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Right answer is (c) X^2 d^2 y/dx^2 + xdy/dx – y = 0

Easiest EXPLANATION: GIVEN, y = Ax + B/x

=> xy = Ax^2 + B……….(1)

Differentiating (1) with respect to x, we get,

d(xy)/dx = d/dx(Ax^2 + B)

or, xdy/dx + y = A * 2x……….(2)

Differentiating again with respect to x, we get,

x*d^2y/dx^2 + dy/dx + dy/dx = A*2……….(3)

Eliminating A from (2) and (3) we get,

x^2 d^2 y/dx^2 + 2xdy/dx = 2Ax[multiplying (3) by x]

or, x^2 d^2 y/dx^2 + 2xdy/dx = xdy/dx + y [using (2)]

or, x^2 d^2 y/dx^2 + xdy/dx – y = 0

Right option is (d) d^2x/dt^2 + α^2x = 0

For EXPLANATION I WOULD say: SINCE, x = a cos(αt + β)

Therefore, dx/dt = a cos(αt + β)

And, d^2x/dt^2 = -a α^2 cos(αt + β)

= -α^2 a cos(αt + β)

Or, d^2x/dt^2 = -α^2 [as a cos(αt + β) = x]

So, d^2x/dt^2 + α^2x = 0

Correct choice is (b) False

The best explanation: The given statement is false.

Given differential EQUATION: \(\frac{d^2 y}{dx^2}\)-3 \(\frac{dy}{dx}\)=0 –(1)

Consider the function y=3 cos⁡x

Differentiating w.r.t x, we GET

\(\frac{dy}{dx}\)=-3 sin⁡x

Differentiating again w.r.t x, we get

\(\frac{d^2 y}{dx^2}\)=-3 cos⁡x

Substituting the VALUES of \(\frac{dy}{dx}\) and \(\frac{d^2 y}{dx^2}\) in equation (1), we get

\(\frac{d^2 y}{dx^2}\)-3 \(\frac{dy}{dx}\)=-3 cos⁡x-3(-3 sin⁡x)

=9 sin⁡x-3 cos⁡x≠0.

Hence, y=3 cos⁡x, is not a SOLUTION of the equation \(\frac{d^2 y}{dx^2}\)-3 \(\frac{dy}{dx}\)=0.

The correct choice is (a) y=5 cos⁡3x

The explanation: Consider the FUNCTION y=5 cos⁡3x

Differentiating w.r.t x, we get

y’=\(\frac{dy}{dx}\)=-15 sin⁡3x

Differentiating again w.r.t x, we get

y”=\(\frac{d^2 y}{dx^2}\)=-30 cos⁡3x

⇒y”+6y=0.

Hence, the function y=5 cos⁡3x is a solution for the DIFFERENTIAL EQUATION y”+6y=0.

The CORRECT choice is (d) y=2x

To EXPLAIN I would say: Consider the function y=2x

Differentiating w.r.t X, we get

y’=\(\frac{DY}{dx}\)=2

Substituting in the equation xy’-y, we get

xy’-y=x(2)-2x=2x-2x=0

Therefore, the function y=2x is a SOLUTION for the differential equation xy’-y=0.

The correct OPTION is (d) 1

For explanation I would say: The given is a polynomial differential equation in \(\frac{d^3 y}{dx^3}\). THEREFORE, its degree will be the power raised to the HIGHEST ORDER derivative \(\frac{d^3 y}{dx^3}\) which is 1.

Correct CHOICE is (a) x(d^2y/ dx) + 2(dy/dx) = xy

To explain: We have xy = ae^x + be^-x ……(1)

Differentiating (1) with RESPECT to x, we get

x(dy/dx) + y = ae^x + be^-x…..(2)

Differentiating (2) now, with respect to x, we get

x(d^2y/ dx) + dy/dx + dy/dx = ae^x + be^-x

From (1),

ae^x + be^-x = xy, so that we get

x(d^2y/ dx) + 2(dy/dx) = xy

which is the required differential equation.

The correct answer is (c) 7y^2=2x^3+5

To EXPLAIN I would say: Given that, \(\frac{dy}{dx}=\frac{3x^2}{7y}\)

Separating the variables, we get

7y dy=3x^2 dx

Integrating both sides, we get

\(7\int y \,dy=3\int x^2 \,dx\)

\(\frac{7y^2}{2}=3(\frac{x^3}{3})+C\)

\(\frac{7y^2}{2}=x^3\)+C –(1)

Given that y=1, when x=1

Substituting the VALUES in equation (1), we get

\(\frac{7(1)^2}{2}=(1)^3+C\)

\(C=\frac{7}{2}-1=\frac{5}{2}\)

Hence, the particular solution of the given differential equation is:

\(\frac{7y^2}{2}=x^3+\frac{5}{2}\)

⇒7y^2=2x^3+5

The correct CHOICE is (c) 2

Explanation: The number of ARBITRARY constants in a GENERAL solution of a n^th order differential EQUATION is n.

