Differential Equations Test

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Differential Equations Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

What is the degree of the differential eqaution : $$\dfrac{d^3y}{dx^3}-6(\dfrac{dy}{dx})^2-4y=0$$

A
1

B
2

C
3

D
None of these

Solution

Given the differential equations is
$$\dfrac{d^3y}{dx^3}-6(\dfrac{dy}{dx})^2-4y=0$$.
The highest order term involved in the equation is $$\dfrac{d^3y}{dx^3}$$.
And the degree of the highest ordered term is $$1$$, so degree of the differential equations is $$1$$.
Question 2
1 / -0

What are the order and degree, respectively, of the differential equation
$${ \left( \cfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ y }^{ 4 }+{ \left( \cfrac { dy }{ dx } \right) }^{ 5 }\quad $$?

A
$$4,5$$

B
$$2,3$$

C
$$3,2$$

D
$$5,4$$

Solution

Order is the highest derivative of the dependent variable with respect to the independent variable and degree is the highest power to which the highest order derivative in the differential equation is raised.

so, $$Order=3$$ and $$Degree=2$$

Therefore, Answer is $$C$$
Question 3
1 / -0

Solution of the differential equation
$$\tan { y } .\sec ^{ 2 }{ x } dx+\tan { x } .\sec ^{ 2 }{ y } dy=0$$ is

A
$$\tan { x }+\tan { y }=k$$

B
$$\tan { x }-\tan { y }=k$$

C
$$\cfrac { \tan { x } }{ \tan { y } } =k$$

D
$$\tan { x }.\tan { y }=k$$

Solution

$$\tan { y } .\sec ^{ 2 }{ x } dx+\tan { x } .\sec ^{ 2 }{ y } dy=0$$
$$\Rightarrow \tan { y } .\sec ^{ 2 }{ x } dx=-\tan { x } .\sec ^{ 2 }{ y } dy$$
$$\Rightarrow \cfrac { \sec ^{ 2 }{ x } dx }{ \tan { x } } =-\cfrac { \sec ^{ 2 }{ y } dx }{ \tan { y } } $$
Integrating, $$\int { \cfrac { \sec ^{ 2 }{ x } dx }{ \tan { x } } } =-\int { \cfrac { \sec ^{ 2 }{ y } dx }{ \tan { y } } } $$
$$\Rightarrow \int { \cfrac { d\left( \tan { x } \right) }{ \tan { x } } } =-\int { \cfrac { d\left( \tan { y } \right) }{ \tan { y } } } $$
$$\Rightarrow \ln { \left| \tan { x } \right| } =-\ln { \left| \tan { x } \right| } +c$$
$$\Rightarrow \ln { \left| \tan { x } \right| } +\ln { \left| \tan { x } \right| } =c$$
$$\Rightarrow \ln { \left| \tan { x } .\tan { y } \right| } =c$$
$$\Rightarrow \tan { x } \tan { y } ={ e }^{ c }$$
$$\Rightarrow \tan { x } \tan { y } =k$$ (where $${ e }^{ c }=k$$)
Question 4
1 / -0

The order of differential equation corresponding to $$y={ c }_{ 1 }\cos { 2x } +{ c }_{ 2 }\cos ^{ 2 }{ x } +{ c }_{ 3 }\sin ^{ 2 }{ x } +{ c }_{ 4 }$$

A
$$2$$

B
$$4$$

C
$$3$$

D
None of these

Solution

$$y={ c }_{ 1 }\cos { 2x } +\cfrac { { c }_{ 2 } }{ 2 } \left( 1+\cos { 2x } \right) +\cfrac { { c }_{ 3 } }{ 2 } \left( 1-\cos { 2x } \right) +{ c }_{ 4 }\quad =\left( \cfrac { { c }_{ 2 } }{ 2 } +\cfrac { { c }_{ 3 } }{ 2 } +{ c }_{ 4 } \right) +\left( { c }_{ 1 }+\cfrac { { c }_{ 2 } }{ 2 } -\cfrac { { c }_{ 3 } }{ 2 } \right) \cos { 2x } =A+B\cos { 2x } $$
Equation has only two independent parameters
Hence, order is $$2$$
Question 5
1 / -0

