Differential Equations Test

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Differential Equations Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The solution of the differential equation $$x \, dx + y\, dy = x^2y\, dy - y^2\, x\, dx$$ is

A
$$x^2-1 = C(1+y^2)$$

B
$$x^2+1 = C(1-y^2)$$

C
$$x^3-1 = C(1+y^3)$$

D
$$x^3+1 = C(1-y^3)$$

Solution

$$x\, dx + y \, dy = x^2 y \, dy - y^2x \, dx$$
$$y\, dy (x^2-1) = x\, dx (1+y^2)$$
$$\displaystyle \int \dfrac{y}{1+y^2}dy - \int \dfrac{x}{x^2-1} dx$$
$$\Rightarrow x^2-1 = C (1+y^2)$$
Question 2
1 / -0

Solve differential equation $$dx=\dfrac{xdv}{v^2-a^2}$$.

A
$$x^{2a}=\dfrac{v-a}{v+a}\times c$$

B
$$x^{2a}=\dfrac{v+a}{v-a}\times c$$

C
$$x=\dfrac{v-a}{v+a}\times c$$

D
None of these

Solution
Question 3
1 / -0

Solution for $$\dfrac{x + y \dfrac{dy}{dx}}{y - x \dfrac{dy}{dx}} = x^2 + 2y^2 + \dfrac{y^4}{x^2}$$ is

A
$$\dfrac{y}{x} - \dfrac{1}{x^2 + y^2} = c$$

B
$$\dfrac{y}{x} + \dfrac{1}{x^2}{y^2} = c$$

C
$$\dfrac{x}{y} + \dfrac{1}{x^2 + y^2 } = c$$

D
$$\dfrac{x}{y} - \dfrac{1}{x^2 + y^2} = c$$

Solution
Question 4
1 / -0

The particular solution of the differential equation $$y(1+\log x)\dfrac{dx}{dy} -x =0$$ when $$x=e, y=e^2$$ is

A
$$y =ex \log x$$

B
$$ey = x \log x$$

C
$$xy = e \log x$$

D
$$y \log x = ex$$

Solution

Correct equation is,
$$y(1+logx)\dfrac{dx}{dy}-xlogx=0$$

$$\to \dfrac{dy}{y}=\dfrac{1+logx}{logx}dx$$$$~~~~~-(1)$$

Let $$xlogx=t$$
$$ (1+logx)dx=dt$$
$$(1) \to \dfrac{dy}{y}=\dfrac{dt}{t}$$
integrating Both the sides
$$ \to \int \dfrac{dy}{y}=\int \dfrac{dt}{t}$$
$$logy=logt+c$$
$$ at x=e and y=e^2$$
$$ loge^2=eloge+c$$
$$2-1=c$$
c=1

$$logy=logt+1$$
$$logy=logt+loge$$
$$logy=log(xlogx)+loge$$
$$y=xelogx$$
Question 5
1 / -0

Solution of differential equation $$xdy-yx=0$$ represents:

A
rectangular hyperbola

B
straight line passing through origin

C
parabola whose vertex is at origin

D
circle whose center is at origin

Solution

$$xdy-ydx=0\Rightarrow \dfrac{dy}{y}=\dfrac{dx}{x}$$
Integrating both sides
$$\ln y=\ln x\Rightarrow y=x$$
Solution of differential equation $$xdy-yx=0$$ reperesnts straight line passing through origin
Question 6
1 / -0

The solution of $$\dfrac{{dy}}{{dx}} = \dfrac{{x{{{\mathop{\rm log x}\nolimits} }^2} + x}}{{\sin y + y\cos y}}$$

