Question Set C

Tamilnadu 12th Standard Maths Ordinary Differential Equations Important Questions Study Material - QB365

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Questions:

PART-B

ANSWER THE FOLLOWING

Show that y = a cos bx is a solution of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ b }^{ 2 }y=0\).

Determine the order and degree (if exists) of the following differential equations:
dy + (xy − cos x)dx = 0

Find the differential equation of the family of parabolas y² = 4ax, where a is an arbitrary constant.

Show that x²+ y² = r², where r is a constant, is a solution of the differential equation \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).

Solve \(\frac { dy }{ dx } +2y={ e }^{ -x }\)

PART-D

ANSWER THE FOLLOWING

Solve the following differential equations:
(ydx-xdy)cot\(\left( \frac { x }{ y } \right) \) = ny² dx

Solve the following differential equations:
\(\\ \\ \\ \frac { dy }{ dx } ={ tan }^{ 2 }(x+y)\)

Solve the differential equation \({ ye }^{ \frac { x }{ y } }dx=\left( { xe }^{ \frac { x }{ y } }+y \right) dy\)

In a bank principal increases at the rate of 5% per year. In how many years Rs.1000 doubled itself.

10.

It is given that the rate at which some bacteria multiply is proportional to the instantaneous number present. If the original number of bacteria doubles in two hours, in how many hours will it be five times.

Multiple Choice Question

11.

The order and degree of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 1/3 }+{ x }^{ 1/4 }=0\) are respectively

(a)

2, 3

(b)

3, 3

(c)

2, 6

(d)

2, 4

12.

The order and degree of the differential equation \(\sqrt { sinx } (dx+dy)=\sqrt { cos x } (dx-dy)\) is

(a)

1, 2

(b)

2, 2

(c)

1, 1

(d)

2, 1

13.

The differential equation of the family of curves y = Ae^x + Be^−x, where A and B are arbitrary constants is

(a)

\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)

(b)

\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)

(c)

\(\frac { { d }y }{ { dx } } +y=0\)

(d)

\(\frac { { d }y }{ { dx } } -y=0\)

14.

The integrating factor of the differential equation \(\frac { dy }{ dx } +y=\frac { 1+y }{ \lambda } \) is

(a)

\(\frac { x }{ { e }^{ \lambda } } \)

(b)

\(\frac { { e }^{ \lambda} }{ x } \)

(c)

\({ \lambda e }^{ x }\)

(d)

e^x

15.

The degree of the differential equation \(y(x)=1+\frac { dy }{ dx } +\frac { 1 }{ 1.2 } { \left( \frac { dy }{ dx } \right) }^{ 2 }+\frac { 1 }{ 1.2.3 } { \left( \frac { dy }{ dx } \right) }^{ 3 }+....\) is

(a)

(b)

(c)

(d)

16.

The solution of the differential equation \(\frac { dy }{ dx } +\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } =0\) is

(a)

y + sin^-1 x = c

(b)

x + sin^-1 y = 0

(c)

y²+ 2 sin^-1 x = c

(d)

x²+ 2 sin^-1y= c

17.

If sin x is the integrating factor of the linear differential equation \(\frac { dy }{ dx } +Py=Q,\) then P is

(a)

log sin x

(b)

cos x

(c)

tan x

(d)

cot x

18.

Integrating factor of the differential equation \(\frac{d y}{d x}=\frac{x+y+1}{x+1}\) is

(a)

\(\frac{1}{x+1}\)

(b)

x+1

(c)

\(\frac { 1 }{ \sqrt { x+1 } } \)

(d)

\({ \sqrt { x+1 } } \)

19.

P is the amount of certain substance left in after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then

(a)

P = Ce^kt

(b)

P = Ce^-kt

(c)

P = Ckt

(d)

Pt = C

20.

The slope at any point of a curve y = f (x) is given by \(\frac{dy}{dx}=3x^2\) and it passes through (-1, 1). Then the equation of the curve is

(a)

y = x³+ 2

(b)

y = 3x²+ 4

(c)

y = 3x⁴+ 4

(d)

y = 3x²+ 5

21.

