Differential Equations Test

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Differential Equations Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The solution of the differential equation $${x}^{2}dy=-2xydx$$ is

A
$$x{y}^{2}=c$$

B
$${x}^{2}{y}^{2}=c$$

C
$${x}^{2}y=c$$

D
$$xy=c$$

Solution

$$x^{2}dy=-2xydx$$
$$xdy=-2ydx$$
$$xdy+2ydx=0$$
$$\displaystyle\int d(xy^{2}) =\int 0$$
$$xy^{2}=C$$
Question 2
1 / -0

If $$f:R \rightarrow R$$ be a continuous function such that $$f(x)=\displaystyle \int^{x}_{1}2tf(t)dt$$, then which of the following does not hold(s) good?

A
$$f(\pi)=e^{\pi^{2}}$$

B
$$f(1)=e$$

C
$$f(0)=1$$

D
$$f(2)=-2$$
Question 3
1 / -0

Solution of the differential equation $$xdy-ydx=\sqrt {x^{2}+y^{2}}dx$$ is

A
$$y+\sqrt {x^{2}+y^{2}}=cx$$

B
$$y+\sqrt {x^{2}+y^{2}}=cx^{2}$$

C
$$y+\sqrt {x^{2}+y^{2}}=C$$

D
$$none\ of\ these$$

Solution

$$xdy-ydx=\sqrt{x^2+y^2}+x$$
$$\dfrac{dy}{dx}=\dfrac{\sqrt{x^2-y^2}+y}{x}$$
$$\Rightarrow F(x,y)=\dfrac{\sqrt{x^2+y^2}+y}{x}$$
$$F(kx\cdot ky)=\dfrac{\sqrt{k^2x^2+k^2y^2+ky}}{kx}=k^o\cdot F(x, y)$$
$$y=vx$$ and $$\dfrac{dy}{x}=v+x\dfrac{dy}{dx}$$
$$v+x\dfrac{dv}{dx}=\dfrac{\sqrt{x^2+v^2x^2}+vx}{x}$$
$$v+x\dfrac{dv}{dx}=\sqrt{1+v^2}+v$$
$$\dfrac{xdv}{dx}=\sqrt{1+v^2}$$
$$\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{k}$$
$$log|v+\sqrt{1-v^2})l=log x+log e$$
$$v=\dfrac{y}{x}$$
$$log\left(\dfrac{y}{x}\right)+\sqrt{1+y^2/x^2}=log cx$$
$$log \dfrac{y+\sqrt{x^2+y^2}}{x}=log cx$$
$$log(y+\sqrt{x^2+y^2})=log cx^2$$
$$y+\sqrt{x^2+y^2}=cx^2$$.
Question 4
1 / -0

Solution of the differential equation $$\dfrac{dy}{dx}+y\sec x=\tan x\left(\le x< \dfrac{\pi}{2}\right)$$ is

A
$$y\left( \sec { x } -\tan { x } \right) =\left( \sec { x } +\tan { x } \right) -x+C$$

B
$$y\left( \sec { x } +\tan { x } \right) =\left( \sec { x } -\tan { x } \right) -x+C$$

C
$$y\left( \sec { x } +\tan { x } \right) =\left( \sec { x } +\tan { x } \right) -x+C$$

D
$$none of these $$

Solution

$$\dfrac{dy}{dx}+y\sec x=\tan x$$
$$\dfrac{dy}{dx}+Py=Q$$
If $$=\displaystyle\int Pdx=e^{\log|\sec x+\tan x|}$$
$$=\sec x+\tan x$$
$$Y(IF)=\displaystyle\int(Q\times IF) dx+C$$
$$Y(\sec x+\tan x)=\displaystyle\int \tan x(\sec x+\tan x)+C$$
$$Y(\sec x+\tan x)=\displaystyle\int \tan x\sec x+\int \tan^{2}x dx$$
$$Y(\sec x+\tan x)=\sec x+\displaystyle\int\sec^{2}-1dx+C$$
$$Y(\sec x+\tan x)=\sec x+\tan x-x+C$$
Question 5
1 / -0

The solution of the differential equation $$\cfrac{dy}{dx}=1+x+y+xy$$ is

A
$$\log{(1+y)}=x+\cfrac{{x}^{2}}{2}+c$$

B
$${(1+y)}^{2}=x+\cfrac{{x}^{2}}{2}+c$$

C
$$\log{(1+y)}=\log{(1+x)}+c$$

D
None of these

Solution

$$\dfrac{dy}{dx}= 1 + x + y + xy = 1 + x + y (1+x)$$
$$\dfrac{dy}{dx}=(1+x)(1+y)$$
$$\dfrac{dy}{1+x}= \displaystyle\int (1+x)dx$$
$$ln(1+y)=x+\dfrac{x^{2}}{2}+C$$
Question 6
1 / -0