A
$$y\sin y = {x^2}\log x + c$$

B
$$y\sin y = {x^2} + c$$

C
$$y\sin y = {x^2} + \log x + c$$

D
$$y\sin y = x \log x + c$$

Solution

$$\dfrac{dy}{dx}=\dfrac{2x\log{x}+x}{\sin{y}+y\cos{y}}$$
$$\Rightarrow\,\left(\sin{y}+y\cos{y}\right)dy=\left(2x\log{x}+x\right)dx$$
$$\Rightarrow\,\sin{y}dy+y\cos{y}dy=2x\log{x}dx+xdx$$
Integrating both sides, we get
$$\Rightarrow\,\displaystyle\int{\sin{y}dy}+\displaystyle\int{y\cos{y}dy}=2\displaystyle\int{x\log{x}dx}+\displaystyle\int{xdx}$$
$$\Rightarrow\,-\cos{y}+\displaystyle\int{y\cos{y}dy}=2\displaystyle\int{x\log{x}dx}+\dfrac{{x}^{2}}{2}+c$$
Consider $$\displaystyle\int{y\cos{y}dy}$$
Let $$u=y\Rightarrow\,du=dy$$
$$dv=\cos{y}dy\Rightarrow\,v=\sin{y}$$
Consider $$\displaystyle\int{x\log{x}dx}$$
Let $$u=\log{x}\Rightarrow\,du=\dfrac{1}{x}dx$$
$$dv=xdx\Rightarrow\,v=\dfrac{{x}^{2}}{2}$$
$$\Rightarrow\,-\cos{y}+y\sin{y}-\displaystyle\int{\sin{y}dy}=2\left[\dfrac{{x}^{2}\log{x}}{2}-\displaystyle\int{\dfrac{{x}^{2}}{2}\times \dfrac{1}{x}dx}\right]+\dfrac{{x}^{2}}{2}+c$$
$$\Rightarrow\,-\cos{y}+y\sin{y}+\cos{y}={x}^{2}\log{x}-\displaystyle\int{xdx}+\dfrac{{x}^{2}}{2}+c$$
$$\Rightarrow\,y\sin{y}={x}^{2}\log{x}+\dfrac{{x}^{2}}{2}-\dfrac{{x}^{2}}{2}+c$$
$$\Rightarrow\,y\sin{y}={x}^{2}\log{x}+c$$
Question 7
1 / -0

The solution of $$\cos \, y \, \log (\sec \, x + \tan \, x) dx = \cos \, x \, \log (\sec \, y + \tan \, y) dy$$ is

A
$$[\log ( \sec \, x + \tan \, x)]^2 - [\log (\sec \, y + \tan \, y)]^2 = c$$

B
$$[\log ( sec \, x + \tan \, x)]^2 + [\log (\sec \, y + \tan \, y)]^2 = c$$

C
$$[\log ( \sec \, x - \tan \, x)]^2 - [\log (\sec \, y - \tan \, y)]^2 = c$$

D
$$[\log ( \sec \, x - \tan \, x)]^2 + [\log (\sec \, y - \tan \, y)]^2 = c$$

Solution

Given,
$$cosylog(secx+tanx)dx=cosxlog(secy+tany)dy$$
or
$$\dfrac{log(secy+tany)dy}{cosy}=\dfrac{log(secx+tanx)dx}{cosx}$$

Integrate Both the sides
$$\int \dfrac{log(secy+tany)dy}{cosy}=\int \dfrac{log(secx+tanx)dx}{cosx}$$$$~~~~~~-(1)$$

Let u=$$log(sey+tany)$$
$$du=\dfrac{1}{cosy}dy$$

and $$log(secx+tanx)=v$$
$$\dfrac{1}{cosx}=dv$$

$$(1)\to \int udu=\int vdv$$
$$\to \dfrac{u^2}{2}=\dfrac{v^2}{2}+c$$
$$u^2=v^2+2c$$
$$[log(secy+tany)]^2-[log(secx+tanx)]^2=c$$
Question 8
1 / -0

The equation of the curve through $$\left(0, \dfrac{\pi}{4} \right)$$ satisfying the differential equation. $$e^x \, \tan \, y \, dx + (1 + e^x) \sec^2 \, ydy = 0$$ is given by

A
$$(1 + e^x ) \tan \, y = 2$$

B
$$1 + e^x = 2 \, \tan \, y$$

C
$$1 + e^x = 2 \, \sec \, y$$

D
$$(1 + e^x ) \tan \, y = 1$$

Solution

$$e^x \tan y dx+(1+e^x)\sec ^2y dy=0$$

$$(1+e^x)\sec ^2y dy=-e^x \tan y dx$$

$$\dfrac{\sec ^2y}{\tan y}dy=\dfrac{-e^x}{1+e^x}dx$$

substitute $$u=\tan y\rightarrow du=\sec ^2y dy $$ and $$v=1+e^x\rightarrow dv =e^x dx$$

$$\int \dfrac{1}{u}du=\int \dfrac{-1}{v}dv$$

$$\ln u = -\ln v$$

$$\ln (\tan y)= -\ln (1+e^x)$$

$$\ln (\tan y)+\ln (1+e^x)=0$$

$$\ln (1+e^x)\tan y =0$$

$$(1+e^x)\tan y =1$$
Question 9
1 / -0

The solution to the differential equation $$\cos\ xdy=y(\sin x-y)dx$$, $$0<x<\dfrac{\pi}{2}$$, is