Therefore, the number of arbitrary constants in the general solution of a second order D.E is 2.

The correct OPTION is (d) x^4-2x^2+4x-4y+C=0

Explanation: Given that, \(\FRAC{dy}{dx}\)=2-x+x^4

Separating the VARIABLES, we get

dy=(2-x+x^3)dx

Integrating on both SIDES, we get

\(\int dy=\int 2-x+x^3 \,dx\)

\(y=2x-\frac{x^2}{2}+\frac{x^4}{4}+C_1\)

4y=4x-2x^2+x^4+4C1

∴x^4-2x^2+4x-4y+4C1=0

x^4-2x^2+4x-4y+C=0 (where 4C1=C)

Correct option is (C) ORDER- 3, DEGREE-3

Easy EXPLANATION: The highest order derivative in the equation is y”’. Hence, the order is 3. The equation is polynomial in y”’. Therefore, the degree of the D.E will be the power of the derivative y”’ i.e. 2.

The correct answer is (a) True

The EXPLANATION: The GIVEN STATEMENT is true. The degree of a DIFFERENTIAL equation is DEFINED only when the differential equation is a polynomial in its derivatives. The degree of a D.E will not be defined if it is not polynomial.

The correct answer is (b) a^2 log (x + √( a^2 + x^2)) + C

To elaborate: The given form of equation can be written as,

dy/dx + 1/√(a^2 + x^2) * y = (√(a^2 + x^2) – x)/√(a^2 + x^2) ……(1)

We have, ∫1/√(a^2 + x^2)dx = log(x + √(a^2 + x^2))

THEREFORE, INTEGRATING factor is,

e^∫1/√(a^2 + x^2) = e^log(x + √(a^2 + x^2))

= x + √(a^2 + x^2)

Therefore, multiplying both sides of (1) by x + √(a^2 + x^2) we get,

x + √(a^2 + x^2dy/dx + (x + √(a^2 + x^2))/ √(a^2 + x^2)*y = (x + √(a^2 + x^2))(√(a^2 + x^2) – x)/√(a^2 + x^2)

or, d/dx[x + √(a^2 + x^2)*y] = (a^2 + x^2) ………..(2)

Integrating both sides of (2) we get,

(x + √(a^2 + x^2) * y = a^2∫dx/√(a^2 + x^2)

= a^2 log (x + √(a^2 + x^2)) + c

The correct ANSWER is (d) not defined

To elaborate: The given differential equation is not POLYNOMIAL. Hence, the DEGREE of the differential equation will not be defined.

Correct CHOICE is (b) 2

Best explanation: The HIGHEST order derivative in the given differential equation is \(\frac{d^2 y}{dx^2}\). THEREFORE, the order of the D.E is 2.

The correct option is (b) d^2y/dx^2 – 4dy/dx + 3y = 0

For explanation: The original given equation is,

y = ae^3x + be^x ……….(1)

DIFFERENTIATING the above equation, we get

dy/dx = 3ae^3x + be^x ……….(2)

d^2y/dx^2 = 9ae^3x + be^x ……….(3)

Now, to obtain the final equation we have to make the RHS = 0,

So, to get RHS = 0, we MULTIPLY, (1) with 3 and (2) with -4,

Thus , 3y = 3ae^3x + 3be^x ……….(4)

And -4(dy/dx) = -12ae^3x – 4 be^x……….(5)

Now, adding the above three EQUATIONS,(i.e. (3) + (4) + (5)) we get,

d^2y/dx^2 – 4dy/dx + 3y = 0

The correct choice is (d) 2

The EXPLANATION: The given DIFFERENTIAL equation is polynomial in \(\FRAC{d^2 y}{dx^2}\). Hence, the DEGREE of the given D.E will be the power raised to the HIGHEST order derivative \(\frac{d^2 y}{dx^2}\) which is 2.

Correct option is (b) ke^2Y/X where, X = x + 2 and Y = y – 2

For explanation I would SAY: Let, X = x + 2 and Y = y – 2

Then, dY/dX = (X + Y)^2/XY

So, let, Y = vX

dY/dX = V + x dv/dX

=>v + Xdv/dX = (v + 1)^2/v

=>v/1 + 2V dv = dX/X

=> (1 – 1/1 + 2v)dv = 2dx/x

=> v – ½ LOG (1 + 2v) = 2 log X + c

This means, X^4(1 + 2Y/X)

= ke^2Y/X where, X = x +2 and Y = y – 2

Correct option is (a) y = x/(1 – a) – 1/a + cx^a

The best explanation: dy/dx – a/x * y = (x + 1)/x …….(1)