The solution of $$\sec ^{2}y\dfrac {dy}{dx}+2x\tan y=x^{3}$$ is

A
$$\tan { y } =\frac { { x }^{ 2 } }{ 2 } +\frac { 1 }{ 2 } +c{ e }^{ -x^{ 2 } }$$

B
$$\tan { y } =\frac { { x }^{ 2 } }{ 2 } -\frac { 1 }{ 2 } +c{ e }^{ -x^{ 2 } }$$

C
$$\tan{ y}$$ $$=\dfrac {x^{2}}{2}-\dfrac {1}{2}+c ({e}^{x})^{2}$$

D
$$\tan y=\dfrac {x^{2}}{2}+\dfrac{1}{2}$$+$$c({e^{x}})^{2}$$

Solution

$$\begin{array}{l} { \sec ^{ 2 } }y\frac { { dy } }{ { dx } } +2x\tan y={ x^{ 3 } }\to (i) \\ let\, \tan y=t\to (ii) \\ different\, \, equation(ii)\, \, w.\, r.\, t\, \ x \\ we\, \, get \\ { \sec ^{ 2 } }y\frac { { dy } }{ { dx } } =\frac { { dt } }{ { dx } } \\ now\, \, equation\, (i)\, \, becomes \\ \frac { { dt } }{ { dx } } +2xt+{ x^{ 3 } } \end{array}$$
This is linear equation
$$\begin{array}{l} I.F=\int _{ e }{ pdx=\int _{ e }{ 2xdx } } ={ e^{ { x^{ 2 } } } } \\ tx{ e^{ { x^{ 2 } } } }=\int { { x^{ 3 } }\cdot { e^{ { x^{ 2 } } } }dx+c } \\ t{ e^{ { x^{ 2 } } } }=\int { x\cdot { x^{ 2 } }\cdot { e^{ { x^{ 2 } } } }dx+c } \, \, \, \, \, \, \, \, \left[ \begin{array}{l} put\, { x^{ 2 } }=u \\ difference\, \, we\, \, get \\ 2x\, dx=du \end{array} \right] \\ t{ e^{ { x^{ 2 } } } }=\frac { 1 }{ 2 } \int { x\cdot { e^{ { x^{ 2 } } } }dx+c } \, \, \, \, \, \, \, \, \left[ { by\, parts\, \, { { int } }eger } \right] \\ t{ e^{ { x^{ 2 } } } }=\frac { 1 }{ 2 } \left[ { x\cdot { e^{ x } }-\int { 1\cdot { e^{ x } }dx } } \right] \\ t{ e^{ { x^{ 2 } } } }=\frac { 1 }{ 2 } \left[ { 4{ e^{ x } }-{ e^{ x } } } \right] +c \\ t{ e^{ { x^{ 2 } } } }=\frac { { { e^{ { x^{ 2 } } } } } }{ 2 } \left( { { x^{ 2 } }-1 } \right) +c \\ \tan y=\frac { 1 }{ 2 } \left[ { { x^{ 2 } }-1 } \right] +c\cdot { e^{ { x^{ 2 } } } } \end{array}$$
Question 6
1 / -0

The solution of $$\dfrac { dx }{ dy } -\dfrac { 2 }{ 3 } xy={ x }^{ 4 }{ y }^{ 3 }$$ is

A
$$\dfrac { 1 }{ { x }^{ 3 } } =\dfrac { 3 }{ 2 } \left( 1-{ y }^{ 2 } \right) +c{ e }^{ -{ y }^{ 2 } }$$

B
$$\dfrac { 1 }{ { x }^{ 3 } } =\dfrac { 3 }{ 2 } \left( 1+{ y }^{ 2 } \right) +c{ e }^{ -{ y }^{ 2 } }$$

C
$$\dfrac { 1 }{ { x }^{ 4 } } =\dfrac { 3 }{ 2 } \left( 1-{ y }^{ 2 } \right) +c{ e }^{ { y }^{ 2 } }$$

D
$$\dfrac { 1 }{ { x }^{ 4 } } =\dfrac { 3 }{ 2 } \left( 1+{ y }^{ 2 } \right) +c{ e }^{ { y }^{ 2 } }$$
Question 7
1 / -0

