Differential Equations Test

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Differential Equations Test - 49

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Question 1
1 / -0

Solution of $$\displaystyle \frac { dy }{ dx } =\frac { y\left( x\log { y } -y \right) }{ x\left( y\log { x } -x \right) } $$ is:

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

B
$$xy=c$$

C
$${ \left( xy \right) }^{ x }=c$$

D
None of these

Solution

$$\displaystyle \frac { dy }{ dx } =\frac { y\left( x\log { y-y } \right) }{ x\left( y\log { x-x } \right) } $$
$$\displaystyle \Rightarrow x\left( y\log { x-x } \right) \frac { dy }{ dx } =y\left( x\log { y-y } \right) $$
$$\displaystyle \Rightarrow \left( \log { x } -\frac { x }{ y } \right) \frac { dy }{ dx } =\log { y } -\frac { y }{ x } $$
$$\displaystyle \Rightarrow \frac { y }{ x } +\log { x } .\frac { dy }{ dx } =\log { y } +\frac { x }{ y } \frac { dy }{ dx } $$
$$\displaystyle \Rightarrow \frac { d }{ dx } \left( x\log { x } \right) =\frac { d }{ dx } \left( x\log { y } \right) $$
$$\Rightarrow y\log { y } =x\log { y } +\log { c } \Rightarrow \log { { x }^{ y } } =\log { { y }^{ x } } +\log { c } \Rightarrow { x }^{ y }=c{ y }^{ x }.$$
Question 2
1 / -0

Through any point $$\left ( x,y \right )$$ of a curve which passes through the origin, lines are drawn parallel to the coordiante axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of

A
circles

B
parabolas

C
hyperbolas

D
straight lines

Solution

Let $$P\left ( x,y \right )$$ be the point on the curve passing through the origin $$O\left ( 0,0 \right )$$,
and let $$PN$$ and $$PM$$ be the lines parallel to the $$x$$- and $$y$$-axes, respectively. If the equation of the curve is $$y= y\left ( x \right )$$,
the area $$POM$$ equals $$\displaystyle \int_{0}^{x}ydx$$ and the area $$PON$$ equals $$xy-\displaystyle \int_{0}^{x}ydx$$.
Assuming that $$2\left ( POM \right )= PON$$, we therefore have $$2\displaystyle \int_{0}^{x}ydx= xy-\displaystyle \int_{0}^{x}ydx\Rightarrow 3\displaystyle \int_{0}^{x}ydx= xy$$.
Differentiating both sides of this gives
$$3y= x\displaystyle \frac{dy}{dx}+y\Rightarrow 2y= x\displaystyle \frac{dy}{dx}\Rightarrow \displaystyle \frac{dy}{y}=2\displaystyle \frac{dx}{x}$$
$$\Rightarrow \log \left | y \right |= 2\log \left | x \right |+C\Rightarrow y= Cx^{2}$$ with $$C$$ being a constant.
This solution represents a parabola.
Question 3
1 / -0

$$\displaystyle y-x\frac{dy}{dx}=b\left ( 1+x^{2}\frac{dy}{dx} \right ).$$
Solve the above differential equation.

A
$$\displaystyle y=k\left ( y-b \right )\left ( 1+bx \right ).$$

B
$$\displaystyle y=k\left ( y+b \right )\left ( 1-bx \right ).$$

C
$$\displaystyle y=k\left ( y+b \right )\left ( 1+bx \right ).$$

D
$$\displaystyle y=k\left ( y-b \right )\left ( 1-bx \right ).$$

Solution

After rearranging the terms, we get
$$y-b=\dfrac{dy}{dx}(bx^{2}+x)$$
$$\dfrac{dx}{x(bx+1)}=\dfrac{dy}{y-b}$$
$$\int \dfrac{bx+1-bx}{x(bx+1)}.dx=ln(y-b)$$
$$\int \dfrac{1}{x}-\dfrac{b}{bx+1}.dx=ln(y-b)$$
$$ln(x)-ln(bx+1)+ln(c)=ln(y-b)$$
$$\dfrac{cx}{bx+1}=y-b$$
Question 4
1 / -0

The differential equation $$\displaystyle \frac{dy}{dx}= \displaystyle \frac{\sqrt{1-y^{2}}}{y}$$ determines a family of circles with:

A
variable radii and a fixed centre $$\left ( 0,1 \right )$$

B
variable radii and a fixed centre $$\left ( 0,-1 \right )$$

C
fixed radius $$1$$ and variable centres along the $$x$$-axis

D
fixed radius $$1$$ and variable centres along the $$y$$-axis

Solution

$$\displaystyle \frac{dy}{dx}= \displaystyle \frac{\sqrt{1-y^{2}}}{y}$$
$$\Rightarrow \displaystyle \frac{y}{\sqrt{1-y^{2}}}\: dy= dx$$
Integrating we have $$-\sqrt{1-y^{2}}= x+C$$
$$\Rightarrow \left ( x+C \right )^{2}+y^{2}= 1$$.
This represents a family of circles with variable centre
$$\left ( -C,0 \right )$$ (which lies on $$x$$-axis) and radius $$1$$
Question 5
1 / -0

The general solution of the equation $$\displaystyle \frac { dy }{ dx } =\frac { { x }^{ 2 } }{ { y }^{ 2 } } $$ is:

A
$$x^3-y^3=c$$

B
$$x^3+y^3=c$$

C
$$x^2+y^2=c$$

D
$$x^2-y^2=c$$

Solution

$$\displaystyle \frac { dy }{ dx } =\frac { { x }^{ 2 } }{ { y }^{ 2 } } \Rightarrow { y }^{ 2 }dy={ x }^{ 2 }dx$$
Integrating both sides
$$\displaystyle \int { { y }^{ 2 }dy } =\int { { x }^{ 2 }dx } \Rightarrow \frac { { y }^{ 3 } }{ 3 } =\frac { { x }^{ 3 } }{ 3 } +\frac { c }{ 3 } $$
$$\Rightarrow { x }^{ 3 }-{ y }^{ 3 }=c$$
Question 6
1 / -0

Find the solution of $$\displaystyle \left ( 3\tan x+4\cot y-7 \right )\sin ^{2}ydx$$ $$\displaystyle -\left ( 4\tan x+7\cot y-5 \right )\cos ^{2}xdy=0$$.

A
$$\displaystyle 3\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

B
$$\displaystyle 8\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

C
$$\displaystyle 3\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

D
$$\displaystyle 8\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

Solution

Given $$\displaystyle \left ( 3\tan x+4\cot y-7 \right )\sin ^{2}y dx -\left ( 4\tan x+7\cot y-5 \right )\cos ^{2}x dy=0$$

Dividing throughout by $$\displaystyle \sin ^{2}y\cos ^{2}x,$$ we get

$$\displaystyle 3\tan x\sec ^{2}x dx-7\sec ^{2}dx-7\cot y cosec^{2}y
dy+5cosec^{2}y dy+4 \left ( \cot y\sec ^{2}x dx-\tan x cosec^{2}y \right )dy=0$$

Integrating on both sides

$$3\int { \tan x\sec ^{ 2 } xdx } -7\int { \sec ^{ 2 } dx } -7\int { \cot ycosec^{ 2 }ydy } +5\int { cosec^{ 2 }ydy } $$
$$+4\int { \left( \cot y\sec ^{ 2 } xdx-\tan xcosec^{ 2 }y \right) dy } =0$$

Put $$\tan x=u \Rightarrow \sec^2 xdx=du$$ for first integral

and $$\cot y=v \Rightarrow -cosec^2ydy=dv$$ for third integral

$$\therefore 3\int { udu } -7\tan { x } -7\int { vdv } -5\cot { y } +4\int { \left( \cot y\sec ^{ 2 } xdx-\tan xcosec^{ 2 }y \right) dy } =0$$

Since, differentiation of $$\cot y \tan x=\cot y\sec ^{ 2 } dx-\tan xcosec^{ 2 }y$$

$$\displaystyle 3\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$
Question 7
1 / -0

If $$\int_{a}^{x} ty (t) dt = x^{2} + y(x)$$ then y as a function of x is:

A
$$y = 2 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

B
$$y = 1 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

C
$$y = 2 - (1 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

D
none

Solution

Given $$\int_{a}^{x} ty(t)dt = x^{2} + y(x)$$ .....(1)
$$\Rightarrow xy = 2x + \displaystyle \frac{dy}{dx}$$
$$ \Rightarrow x(y - 2) = \displaystyle \frac{dy}{dx}$$
$$\Rightarrow \int xdx = \int \displaystyle \frac{dy}{y - 2}$$
$$ \Rightarrow \displaystyle \frac{x^{2}}{2} = ln | y - 2 | + lnc$$
$$\Rightarrow e^{\displaystyle \frac{x^{2}}{2}} = c(y - 2)$$ ....(2)