If cosx is an integrating factor of the differential equation \(\frac{dy}{dx}+Py= Q\), then P = ___________

(a)

-cot x

(b)

cot x

(c)

tan x

(d)

-tan x

22.

The I.F. of cosec x \(\frac{dy}{dx}+y\) sec²x = 0 is ___________

(a)

e^{sec x}

(b)

e^tan x

(c)

e^{sec x tan x}

(d)

e^{sec2 x}

23.

The general solution of \(4\frac{d^2 y}{dx^2}\) + y = 0 is _________

(a)

\(y={ e }^{ \frac { x }{ 2 } }\left[ A\ cos\frac { x }{ 2 } +B\ sin\frac { x }{ 2 } \right] \)

(b)

\(y={ e }^{ \frac { x }{ 2 } }\left[ A\ cos\frac { x }{ 2 } -B\ sin\frac { x }{ 2 } \right] \)

(c)

\(y=Acos\frac { x }{ 2 } +Bsin\frac { x }{ 2 } \)

(d)

\(y={ Ae }^{ \frac { x }{ 2 } }+B{ e }^{ \frac { -x }{ 2 } }\)

24.

The I.F of \(\frac{dy}{dx}-y\) tan x = cos x is _________

(a)

sec x

(b)

cos x

(c)

e^{tan x}

(d)

cot x

25.

On finding the differential equation corresponding to y = e^mx where m is the arbitrary constant, then m is ________.

(a)

\(\frac { y }{ { y }^{ 1 } } \)

(b)

\(\frac { { y }^{ 1 } }{ y } \)

(c)

(d)

26.

The population p of a certain bacteria decreases at a rate proportional to the population p. The differential equation corresponding to the above statement is __________.

(a)

\(\frac{dp}{dt}=\frac{k}{p}\)

(b)

\(\frac{dp}{dt}=kt\)

(c)

\(\frac{dp}{dt}=kp\)

(d)

\(\frac{dp}{dt}=-kp\)

27.

The general solution of x \(\frac{dy}{dx}\) = y is _________.

(a)

y = cx

(b)

x²+ y²= c

(c)

x²- y²= c

(d)

y = c^x

28.

The differential equation associated with the family of concentric circles having their centres at the origin is _________.

(a)

\(\frac { dy }{ dx } =\frac { -x }{ y } \)

(b)

\(\frac { dy }{ dx } =\frac { -y }{ x } \)

(c)

\(\frac { dy }{ dx } =\frac { x }{ y } \)

(d)

\(\frac { dy }{ dx } =\frac { y }{ x } \)

29.

The solution of the differential equation \(\frac{d y}{d x}=e^{x+y}\) is __________

(a)

e^x + e^y = c

(b)

e^x + e^-y = c

(c)

e^x - e^-y = c

(d)

none of these

30.

The solution of the \(\operatorname{DE} \ y \frac{d x}{d y}=\cot x \text { is }\) __________

(a)

sec x = cy

(b)

sec y = cx

(c)

sec y = c

(d)

sec x = c

31.

The order and degree of the differential equation \(\frac{d^{3} y}{d x^{3}}+6 \frac{d y}{d x}+3 y=0 \text { is }\)__________

(a)

3, 1

(b)

1, 3

(c)

1, 1

(d)

none of these

32.

The solution of the differential equation x dy + y dx = 0 is __________

(a)

x - y = c

(b)

x + y = c

(c)

xy = c

(d)

none of these

33.

The solution of the differential equation is \(\frac{d y}{d x}=1-y-x+x y \text { is }\)

(a)

\(\log (1-y)=x-\frac{x^{2}}{2}+c \)

(b)

\(\log (1+y)=x-\frac{x^{2}}{2}+c \)

(c)

\(e^{y}=x-\frac{x^{3}}{3}+c \)

(d)

none of these

34.