Solution of $${y^2}dx + ({x^2} - xy + {y^2})dy = 0$$

A
$$\tan{^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$$

B
$$2\tan{^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$$

C
$$\log y\left( {y + \sqrt {{x^2} + {y^2}} } \right) + \log y + c = 0$$

D
$$\log y\left( {y - \sqrt {{x^2} + {y^2}} } \right) + \log y + c = 0$$

Solution

$${y^2}dx + \left( {{x^2} - xy + 42} \right)dy = 0$$
$$\displaystyle {{dx} \over {dy}} = {{ - \left( {{x^2} - xy + {y^2}} \right)} \over {{y^2}}}$$
$$\displaystyle {{dx} \over {dy}} = - {{{x^2}} \over {{y^2}}} + {x \over y} - 1$$
Put $$ v = x/y\implies x=vy$$
$$dx/dy = v + ydv/dy$$
$$ \Rightarrow v + ydv/dy = - {v^2} + v - 1$$
$$ \Rightarrow ydv/dy = - \left( {{v^2} + 1} \right)$$
$$\displaystyle {{dv} \over {{v^2} + 1}} = - {1 \over y}dy$$
Integrating both side
$$ \Rightarrow {\tan ^{ - 1}}v = - \log y + c$$
$$ \Rightarrow {\tan ^{ - 1}}x/y = - \log y + c\,\,\,\,c = - c$$
$$ \Rightarrow {\tan ^{ - 1}}\left( {x/y} \right) + \log y + c = 0$$
Question 8
1 / -0

If tangent at point p, with parameter t, on the curve $$x=4t^2+3$$,$$y=8t^3-1,t \epsilon R,$$ meets the curve again at point Q, then the coordinates of Q are:-

A
$$t^2$$+3,-$$t^3$$−1

B
4$$t^2$$+3,−8$$t^2$$−1

C
$$t^2$$+3,$$t^3$$−1

D
16$$t^2$$+3,−64$$t^3$$−1

Solution

Given,

$$P(x=4t^2+3,y=8t^3-1)$$

$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dy}{dt}}$$

$$=\dfrac{24t^2}{8t}=3t$$

Tangent at P is given by, $$(y-8t^3+1)=3t(x-4t^2-3)$$

Let,

$$Q(4t_1^2+3,8t_1^3-1)$$

Therefore Q satisfies the equation of tangent at P,

$$8t_1^3-1-8t^3+1=3t(4t_1^2+3-4t^2-3)$$

$$8t_1^3-8t^3=3t(4t_1^2-4t^2)$$

$$8(t_1-t)(t^2+tt_1+t_1^2)=3t \times 4(t_1-t)(t_1+t)$$

$$2(t^2+tt_1+t_1^2)=3t(t_1+t)$$

$$2t_1^2-tt_1+t^2=0$$

$$(t_1-t)(2t_1+t)=0$$

$$t_1=-\dfrac{t}{2}$$

$$\therefore Q=(4t_1^2+3,8t_1^3-1)=(t^2+3,-t^3-1)$$
Question 9
1 / -0

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

A
$$\dfrac{1}{y} = cx - xlogx$$

B
$$\dfrac{1}{x} = cy - ylogy$$

C
$$\dfrac{1}{x} = cx + xlogy$$

D
$$\dfrac{1}{y} = cx - ylogx$$

Solution

$$ \dfrac{dx}{dy}+\dfrac{x}{y} = x^{2} $$
$$ \dfrac{1}{x^{2}}\dfrac{dx}{dy}+\dfrac{1}{xy} = 1 $$
Let $$ \dfrac{1}{x} = 1 $$
$$ \dfrac{-1}{x^{2}}\dfrac{dx}{dy} = \dfrac{dt}{dy} $$
$$ \dfrac{-dt}{dy}+\dfrac{t}{y} = 1 $$
$$ \dfrac{dt}{dy}+(\dfrac{-1}{y})t = -1 $$
I.f $$\displaystyle e\int \dfrac{1}{y}dy $$
I.f $$ = \dfrac{1}{y} $$
$$\displaystyle t.\dfrac{1}{y} = \int -1\times \dfrac{1}{y} = -\int \dfrac{1}{y} $$
$$ \dfrac{1}{xy} = -lny+c $$
$$ \dfrac{1}{x} = -y\,ln\,y+cy $$
$$ \dfrac{1}{x} = cy-y\,ln\,y $$
Question 10
1 / -0