A continuously differentiable function $$\phi(x)$$ in $$(0,\pi)$$ satisfying $$y'=1+{y}^{2},y(0)=0=y(\pi)$$ is

A
$$\tan{x}$$

B
$$x(x-\pi)$$

C
$$(x-\pi)(1-{e}^{x})$$

D
Not possible

Solution

$$\dfrac{dy}{dx}=1+y^{2}$$
$$\displaystyle \int \frac{dy}{1+y^{2}}=\int dx$$
$$\tan^{-1}y=x+c$$
at $$x=0;y=0$$
$$c=0$$
$$y=\tan x $$
Question 7
1 / -0

Solution of the differential equation $$(1+x^{2})dy+2xy dx=\cot x dx$$ is

A
$$y=\log|\sin x|(1+x^{2})+C(1+x^{2})^{-1}$$

B
$$y=\log|\sin x|(1+x^{2})^{-1}+C(1+x^{2})$$

C
$$y=\log|\sin x|(1+x^{2})+C(1+x)$$

D
$$y=\log|\sin x|(1+x^{2})^{-1}+C(1+x^{2})^{-1}$$

Solution

$$(1+x^2)dy+2xydx=\cot xdx$$
Put in form $$\dfrac{dy}{dx}+PY=Q$$
$$(1+x)^2dy+2xydx=\cot xdx$$
dividing both sides by $$(1+x^2)$$
$$\dfrac{dy}{dx}+\dfrac{2xy}{1x^2}\dfrac{dx}{dx}=\dfrac{\cot x}{(1+x^2)}\dfrac{dx}{dx}$$
$$\dfrac{dy}{dx}+\left(\dfrac{2x}{1+x^2}\right)y=\dfrac{\cot x}{1+x^2}$$
Find P and Q
$$\dfrac{dy}{dx}+PY=Q$$
$$P=\dfrac{2x}{1+x^2}$$
$$Q=\dfrac{\cot x}{1+x^2}$$
If$$=e^{\displaystyle\int \dfrac{2x}{1+x^2}}$$ $$t=1+x^2$$
If $$=e^{\displaystyle\int \dfrac{dt}{t}}=e^{log t}=t=1+x^2$$
Solution: $$Y\times IF=\displaystyle\int Q\times IFdx+C$$
$$Y(1+x^2)=\displaystyle\int \dfrac{\cot x}{1+x^2}\times (1+x^2)dx+c$$
$$Y(1+x^2)=\displaystyle\int \cot xdx+c$$
$$Y(1+x^2)=log|\sin x|+c$$
$$Y=(1+x^2)^{-1}log|\sin x|+c(1+x^2)^{-1}$$.
Question 8
1 / -0

The solution of the differential equation $$x^3 \dfrac{dy}{dx} + 4x^2 \tan y = e^x \sec y$$ satisfying $$y(1) = 0$$ is

A
$$\tan \, y = (x - 2) e^x \log \, x$$

B
$$\sin \, y = e^x (x - 1) x^{-4}$$

C
$$\tan \, y = (x - 1) e^x x^{-3}$$

D
$$\sin \, y = e^x (x - 1) x^{-3}$$

Solution

Given,
$$x^3\dfrac{dy}{dx}+4x^2tany=e^xsecy$$

Dividing whole equation by $$x^3secy$$
$$cosy\dfrac{dy}{dx}+\dfrac{4}{x}siny=\dfrac{e^x}{x^3}$$
let siny=t
$$cosydy=dt$$
Put in above equation
$$\dfrac{dt}{dx}+\dfrac{4}{x}t=\dfrac{e^x}{x^3}$$
P=$$\dfrac{4}{x},Q=\dfrac{e^x}{x^3}$$

I.f.=$$e^{\int P.dx}$$
=$$e^{\int \dfrac{4}{x}dx}$$

=$$e^{4logx}$$
=$$e^{logx^4}$$
=$$x^4$$
Complete solution,
$$t\times x^4=\int x^4\times \dfrac{e^x}{x^3}dx+c$$
$$t\times x^4=\int xe^xdx+c$$
$$t\times x^4=xe^x-e^x+c$$
$$siny\times x^4=xe^x-e^x+c$$
at x=1,y=0
$$0=1.e-e+c$$
c=0
$$siny \times x^4=e^x(x-1)+0$$
$$siny=x^{-4}e^x(x-1)$$
Question 9
1 / -0