The solution of the differential equation x cos y dy - (xe^x log x 4 e^x) dx is __________

(a)

sin y = e^x log x + c

(b)

sin y = e^x + log y + c

(c)

sin y = e^x + log x + c

(d)

none of these

35.

The solution of the differential equation \(\frac{d y}{d x}=e^{x}+2 \text { is }\)__________

(a)

\(y=e^{x}+C\)

(b)

\(y=2 x+e^{x}+C\)

(c)

\(y=2 x e^{x}+C\)

(d)

\(y=e^{x}+2 C x\)

PART-C

ANSWER THE FOLLOWING

36.

Form the differential equation for y = e^-2x [A cos 3x-B sin 3x]

37.

Solve: \(\frac{dy}{dx}+y=cos x\)

38.

Obtain the D.E of all circles of radius ‘r’

39.

Find the differential equation of the family of curves \(y=A e^{-x}+B e^{x}\) where A and B are arbitrary constants.

40.

Form the differential equation of y = e^3x (C cos 2x + D sin 2x), where C and D are atbitrary constants.

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Answers Key:

PART-B

ANSWER THE FOLLOWING

Given y = a cos bx ...(1)
Differentiating equation (1) w.r.t 'x', we get
\(\frac{d y}{d x}=\mathrm{a}(-\sin \mathrm{b} x) \mathrm{b}=-\mathrm{ab} \sin \mathrm{b} x\)
Again differentiating, we get
\(\frac{d^2 y}{d x^2} =-\mathrm{ab} \cos \mathrm{b} x \cdot \mathrm{b} \)
\(\frac{d^2 y}{d x^2} =-\mathrm{ab}^2 \cos \mathrm{b} x=-\mathrm{b}^2(\mathrm{a} \cos \mathrm{b} x) \)
\(\frac{d^2 y}{d x^2} =-\mathrm{b}^2 \mathrm{y} \)
\(\frac{d^2 y}{d x^2}+\mathrm{b}^2 \mathrm{y} =0\)
Therefore, y = a cos bx is a solution of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ b }^{ 2 }y=0\)

dy + (xy − cos x)dx = 0 is a first order differential equation with degree 1
since the equation can be rewritten as
\(\frac{dy}{dx}\) + xy - cos x = 0

The equation of the family of parabolas is given by y² ax = 4, a is an arbitrary constant. ... (1)
Differentiating both sides of (1) with respect to x , we get 2y\(\frac{dy}{dx}=4a\Rightarrow a=\frac{y}{2}\frac{dy}{dx}\)
Substituting the value of a in (1) and simplifying, we get \(\frac{dy}{dx}=\frac{y}{2x}\) as the required differential equation.

Given that x² + y² = r², r∈R ...(1)
The given equation contains exactly one arbitrary constant.
So, we have to differentiate the given equation once. Differentiate (1) with respect to x, we get
2x +2y\(\frac{dy}{dx}\) = 0 which implies \(\frac{dy}{dx}\) = \(-\frac{x}{y}\)
Thus, x² + y²= r² satisfies the differential equation \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).
Hence, x²+ y²= r² is a solution of the differential equation \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

PART-D

ANSWER THE FOLLOWING

\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } .cot\left( \frac { x }{ y } \right) =xdx\)
put \(\frac { x }{ y } =t\)
\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } =dt\)
Substituting these values in equation (1), we get
dt cot(t) = x dx
cot t dt = ndx
Taking integration on both sides, we get
\(\Rightarrow \int { cot(t)dt=n\int { dx } } \)
\( \int \cot t \mathrm{dt} =n \int d x \)
\(\log (\sin \mathrm{t}) =\mathrm{n} x+\mathrm{C}_1 \)
\(\sin \mathrm{t} =\mathrm{e}^{\mathrm{nx}+\mathrm{c}_1} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{e}^{n x} \mathrm{e}^{\mathrm{C}_r} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{C}^{\mathrm{nx}}\)
\(\\ \Rightarrow sin\left( \frac { x }{ y } \right) ={ e }^{ nx+c }\left[ \because t=\frac { x }{ y } \right] \)