Now, put $$x = a$$ in eqn (1)
$$ \Rightarrow y = -a^{2}$$
So, by eqn (2)
$$\therefore e^{ \frac { a^{ 2 } }{ 2 } }=c(-a^{ 2 }-2)$$
$$\Rightarrow c= -\dfrac { e^{ \frac { a^{ 2 } }{ 2 } } }{ (a^{ 2 }+2) } $$
Put this value in (2)
$$\therefore e^{\displaystyle \frac{x^{2}}{2}} = - \displaystyle \frac{e^{\frac{a^{2}}{2}}}{(a^{2} + 2)} (y - 2)$$
$$\Rightarrow -y + 2 = (a^{2} + 2) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$
$$\Rightarrow y = 2 - (2 + a^{2}) e^{\dfrac{x^{2} -a^{2}}{2}}$$
Question 8
1 / -0

Find the curve for which the sum of the lengths of the tangent and subtangent at any of its point is proportional to the product of the coordinates of the point of tangency, the proportionality factor is equal to k.

A
$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2-1)|$$

B
$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2-1)|$$

C
$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2+1)|$$

D
$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2+1)|$$

Solution

Length of tangent at any point of curve $$= y\sqrt {1+\, \left ( \displaystyle \frac {dx}{dy} \right )^2}\,$$
Length of subtangent $$= y\, \displaystyle \frac {dx}{dy}\,$$
$$y\sqrt {1+\, \left ( \displaystyle \frac {dx}{dy} \right )^2}\, +\, y\, \displaystyle \frac {dx}{dy}\, =\, kxy$$
$$\Rightarrow\, \sqrt {1+\, \left ( \displaystyle \frac {dx}{dy} \right )^2}\, =\, kx\, -\, \displaystyle \frac {dx}{dy}$$
$$\Rightarrow\,

1\, +\, \left ( \displaystyle \frac {dx}{dy} \right )^2\, =\, k^2x^2\,

+\, \left ( \displaystyle \frac {dx}{dy} \right )^2\,

-2kx\, \displaystyle \frac {dx}{dy}$$
$$\Rightarrow\,

2kx\, \displaystyle \frac {dx}{dy}\, =\, k^2x^2\, -\, 1\quad

\Rightarrow\, \displaystyle \int \frac {2kxdx}{k^2x^2-1}\, =\, \int dy$$
$$\Rightarrow\, =\, \displaystyle \frac {1}{k}\, ln\, |c(k^2x^2-1)|$$
Question 9
1 / -0

Solve:
$$(1-x^2)\, (1-y)\, dx\, =\, xy\, (1+y)\, dy$$

A
$$\ln(x) (1-y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$

B
$$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$

C
$$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, +\, 2y\, +\, \displaystyle \frac {1}{2}\, y^2$$

D
$$\ln(x) (1+y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$

Solution

$$(1-x^2)\, (1-y)\, dx\, =\, xy\, (1+y)\, dy$$

$$\displaystyle \frac { 1-x^{ 2 } }{ x } dx=\frac { { y }^{ 2 }+y }{ 1-y } dy$$
Integrating both sides, we get
$$\displaystyle \log { x } -\frac { x^{ 2 } }{ 2 } =-\int { \left(y+2+ \frac { 2 }{ y-1 } \right)dy}$$

$$\displaystyle \Rightarrow \log { x } -\frac { x^{ 2 } }{ 2 } =-\frac { y^{ 2 } }{ 2 } -2y+2\log { (y-1) } $$

$$\log (x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$
Question 10
1 / -0

Solve:
$$\displaystyle \frac {dy}{dx}\, +\, \displaystyle \frac {\sqrt {(x^2-1)\, (y^2-1)}}{xy}\, =0$$

A
$$\sqrt {x^2+1}\, -\, \text{cosec}^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$$

B
$$\sqrt {x^2-1}\, -\, \sec^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$$

C
$$\sqrt {x^2-1}\, -\, \text{cosec}^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$$

D
$$\sqrt {x^2+1}\, -\, \sec^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$$

