Differential Equations Test

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Differential Equations Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The general solution of $$xy^{5}=y_{4}(y_{n}=\displaystyle\frac{d^{n}y}{dx^{n}})$$ is given by:

A
$${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{3}+C_{3}x^{2}+C_{4}x+C_{5}$$

B
$${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{4}+C_{3}x^{2}+C_{4}x+C_{5}$$

C
$${ y }^{ -1 }=C_{1}+C_{2}x+C_{3}x^{2}+C_{4}x^{3}+C_{5}x^{4}$$

D
none of these

Solution

$$\displaystyle x{ y }^{ 5 }=\frac { { d }^{ 4 }y }{ d{ x }^{ 4 } } \Rightarrow \frac { { d }^{ 4 }y }{ d{ x }^{ 4 } } { y }^{ -5 }=x$$
$$\displaystyle \Rightarrow \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \frac { { y }^{ -4 } }{ -4 } =\frac { { x }^{ 2 } }{ 2 } +c$$
$$\displaystyle \Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \frac { { y }^{ -3 } }{ 12 } =\frac { { x }^{ 3 } }{ 6 } +cx+d$$
$$\displaystyle \Rightarrow \frac { dy }{ dx } \frac { { y }^{ -2 } }{ -24 } =\frac { { x }^{ 4 } }{ 24 } +\frac { c{ x }^{ 2 } }{ 2 } +dx+e$$
$$\displaystyle \Rightarrow \frac { { y }^{ -1 } }{ 24 } =\frac { { x }^{ 5 } }{ 120 } +\frac { c{ x }^{ 3 } }{ 6 } +\frac { d{ x }^{ 2 } }{ 2 } +ex+f$$
$$\Rightarrow { y }^{ -1 }={ C }_{ 1 }{ x }^{ 5 }+{ C }_{ 2 }{ x }^{ 3 }+{ C }_{ 3 }{ x }^{ 2 }+{ C }_{ 4 }x+{ C }_{ 5 }$$
Question 2
1 / -0

Solve:
$$\displaystyle \frac {dy}{dx}\, +\, \sin\, \displaystyle \frac {x+y}{2}\, =\, \sin\, \displaystyle \frac {x-y}{2}$$

A
$$\ln \left |\tan \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{4}$$

B
$$\ln \left |\tan \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{2}$$

C
$$\ln \left |\cot \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{2}$$

D
$$\ln \left |\cot \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{4}$$

Solution

$$\displaystyle \frac {dy}{dx}\, +\, \sin \, \displaystyle \frac {x+y}{2}\, =\, \sin \, \displaystyle \frac {x-y}{2}$$
we know $$\sin (A-B)-\sin (A+B)=-\cos A\sin B$$
$$\Rightarrow\, \displaystyle \frac {dy}{dx}\, =\, -\, 2\, \cos \, \displaystyle \frac {x}{2}\, \sin \displaystyle \frac {y}{2}$$
$$\Rightarrow\, \text{cosec}\, \displaystyle \frac {y}{2}\, dy\, =\, -2\, \cos \, \displaystyle \frac {x}{2}\, dx$$
On integration we get
$$\Rightarrow\, 2\, \left (\ln \left | \tan \, \displaystyle \frac {y}{4} \right | \right )\, =\, 2\, \left (-2\, \sin \, \displaystyle \frac {x}{2}\right )\, +\, c$$
$$\Rightarrow\, \ln \left |\tan \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \sin \, \displaystyle \frac {x}{2}$$
Question 3
1 / -0

Solve:
$$\displaystyle \frac {dy}{dx}\, =\, \displaystyle \frac {x(2\ln\, x\, +\, 1)}{\sin\, y\, +\, y\cos\, y}$$