\(\\ \\ \\ \frac { dy }{ dx } ={ tan }^{ 2 }(x+y)...(1)\)
Take x + y = t
\(\Rightarrow 1+\frac { dy }{ dx } =\frac { dt }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { dt }{ dx } -1\)
∴ (1) becomes,
\(\frac { dt }{ dx } -1={ tan }^{ 2 }t\)
\(\Rightarrow \frac { dt }{ dx } ={ tan }^{ 2 }t1\)
\(\Rightarrow \frac { dt }{ dx } ={ sec }^{ 2 }(t)\)
\(\Rightarrow \frac { dt }{ { sec }^{ 2 }t } =dx\)
\(\Rightarrow { cos }^{ 2 }t\quad dt=dx\)
\(\left(\frac{1+\cos 2 t^{\circ}}{2}\right) d t=\mathrm{d} x \quad\left(\because \cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\right)\)
\(\left[ cos\quad 2x=2{ cos }^{ 2 }x-1{ cos }^{ 2 }x=\frac { 1+cos2x }{ 2 } \right] \)
Taking integration on both sides, we get
\(\Rightarrow \left( \frac { 1+cos2\quad t }{ 2 } \right) dt=dx\)
\(\Rightarrow \frac { 1 }{ 2 } \int { (1+cos2t)dt=\int { dx } } \)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { sin2t }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { 2sintcost }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } [t+sin\ t\ cost]=x+c\ [\because t=x+y]\)
\(\Rightarrow \frac { 1 }{ 2 } [x+y+sin(x+y)cos(x+y)=x+c\)

\( y e^{\left(\frac{1}{r}\right)} \cdot d x =\left(x e^{\frac{1}{y}}+y\right) d y \)
\(\frac{d x}{d y} =\frac{x \cdot e^{\left(\frac{6}{y}\right)}+y}{y e^{\left(\frac{x}{y}\right)}} \)
\(\frac{d x}{d y} =\left(\frac{x}{y}\right)+\frac{1}{e^{\left(\frac{6}{y}\right)}}\)
Put x = \( \mathrm{vy}\) \( \Rightarrow\left(\frac{x}{y}\right)=\mathrm{v}\) and \(\frac{d x}{d y}=\cdot v+y \cdot \frac{d v}{d y} \)
\((1) \Rightarrow \ v+y \cdot \frac{d v}{d y}=v+\frac{1}{e^y} \)
\(\mathrm{e}^v \cdot \mathrm{dv}=\frac{d y}{y}\)
Integrating on both sides,
ie) \(\int e^v \cdot d v =\int \frac{d y}{y} \)
\(e^v =\log |y|+\log |c| \)
\(e^{\left(\frac{x}{y}\right)} =\log |c y|\)

20log_e2years

10.

\(\frac { 2log5 }{ lof2 } hours\)

Multiple Choice Question

11.

(a)
2, 3

12.

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

13.

(b)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)

14.

(d)
e^x

15.

16.

(a)
y + sin^-1 x = c

17.

(d)
cot x

18.

(a)
\(\frac{1}{x+1}\)

19.

(b)
P = Ce^-kt

20.

(a)
y = x³+ 2

21.

(d)
-tan x

22.

(a)
e^{sec x}

23.

24.

(b)
cos x

25.

(b)
\(\frac { { y }^{ 1 } }{ y } \)

26.

(d)
\(\frac{dp}{dt}=-kp\)

27.

(a)
y = cx

28.

(a)
\(\frac { dy }{ dx } =\frac { -x }{ y } \)

29.

(b)
e^x + e^-y = c

30.

(a)
sec x = cy

31.

(a)
3, 1

32.

33.

(a)
\(\log (1-y)=x-\frac{x^{2}}{2}+c \)

34.

(a)
sin y = e^x log x + c

35.

(b)
\(y=2 x+e^{x}+C\)

PART-C

ANSWER THE FOLLOWING

36.