Solution

Given differential eqn can be written as
$$\displaystyle \frac { dy }{ dx } \, =-\, \frac { \sqrt { (x^{ 2 }-1)\, (y^{ 2 }-1) } }{ xy } \, $$

$$\Rightarrow \displaystyle \frac { y }{ \sqrt { y^{ 2 }-1 } } dy\, =-\, \frac { \sqrt { x^{ 2 }-1 } }{ x } dx\, $$
Integrating both sides
$$\displaystyle \int { \frac { y }{ \sqrt { y^{ 2 }-1 } } dy\, } =-\, \int { \frac { \sqrt { x^{ 2 }-1 } }{ x } dx\, } $$

Put $$y^2-1=t$$
$$\Rightarrow 2ydy=dt$$

Also, put $$\sqrt { x^{ 2 }-1 } =u$$
$$\Rightarrow x^2-1=u^2$$
$$\Rightarrow xdx=udu$$

$$\displaystyle \int { \frac { 1 }{ 2\sqrt { t } } dt\, } =-\, \int { \frac { u }{ 1+{ u }^{ 2 } } udu\, } $$

$$\displaystyle \Rightarrow \int { \frac { 1 }{ 2\sqrt { t } } dt\, } =-\, \int { \frac { { u }^{ 2 } }{ 1+{ u }^{ 2 } } du\, } $$

$$ \displaystyle \Rightarrow \sqrt { t } =-\, \int { (1-\frac { 1 }{ 1+{ u }^{ 2 } } )du\, } $$

$$ \displaystyle \Rightarrow \sqrt { { y }^{ 2 }-1 } =-u+\tan ^{ -1 }{ u } +C$$

$$\displaystyle \Rightarrow \sqrt { { y }^{ 2 }-1 } =-\sqrt { { x }^{ 2 }-1 } +\tan ^{ -1 }{ \sqrt { { x }^{ 2 }-1 } } +C$$

$$\displaystyle \Rightarrow \sqrt { { y }^{ 2 }-1 } =-\sqrt { { x }^{ 2 }-1 } +\sec ^{ -1 }{ x} +C$$ ($$\tan ^{ -1 }{ \sqrt { { x }^{ 2 }-1 } }=z \Rightarrow \sqrt { { x }^{ 2 }-1 } =\tan z \Rightarrow \sec z=x$$ )

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Differential Equations Test - 49

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Differential Equations Test - 49

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SHARING IS CARING

Question 1

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

$$xy=c$$

$${ \left( xy \right) }^{ x }=c$$

None of these

Solution

Question 2

circles

parabolas

hyperbolas

straight lines

Solution

Question 3

$$\displaystyle y=k\left ( y-b \right )\left ( 1+bx \right ).$$

$$\displaystyle y=k\left ( y+b \right )\left ( 1-bx \right ).$$

$$\displaystyle y=k\left ( y+b \right )\left ( 1+bx \right ).$$

$$\displaystyle y=k\left ( y-b \right )\left ( 1-bx \right ).$$

Solution

Question 4

variable radii and a fixed centre $$\left ( 0,1 \right )$$

variable radii and a fixed centre $$\left ( 0,-1 \right )$$

fixed radius $$1$$ and variable centres along the $$x$$-axis

fixed radius $$1$$ and variable centres along the $$y$$-axis

Solution

Question 5

$$x^3-y^3=c$$

$$x^3+y^3=c$$

$$x^2+y^2=c$$

$$x^2-y^2=c$$

Solution

Question 6

$$\displaystyle 3\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

$$\displaystyle 8\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

$$\displaystyle 3\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

$$\displaystyle 8\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$

Solution

Question 7

$$y = 2 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

$$y = 1 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

$$y = 2 - (1 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

none

Solution

Question 8

$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2-1)|$$

$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2-1)|$$

$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2+1)|$$

$$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2+1)|$$

Solution

Question 9

$$\ln(x) (1-y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$

$$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$

$$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, +\, 2y\, +\, \displaystyle \frac {1}{2}\, y^2$$

$$\ln(x) (1+y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$

Solution

Question 10

$$\sqrt {x^2+1}\, -\, \text{cosec}^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$$

$$\sqrt {x^2-1}\, -\, \sec^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$$

$$\sqrt {x^2-1}\, -\, \text{cosec}^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$$

$$\sqrt {x^2+1}\, -\, \sec^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$$

Solution

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