A
$$y\, \sin\, y\, =\, x\, \ln\, x\, +\, c$$

B
$$y\, \cos\, y\, =\, x^2\, \ln\, x\, +\, c$$

C
$$y\, \cos\, y\, =\, x\, \ln\, x\, +\, c$$

D
$$y\, \sin\, y\, =\, x^2\, \ln\, x\, +\, c$$

Solution

Given $$\displaystyle \frac {dy}{dx}\, =\, \displaystyle \frac {x(2\ln\, x\, +\, 1)}{\sin\, y\, +\, \cos\, y}$$
$$\dfrac { dy }{ dx } (\sin\, y\, +\, y\cos\, y)\, =\, 2x\ln\, x\, +\, x$$
$$\dfrac { dy }{ dx } \times \sin\, y\, +\, y\times \dfrac { d(\sin y) }{ dx } \, =\dfrac { \, { d(x }^{ 2 }) }{ dx } \times \ln\, x\, +\, { x }^{ 2 }\times \dfrac { d(\ln x) }{ dx } $$
On integration
$$y\sin y={ x }^{ 2 }\ln x+c$$
Question 4
1 / -0

Solve the following differential equation:

$$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$$

A
$$\tan^{-1} y=x+\dfrac{x^3}{3}+C$$

B
$$\tan^{-1}y=\dfrac{x^2}{2}+x+C$$

C
$$\sec^{-1}y=x+C$$

D
None of these

Solution

The given differential eqaution :
$$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$$

$$\dfrac{dy}{1+y^2}=(1+x^2) \ dx$$

Integrating both sides,
$$\int \dfrac{dy}{1+y^2}=\int (1+x^2) \ dx$$

$$\tan^{-1}y = x+\dfrac{x^3}{3}+C$$
Question 5
1 / -0

The equation of the curve in which the portion of the tangent included between the coordinate axes is bisected at the point of contact is:

A
a parabola

B
an ellipse

C
a hyperbola

D
a circle

Solution

Let the equation of curve be $$y=f(x)$$
Let $$P(x,y)$$ be any point on the curve.
Equation of tangent at $$P(x,y)$$ is
$$Y-y=y'(X-x)$$
$$\Rightarrow y'X-Y=xy'-y$$
$$\Rightarrow \displaystyle \dfrac{X}{\dfrac{xy'-y}{y'}}+\dfrac{Y}{y-xy'}=1$$
So, the intercepts of the tangent on the axis are
$$\dfrac{xy'-y}{y'}$$ and $$y-xy'$$
So, the point $$A$$ on $$x$$-axis is $$(\dfrac{xy'-y}{y'} ,0)$$ and $$B$$ on $$y$$-axis is $$(0,y-xy')$$.
Now, $$P$$ is mid-point of $$AB$$
So coordinates of $$P$$ are $$(\dfrac{xy'-y}{2y'}, \dfrac{y-xy'}{2})$$
$$\Rightarrow x=\dfrac{xy'-y}{2y'}$$ and $$y=\dfrac{y-xy'}{2}$$
$$\Rightarrow \displaystyle xy'-y=2y'x $$ and $$xy'-y=-2y$$
$$\Rightarrow 2y'x=-2y$$
$$\Rightarrow \dfrac{dy}{y}=-\dfrac{dx}{x}$$
Integrating both sides, we get
$$\log y=-\log x +\log C$$
$$\Rightarrow \log xy=\log C$$
$$\Rightarrow xy=C$$
Question 6
1 / -0

If $$y=\sin ^{ -1 }{ \left[ \sqrt { x-ax } -\sqrt { a-ax } \right] } $$, then $$\cfrac { dy }{ dx } $$ is equal to

A
$$\cfrac { 1 }{ \sin { \sqrt { a-ax } } } $$

B
$$\sin { \sqrt { x } } .\sin { \sqrt { a } } $$

C
$$\cfrac { 1 }{ 2\sqrt { x } \sqrt { 1-x } } $$

D
$$0$$

Solution

Given $$y=\sin ^{ -1 }{ \left[ \sqrt { x-ax } -\sqrt { a-ax } \right] } $$

Put $$x=\sin ^{ 2 }{ \theta } ,a=\sin ^{ 2 }{ \phi } $$ then

$$y=\sin ^{ -1 }{ \left[ \sqrt { \sin ^{ 2 }{ \theta } -\sin ^{ 2 }{ \phi } \sin ^{ 2 }{ \theta } } -\sqrt { \sin ^{ 2 }{ \phi } -\sin ^{ 2 }{ \phi } \sin ^{ 2 }{ \theta } } \right] } $$