Given y = e-2x[A cos 3x- B sin 3x]
⇒ ye²x = A cos 3x-B sin 3x
Differentiating,y¹e²+2y e^2x= -3A sin 3x-3B
cos 3x
Differentiating again we get,
y"e^2x+2(2y')e^2x+4ye^2x= -9(A cos 3x-B sin 3x)
⇒ e^2x( y"+4y'+4y) = -9(A cos 3x - B sin 3x)
⇒ z y"+4y'+4y = -9(A cos 3x-B sin 3x)
⇒ y"+4y'+4y = -9(using (1))
⇒ y"+4y'+13y = 0
is the required differential equation

37.

Given \(\frac { dy }{ dx } +y=cosx\)
This is a linear differential equation
Here p = 1, Q = cos x
\(\therefore \int { p\ dx } =\int { dx } =x\)
\(I.F={ e }^{ \int { p\ dx } }={ e }^{ x }\)
The solution is
\({ y }^{ \int { p\ dx } }=\int { Q{ e }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow { ye }^{ x }=\int { cosx.{ e }^{ x }dx+c } \)
\(\Rightarrow { ye }^{ x }=\frac { { e }^{ x } }{ 2 } \left( cosx+sinx \right) +c\)
\(\Rightarrow y=\frac { 1 }{ 2 } \left( cosx+sinx \right) +{ ce }^{ x }\)
\(\therefore \int { { e }^{ ax }cos\ bx\ dx=\frac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \left[ acos\ bx+sin\ ax \right] } \)

38.

\(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 2 }={ y }^{ 2 }\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 2 }\)

39.

The equation of the given family of curves is
\(y=A e^{-x}+B e^{x}\) .............(1)
where A and B are arbitrary constant
Differentiating equation (1) with respect to 'x', we get
\( \frac{d y}{d x} =\mathrm{Ae}^{-x}(-1)+\mathrm{Be}^{x}(1) \)
\( =-\mathrm{Ae}^{-x}+\mathrm{Be}^{x}\)
Again differentiating equation (2) with respect to 'x', we get
\( \frac{d^{2} y}{d x^{2}}=-A \mathrm{e}^{-x}(-1)+\mathrm{Be}^{x} \)
\( \frac{d^{2} y}{d x^{2}}=A \mathrm{e}^{-x}+\mathrm{Be}^{x 14.3} \)
\( \frac{d^{2} y}{d x^{2}}=\mathrm{y} \) = 0 is a required differential equation

40.

y = e^3x (C cos 2x + D sin 2x) .............. (1)
\( \Rightarrow \mathrm{y} \mathrm{}^{-3 x}(-3)+\mathrm{e}^{-3 x} \mathrm{y}^{\prime} \)
\( =-\mathrm{C} 2 \sin 2 x+2 \mathrm{D} \cos 2 x \)
\(\mathrm{e}^{-3 x}\left(-3 \mathrm{y}+\mathrm{y}^{\prime}\right) =-2[\mathrm{C} \sin 2 x-\mathrm{D} \cos 2 x] \)
\(\Rightarrow \mathrm{e}^{-3 x}\left(-3 \mathrm{y}^{\prime}+\mathrm{y}^{\prime \prime}\right) +\left(-3 \mathrm{y}+\mathrm{y}^{\prime}\right)\left(\mathrm{e}^{-3 x} \mathrm{x}-3\right) \)
\( = -2[2 \mathrm{C} \cos 2 x+2 \mathrm{D} \sin 2 x] \)
\( \mathrm{e}^{-3 x}\left[-3 \mathrm{y}^{\prime}+\mathrm{y}^{\prime \prime}+9 \mathrm{y}-3 \mathrm{y}^{\prime}\right]\)
= -4 [C cos 2x + D sin 2x]
\( y^{\prime \prime}-6 y^{\prime}+9 y=-4 e^{3 x}[C \cos 2 x+D \sin 2 x] \)
\( y^{\prime \prime}-6 y^{\prime}+9 y=-4 y \)
\( y^{\prime \prime}-6 y^{\prime}+13 y=0\)