If $$y = cos^{-1} x$$ then $$(1 - x^2)y'' -xy'=?$$ where $$y'= \dfrac{dy}{dx}$$

A
$$1$$

B
$$0$$

C
$$2x$$

D
$$2y$$

Solution

$$\begin{array}{l} y={ \cos ^{ -1 } }x \\ \frac { { dy } }{ { dx } } =-\frac { 1 }{ { \sqrt { 1-{ x^{ 2 } } } } } \\ \left( { 1-{ x^{ 2 } } } \right) y_{ 1 }^{ 2 }-1=0 \\ { y_{ 1 } }=\frac { { dy } }{ { dx } } \, \, \, \, \, ,{ y_{ 2 } }=\frac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } \\ -2xy_{ 1 }^{ 2 }+\left( { 1-{ x^{ 2 } } } \right) 2{ y_{ 1 } }{ y_{ 2 } }=0 \\ -x{ y_{ 1 } }+\left( { 1-{ x^{ 2 } } } \right) { y_{ 2 } }=0 \\ \left( { 1-{ x^{ 2 } } } \right) { y_{ 2 } }-x{ y_{ 1 } }=0 \\ \left( { 1-{ x^{ 2 } } } \right) y''-xy'=0 \\ Option\, \, B\, \, \, is\, \, correct\, \, answer. \end{array}$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 35

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Differential Equations Test - 35

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SHARING IS CARING

Question 1

1

2

3

None of these

Solution

Question 2

$$4,5$$

$$2,3$$

$$3,2$$

$$5,4$$

Solution

Question 3

$$\tan { x }+\tan { y }=k$$

$$\tan { x }-\tan { y }=k$$

$$\cfrac { \tan { x } }{ \tan { y } } =k$$

$$\tan { x }.\tan { y }=k$$

Solution

Question 4

$$2$$

$$4$$

$$3$$

None of these

Solution

Question 5

$$\tan { y } =\frac { { x }^{ 2 } }{ 2 } +\frac { 1 }{ 2 } +c{ e }^{ -x^{ 2 } }$$

$$\tan { y } =\frac { { x }^{ 2 } }{ 2 } -\frac { 1 }{ 2 } +c{ e }^{ -x^{ 2 } }$$

$$\tan{ y}$$ $$=\dfrac {x^{2}}{2}-\dfrac {1}{2}+c ({e}^{x})^{2}$$

$$\tan y=\dfrac {x^{2}}{2}+\dfrac{1}{2}$$+$$c({e^{x}})^{2}$$

Solution

Question 6

$$\dfrac { 1 }{ { x }^{ 3 } } =\dfrac { 3 }{ 2 } \left( 1-{ y }^{ 2 } \right) +c{ e }^{ -{ y }^{ 2 } }$$

$$\dfrac { 1 }{ { x }^{ 3 } } =\dfrac { 3 }{ 2 } \left( 1+{ y }^{ 2 } \right) +c{ e }^{ -{ y }^{ 2 } }$$

$$\dfrac { 1 }{ { x }^{ 4 } } =\dfrac { 3 }{ 2 } \left( 1-{ y }^{ 2 } \right) +c{ e }^{ { y }^{ 2 } }$$

$$\dfrac { 1 }{ { x }^{ 4 } } =\dfrac { 3 }{ 2 } \left( 1+{ y }^{ 2 } \right) +c{ e }^{ { y }^{ 2 } }$$

Question 7

$$\tan{^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$$

$$2\tan{^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$$

$$\log y\left( {y + \sqrt {{x^2} + {y^2}} } \right) + \log y + c = 0$$

$$\log y\left( {y - \sqrt {{x^2} + {y^2}} } \right) + \log y + c = 0$$

Solution

Question 8

If tangent at point p, with parameter t, on the curve $$x=4t^2+3$$,$$y=8t^3-1,t \epsilon R,$$ meets the curve again at point Q, then the coordinates of Q are:-

$$t^2$$+3,-$$t^3$$−1

4$$t^2$$+3,−8$$t^2$$−1

$$t^2$$+3,$$t^3$$−1

16$$t^2$$+3,−64$$t^3$$−1

Solution

Question 9

$$\dfrac{1}{y} = cx - xlogx$$

$$\dfrac{1}{x} = cy - ylogy$$

$$\dfrac{1}{x} = cx + xlogy$$

$$\dfrac{1}{y} = cx - ylogx$$

Solution

Question 10

$$1$$

$$0$$

$$2x$$

$$2y$$

Solution

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