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

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. 

Calculation:

Diff w.r.t x

⇒ \(g(x) = 1 - x g(x)\)

⇒ \(g(x) = \frac{1}{1 + x}\)

\(\text{so } \frac{dy}{dx} - y \tan x = 2 \sec x g(x) = 2 \sec x \left(\frac{1}{1+x}\right)\)

⇒ \(\frac{dy}{dx} - y \tan x = \frac{2 \sec x}{1 + x}\)

\(\text{IF} = e^{\int -\tan x \, dx} = e^{\log |\cos x|} = |\cos x| = \cos x \quad \left(x \in \left[0, \frac{\pi}{2}\right)\right) \)

Solution of D.E

⇒ \(y \cos x = \int \frac{2 \sec x}{1 + x} \cdot \cos x \, dx + c\)

⇒ \(y \cos x = \int \frac{2}{1 + x} \, dx + c\)

⇒ \(y \cos x = 2 \log |1 + x| + c\)

⇒ \(y(0) = 0 \implies 0 \cdot \cos 0 = 2 \log |1 + 0| + c\)

⇒ \(0 = 2 \log 1 + c \implies 0 = 0 + c \implies c = 0\)

⇒ \(y \cos x = 2 \log (1 + x)\)

⇒ \(y = \frac{2 \log(1 + x)}{\cos x} = 2 \sec x \log(1 + x)\)

⇒ \(y\left(\frac{\pi}{3}\right) = 2 \sec \left(\frac{\pi}{3}\right) \log\left(1 + \frac{\pi}{3}\right)\)

⇒ \(y\left(\frac{\pi}{3}\right) = 2 \cdot 2 \cdot \log \left(\frac{3 + \pi}{3}\right) = 4 \log \left(\frac{3 + \pi}{3}\right)\)

Hence, the Correct answer is Option 2

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

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.

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. 

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

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

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