x - c = 0

Please note: c is constant here, so 2c can be also considered as a constant.