$$=\sin ^{ -1 }{ \left[ \sqrt { \sin ^{ 2 }{ \theta } (1-\sin ^{ 2 }{ \phi } ) } -\sqrt { \sin ^{ 2 }{ \phi } (1-\sin ^{ 2 }{ \theta } ) } \right] } $$

$$=\sin ^{ -1 }{ \left[ \sin { \theta } \cos { \phi } -\sin { \phi } \cos { \theta } \right] } $$

$$=\sin ^{ -1 }{ \left[ \sin { \left( \theta -\phi \right) } \right] } =\theta -\phi $$

$$=\sin ^{ -1 }{ \sqrt { x } } -\sin ^{ -1 }{ \sqrt { a } } $$

$$\therefore \cfrac { dy }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ \left( \sqrt { x } \right) }^{ 2 } } } .\cfrac { 1 }{ 2\sqrt { x } } =\cfrac { 1 }{ \sqrt { 1-x } } .\cfrac { 1 }{ 2\sqrt { x } } $$
Question 7
1 / -0

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

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

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

C
$$\cfrac { 1 }{ { 2 }^{ x } } -\cfrac { 1 }{ { 2 }^{ y } } =c$$

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

Solution

Given $$\cfrac { dy }{ dx } ={ 2 }^{ y-x }$$

$$\Rightarrow { 2 }^{ -y }dy={ 2 }^{ -x }dx$$

On integrating both sides, we get

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

$$\Rightarrow -\cfrac { { 2 }^{ -y } }{ \log { 2 } } =\cfrac { -{ 2 }^{ -x } }{ \log { 2 } } +{ c }_{ 1 }$$

$$\Rightarrow -{ 2 }^{ -y }=-{ 2 }^{ -x }+{ c }_{ 1 }\log { 2 } $$

$$\therefore \quad \cfrac { 1 }{ { 2 }^{ x } } -\cfrac { 1 }{ { 2 }^{ y } } =c\quad \left( c={ c }_{ 1 }\log { 2 } \right) $$
Question 8
1 / -0

A curve, whose concavity is directly proportional to the logarithm of its x-coordinate at any point of curve, is given by:

A
$${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } -3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$$

B
$${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } +3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$$

C
$${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } \right) +{ c }_{ 2 }$$

D
None of these

Solution

Let $$P(x,y)$$ be any point on the curve
According to given conditon,
$$\displaystyle \frac{d^2 y}{dx^2}=k\log x$$ ....(1)
where $$k$$ is constant of proportionality.
Integrating both sides of eqn (1),
$$\displaystyle \frac{dy}{dx}=k\int \log x.1 dx$$
$$\displaystyle \frac{dy}{dx}=k(x\log x-x)+c_2$$
Again integrating, we get
$$y=k \int {x\log x} -k\int xdx +\int c_2 dx$$

$$\Rightarrow \displaystyle y=k [\log x \frac{x^2}{2}-\int \frac{x^2}{2} \frac{1}{x}dx]-k\frac{x^2}{2}+c_2 x+c_3$$

$$\Rightarrow \displaystyle y=\frac { k }{ 4 } [2{ x }^{ 2 }\log x-x^{ 2 }-2x^{ 2 }]+c_{ 2 }x+c_{ 3 }$$

$$\Rightarrow \displaystyle y={ c }_{ 1 }[2{ x }^{ 2 }\log x-3x^{ 2 }]+c_{ 2 }x+c_{ 3 }$$
Question 9
1 / -0

The order of the differential equation whose solution is $$y=a\cos { x } +b\sin { x } +c{ e }^{ -x }$$, is

A
$$3$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

Given DE has 3 different independent constants $$a,b,c$$

Therefore order of given differential equation is $$3$$
Question 10
1 / -0

$$\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } } \right] $$ is equal to