The solution of $$\cfrac{dy}{dx}=\cfrac{1+{y}^{2}}{\sec{x}}$$ is

A
$$\tan ^{ -1 }{ y } =\cos { x } +c$$

B
$$\tan ^{ -1 }{ y } =\sin { x } +c$$

C
$${ y }^{ 3 }=co\sec { x } +c$$

D
$$\log { \left( 1+{ y }^{ 2 } \right) } =\cos { x } +c$$

Solution

$$\dfrac{dy}{dx} = \dfrac{1+y^{2}}{\sec x}$$
$$\dfrac{dy}{1+y^{2}} = \dfrac{1+y^{2}}{\sec x}$$
$$\displaystyle\int \dfrac{dy}{1+y^{2}} = \displaystyle\int \cos x dx$$
$$\tan^{-1}y= \sin x+c$$
Question 10
1 / -0

The solution of $$ydx-xdy=0$$ is

A
$${y}^{2}=cx$$

B
$$y=c{x}^{3}$$

C
$$y=cx$$

D
$${x}^{2}=cy$$

Solution

$$y\ dx=x\ dy=0$$
$$y\ dx=x\ dy$$
$$\displaystyle \int \dfrac {dx}{x} = \displaystyle \int \dfrac {dy}{y}$$
$$\ln x=\ln y+\ln c$$
$$\ln \left (\dfrac {x}{y}\right)=\ln c\ \Rightarrow \dfrac {x}{y}=c$$
$$y=\dfrac {x}{c}=xc'$$
$$y=xc'$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 39

Result

Differential Equations Test - 39

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SHARING IS CARING

Question 1

$$x{y}^{2}=c$$

$${x}^{2}{y}^{2}=c$$

$${x}^{2}y=c$$

$$xy=c$$

Solution

Question 2

$$f(\pi)=e^{\pi^{2}}$$

$$f(1)=e$$

$$f(0)=1$$

$$f(2)=-2$$

Question 3

$$y+\sqrt {x^{2}+y^{2}}=cx$$

$$y+\sqrt {x^{2}+y^{2}}=cx^{2}$$

$$y+\sqrt {x^{2}+y^{2}}=C$$

$$none\ of\ these$$

Solution

Question 4

$$y\left( \sec { x } -\tan { x } \right) =\left( \sec { x } +\tan { x } \right) -x+C$$

$$y\left( \sec { x } +\tan { x } \right) =\left( \sec { x } -\tan { x } \right) -x+C$$

$$y\left( \sec { x } +\tan { x } \right) =\left( \sec { x } +\tan { x } \right) -x+C$$

$$none of these $$

Solution

Question 5

$$\log{(1+y)}=x+\cfrac{{x}^{2}}{2}+c$$

$${(1+y)}^{2}=x+\cfrac{{x}^{2}}{2}+c$$

$$\log{(1+y)}=\log{(1+x)}+c$$

None of these

Solution

Question 6

$$\tan{x}$$

$$x(x-\pi)$$

$$(x-\pi)(1-{e}^{x})$$

Not possible

Solution

Question 7

$$y=\log|\sin x|(1+x^{2})+C(1+x^{2})^{-1}$$

$$y=\log|\sin x|(1+x^{2})^{-1}+C(1+x^{2})$$

$$y=\log|\sin x|(1+x^{2})+C(1+x)$$

$$y=\log|\sin x|(1+x^{2})^{-1}+C(1+x^{2})^{-1}$$

Solution

Question 8

$$\tan \, y = (x - 2) e^x \log \, x$$

$$\sin \, y = e^x (x - 1) x^{-4}$$

$$\tan \, y = (x - 1) e^x x^{-3}$$

$$\sin \, y = e^x (x - 1) x^{-3}$$

Solution

Question 9

$$\tan ^{ -1 }{ y } =\cos { x } +c$$

$$\tan ^{ -1 }{ y } =\sin { x } +c$$

$${ y }^{ 3 }=co\sec { x } +c$$

$$\log { \left( 1+{ y }^{ 2 } \right) } =\cos { x } +c$$

Solution

Question 10

$${y}^{2}=cx$$

$$y=c{x}^{3}$$

$$y=cx$$

$${x}^{2}=cy$$

Solution

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