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

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:

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:

$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}+\mathrm{a}\sin \mathrm{x}=0$

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: $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{y}{\varphi }^{\prime }\left(\mathrm{x}\right)-{\mathrm{y}}^2}{\varphi \left(\mathrm{x}\right)}$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{y}{\varphi }^{\prime }\left(\mathrm{x}\right)-{\mathrm{y}}^2}{\varphi \left(\mathrm{x}\right)}=\frac{\mathrm{y}{\varphi }^{\prime }\left(\mathrm{x}\right)}{\varphi \left(\mathrm{x}\right)}-\frac{\mathrm{y}.\mathrm{y}}{\varphi \left(\mathrm{x}\right)}$

$\mathrm{L}\mathrm{e}\mathrm{t},\frac{\mathrm{y}}{\varphi \left(\mathrm{x}\right)}=\mathrm{z}$     ...1)

$\therefore \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{y}{\varphi }^{\prime }\left(\mathrm{x}\right)}{\varphi \left(\mathrm{x}\right)}-\frac{\mathrm{y}.\mathrm{y}}{\varphi \left(\mathrm{x}\right)}={\varphi }^{\prime }\left(\mathrm{x}\right)\mathrm{z}-\varphi \left(\mathrm{x}\right){\mathrm{z}}^2$

Now, $\mathrm{y}=\varphi \left(\mathrm{x}\right)\mathrm{z}$ 

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\varphi \left(\mathrm{x}\right)\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}+{\varphi }^{\prime }\left(\mathrm{x}\right)\mathrm{z}$

$\therefore {\varphi }^{\prime }\left(\mathrm{x}\right)\mathrm{z}-\varphi \left(\mathrm{x}\right){\mathrm{z}}^2=\varphi \left(\mathrm{x}\right)\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}+{\varphi }^{\prime }\left(\mathrm{x}\right)\mathrm{z}$

$\Rightarrow -{\mathrm{z}}^2=\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}$

$\Rightarrow -\mathrm{d}\mathrm{x}=\frac{\mathrm{d}\mathrm{z}}{{\mathrm{z}}^2}$

Integrating both sides, we get

$\Rightarrow \mathrm{x}+\mathrm{c}=\frac{1}{\mathrm{z}}$      

$\mathrm{B}\mathrm{u}\mathrm{t}\mathrm{z}=\frac{\mathrm{y}}{\varphi \left(\mathrm{x}\right)}$     ...(From (1)

$\therefore \mathrm{x}+\mathrm{c}=\frac{\varphi \left(\mathrm{x}\right)}{\mathrm{y}}$

$\Rightarrow \mathrm{y}=\frac{\varphi \left(\mathrm{x}\right)}{\mathrm{x}+\mathrm{c}}$

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. $=e^{\int {\displaystyle \frac{x}{x^2-1}}dx}=e^{{\displaystyle \frac{1}{2}}ln|x^2-1|}=\sqrt{1-x^2}$

$\Rightarrow$ solution is

$y\sqrt{1-x^2}=\int {\displaystyle \frac{x(x^3+2)}{\sqrt{1-x^2}}}\cdot \sqrt{1-x^2}dx$

or $y\sqrt{1-x^2}=\int (x^4+2x)dx=\sqrt{x^5}5+x^2+c$

$f(0)=0\Rightarrow c=0$

$\Rightarrow f(x)\sqrt{1-x^2}={\displaystyle \frac{x^5}{5}}+x^2$

Now,

${\int }_{-\sqrt{3}/2}^{\sqrt{3}/2}f(x)dx={\int }_{\sqrt{3}/2}^{\sqrt{3}/2}d{\displaystyle \frac{x^2}{\sqrt{1-x^2}}}dx$ (Using property)

$=2{\int }_0^{\sqrt{3}/2}{\displaystyle \frac{x^2}{\sqrt{1-x^2}}}dx=2{\int }_0^{\pi /3}{\displaystyle \frac{si{n}^2\theta }{cos\theta }}cos\theta d\theta$ (Taking $x=sin\theta$)

$=2{\int }_0^{\pi /3}si{n}^2\theta d\theta =2{[\frac{\theta }{2}-\frac{sin2\theta }{4}]}_0^{\frac{\pi }{3}}=2\left(\frac{\pi }{6}\right)-2\left(\frac{\sqrt{3}}{8}\right)=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$

Concept:

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 .