A
$$\tan x=(\sec x+c)y$$

B
$$\sec x=(\tan x+c)y$$

C
$$y\sec x=\tan x+c$$

D
$$y\tan x=\sec x+c$$

Solution

$$cosx\, dy=y(sinx-y)dx$$ ___(1)
$$\dfrac{dy}{dx}=\dfrac{y(sinx-y)}{cos x}= ytanx-y^{2}secx$$
$$\dfrac{dy}{dx}-ytanx=-y^{2}secx$$ ___(2)
[Divide by $$y^2$$]
$$y^{-2}\dfrac{dy}{dx}-y^{-1}tanx= -secx$$ ___ (3)
Let $$+y^{-1}=z\Rightarrow -y^{-2}\dfrac{dy}{dx}=\dfrac{dz}{dx}$$
$$-\dfrac{dz}{dx}-ztanx= -secx$$ ___ (4) $$=\dfrac{dz}{dx}+ztanx=secx$$
$$I.F=e^{\displaystyle + \int tanx}= e^{+log\,secx}= secx$$
$$\displaystyle z\times I.F= \int secx*I.F\, dx$$
$$\displaystyle zsecx=\int sec^{2}x\, dx$$
$$\displaystyle \frac{1}{y}secx= tan x+c$$
$$secx=y(tan x+c)$$
Question 10
1 / -0

If $$\phi \left( x \right) =\phi \prime \left( x \right) $$ and $$\phi \left( 1 \right) =2$$, then $$\phi \left( 3 \right) $$ equals

A
$${ e }^{ 2 }$$

B
$$2{ e }^{ 2 }$$

C
$$3{ e }^{ 2 }$$

D
$$2{ e }^{ 3 }$$

Solution

$$\phi(x)=\phi(x)$$ and $$\phi(1)=2$$
Let $$y=\phi(x)$$
$$y=\dfrac{dy}{dx}$$
$$\displaystyle \int dx=\int \dfrac{dy}{y}$$
$$x+c=ln \,y\Rightarrow y=e^{x+c}$$
Given
$$\phi(1)=2\Rightarrow 2=e^{1+c}$$
$$\Rightarrow ln^2=1+c\Rightarrow c=ln^2-1$$
$$c=ln2-ln e$$ $$[\therefore ln_e\, e=1]$$
$$c=ln(\dfrac{2}{e})$$
$$\therefore y=e^{x+ln(\dfrac{2}{e})}$$
$$\phi(3)=?\,\,\,\,\,\,\,\,\,y=e^{3+\ln(\dfrac{2}{e})}=e^3.e^{ln(\dfrac{2}{e})}$$
$$=e^3\times \dfrac{2}{e}$$ $$[\because a^{log\,a^b}=b]$$
$$=\phi(3)=2e^2$$
options $$B$$

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Differential Equations Test - 37

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Differential Equations Test - 37

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SHARING IS CARING

Question 1

$$x^2-1 = C(1+y^2)$$

$$x^2+1 = C(1-y^2)$$

$$x^3-1 = C(1+y^3)$$

$$x^3+1 = C(1-y^3)$$

Solution

Question 2

$$x^{2a}=\dfrac{v-a}{v+a}\times c$$

$$x^{2a}=\dfrac{v+a}{v-a}\times c$$

$$x=\dfrac{v-a}{v+a}\times c$$

None of these

Solution

Question 3

$$\dfrac{y}{x} - \dfrac{1}{x^2 + y^2} = c$$

$$\dfrac{y}{x} + \dfrac{1}{x^2}{y^2} = c$$

$$\dfrac{x}{y} + \dfrac{1}{x^2 + y^2 } = c$$

$$\dfrac{x}{y} - \dfrac{1}{x^2 + y^2} = c$$

Solution

Question 4

$$y =ex \log x$$

$$ey = x \log x$$

$$xy = e \log x$$

$$y \log x = ex$$

Solution

Question 5

rectangular hyperbola

straight line passing through origin

parabola whose vertex is at origin

circle whose center is at origin

Solution

Question 6

$$y\sin y = {x^2}\log x + c$$

$$y\sin y = {x^2} + c$$

$$y\sin y = {x^2} + \log x + c$$

$$y\sin y = x \log x + c$$

Solution

Question 7

$$[\log ( \sec \, x + \tan \, x)]^2 - [\log (\sec \, y + \tan \, y)]^2 = c$$

$$[\log ( sec \, x + \tan \, x)]^2 + [\log (\sec \, y + \tan \, y)]^2 = c$$

$$[\log ( \sec \, x - \tan \, x)]^2 - [\log (\sec \, y - \tan \, y)]^2 = c$$

$$[\log ( \sec \, x - \tan \, x)]^2 + [\log (\sec \, y - \tan \, y)]^2 = c$$

Solution

Question 8

$$(1 + e^x ) \tan \, y = 2$$

$$1 + e^x = 2 \, \tan \, y$$

$$1 + e^x = 2 \, \sec \, y$$

$$(1 + e^x ) \tan \, y = 1$$

Solution

Question 9

$$\tan x=(\sec x+c)y$$

$$\sec x=(\tan x+c)y$$

$$y\sec x=\tan x+c$$

$$y\tan x=\sec x+c$$

Solution

Question 10

$${ e }^{ 2 }$$

$$2{ e }^{ 2 }$$

$$3{ e }^{ 2 }$$

$$2{ e }^{ 3 }$$

Solution

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