A
$$\cfrac { 1 }{ 1+{ x }^{ 2 } } $$

B
$$\cfrac { 2 }{ 1+{ x }^{ 2 } } $$

C
$$\cfrac { { x }^{ 2 } }{ 2\sqrt { 1+{ x }^{ 2 } } \sqrt { 1+{ x }^{ 2 }-1 } } $$

D
$$\cfrac { 1 }{ 2\left( 1+{ x }^{ 2 } \right) } $$

Solution

Let $$y=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } } \right] $$

Put $$x=\tan { \theta } $$

$$\Rightarrow \theta =\tan ^{ -1 }{ x } $$

Then,
$$y=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { 1+\tan ^{ 2 }{ \theta } } -1 }{ \tan { \theta } } } \right] $$

$$=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { \sec ^{ 2 }{ \theta } } -1 }{ \tan { \theta } } } \right] $$

$$=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \cfrac { 1 }{ \cos { \theta } } -1 }{ \cfrac { \sin { \theta } }{ \cos { \theta } } } } \right] $$

$$=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \left( \cfrac { 1-\cos { \theta } }{ sin\theta } \right) } \right] $$

$$=\cfrac { d }{ dx } \tan ^{ -1 }{ \left( \tan { \cfrac { \theta }{ 2 } } \right) } $$

$$=\cfrac { d }{ dx } \left( \cfrac { \theta }{ 2 } \right) =\cfrac { 1 }{ 2 } \cfrac { d }{ dx } \left( \tan ^{ -1 }{ x } \right) $$

$$=\cfrac { 1 }{ 2 } \cfrac { 1 }{ \left( 1+{ x }^{ 2 } \right) } $$

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Differential Equations Test - 50

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Differential Equations Test - 50

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Question 1

$${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{3}+C_{3}x^{2}+C_{4}x+C_{5}$$

$${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{4}+C_{3}x^{2}+C_{4}x+C_{5}$$

$${ y }^{ -1 }=C_{1}+C_{2}x+C_{3}x^{2}+C_{4}x^{3}+C_{5}x^{4}$$

none of these

Solution

Question 2

$$\ln \left |\tan \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{4}$$

$$\ln \left |\tan \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{2}$$

$$\ln \left |\cot \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{2}$$

$$\ln \left |\cot \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{4}$$

Solution

Question 3

$$y\, \sin\, y\, =\, x\, \ln\, x\, +\, c$$

$$y\, \cos\, y\, =\, x^2\, \ln\, x\, +\, c$$

$$y\, \cos\, y\, =\, x\, \ln\, x\, +\, c$$

$$y\, \sin\, y\, =\, x^2\, \ln\, x\, +\, c$$

Solution

Question 4

$$\tan^{-1} y=x+\dfrac{x^3}{3}+C$$

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

$$\sec^{-1}y=x+C$$

None of these

Solution

Question 5

a parabola

an ellipse

a hyperbola

a circle

Solution

Question 6

$$\cfrac { 1 }{ \sin { \sqrt { a-ax } } } $$

$$\sin { \sqrt { x } } .\sin { \sqrt { a } } $$

$$\cfrac { 1 }{ 2\sqrt { x } \sqrt { 1-x } } $$

$$0$$

Solution

Question 7

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

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

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

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

Solution

Question 8

$${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } -3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$$

$${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } +3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$$

$${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } \right) +{ c }_{ 2 }$$

None of these

Solution

Question 9

$$3$$

$$1$$

$$2$$

$$4$$

Solution

Question 10

$$\cfrac { 1 }{ 1+{ x }^{ 2 } } $$

$$\cfrac { 2 }{ 1+{ x }^{ 2 } } $$

$$\cfrac { { x }^{ 2 } }{ 2\sqrt { 1+{ x }^{ 2 } } \sqrt { 1+{ x }^{ 2 }-1 } } $$

$$\cfrac { 1 }{ 2\left( 1+{ x }^{ 2 } \right) } $$

Solution

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