Multiplying both sides of equation (1) by

e^∫-a/xdx

= e^-a log x

= e^log x^-a

= x^-a

We get, x^-ady/dx – x^-a (a/x)y = x^-a (x + 1)/x

Or, d/dx(y . x^-a) = x^-a + x^-a – 1 …….(2)

INTEGRATING both sides of (2) we get,

y. x^-a = x^-a + 1/(-a + 1) + x^-a – 1 + 1/(-a -1 + 1) + C

= x^-a.x/(1 – a) + x^-a/-a + c

Or, y = x/(1 – a) – 1/a + cx^a

Right OPTION is (B) x^4+8x+y^4-16y+C=0

The EXPLANATION: GIVEN that, \(\frac{dy}{dx}=\frac{2+x^3}{4-y^3}\)

Separating the variables, we get

(4-y^3)dy=(2+x^3)dx

Integrating both sides, we get

\(\INT 4-y^3 \,dy=\int 2+x^3 \,dx\)

\(4y-\frac{y^4}{4}=2x+\frac{x^4}{4}+C_1\)

x^4+8x+y^4-16y+C=0 (where 4C1=C)

The correct choice is (c) Ae^4x + Be^-x

To elaborate: d^2y/dx^2 – 3dy/dx + 4y = 0…..(1)

LET, y = e^mx be a trial solution of (1), then,

=> dy/dx = me^mx and d^2y/dx^2 = m^2e^mx

Clearly, y = e^mx will satisfy equation (1). Hence, we have,

m^2e^mx – 3M * e^mx – 4e^mx = 0

=>m^2 – 3m – 4 = 0 (as e^mx ≠ 0)…….(2)

=> m^2 – 4m + m – 4 = 0

=> m(m – 4) + 1(m – 4) = 0

Or, (m – 4)(m + 1) = 0

Thus, m = 4 or m = -1

Clearly, the roots of the auxiliary equation (2) are real and unequal.

Therefore, the required general solution of (1) is

y = Ae^4x + Be^-x where A and B are constants.

The correct CHOICE is (c) dy/dx = [(xy – 2) ± √(1 – xy)]/ x^2

To explain: The equation of tangent to the curve y = f(x), at POINT (x, y), is

Y – y = dy/dx * (X – x) …..(1)

Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))

Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))

 ALSO, the AREA of the triangle formed by (1) with the coordinate axes is 2, so that,

(x – y/(dy/dx))* (y – x/(dy/dx)) = 4

Or, (y – x/(dy/dx))^2 – 4dy/dx = 0

Or, x^2(dy/dx)^2 – 2(xy – 2)dy/dx + y^2 = 0

Solving for dy/dx we get,

dy/dx = [(xy – 2) ± √(1 – xy)]/ x^2

Right choice is (b) (log⁡y)^2-(log⁡x)^2=0

Easy explanation: \(\frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}\)

SEPARATING the variables, we get

\(\frac{5 \,log⁡y}{y} dy=\frac{9 \,log⁡x}{x} dx\)

INTEGRATING both sides, we get

\(5\int \frac{log⁡y}{y} \,dy=9\int \frac{log⁡x}{x} \,dx\) –(1)

First, for integrating \(\frac{log⁡y}{y}\)

Let log⁡y=t

Differentiating w.r.t y, we get

\(\frac{1}{y}\) dy=dt

∴\(\int \frac{log⁡y}{y} \,dy=\int t \,dt\)

=\(\frac{t^2}{2}=\frac{log⁡y^2}{2}\)

Similarly integrating \(\frac{log⁡x}{x}\)

Let log⁡x=t

Differentiating w.r.t x, we get

\(\frac{1}{x}\) dx=dt

∴\(\int \frac{log⁡x}{x} dy=\int t \,dt\)

=\(\frac{t^2}{2}=\frac{(log⁡x)^2}{2}\)

Hence, equation (1), becomes

\(\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C\) –(2)

GIVEN y=2, we get x=2

Substituting the values in equation (2), we get

\(\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C\)

C=0

Therefore, the equation becomes \((log⁡y)^2=(log⁡x^2)\)

∴(log⁡y)^2-(log⁡x)^2=0.

Correct answer is (b) 1

For EXPLANATION I WOULD say: The given equation is a polynomial differential equation. Therefore, the degree of the equation will the power of the highest DERIVATIVE \(\frac{d^2 y}{dx^2}\) i.e. 1.

The CORRECT answer is (B) x^2 + y^2 = 4tan^-1(√(y/x)) + c

Best explanation: The GIVEN equation can be written as on simplifying,

xdx + ydy + (YDX – XDY)/(√xy)(x + y) = 0

the above equation can be written as,

 or, 1/2d(x^2 + y^2) = (xdy – ydx)/(x^2 (√y/x) (1 + y/x))

= 2/(1 + y/x)* d(√y/x)

Thus, x^2 + y^2 = 4tan^-1(√(y/x)) + c

Correct choice is (C) 29 cm/sec

To elaborate: Let X cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.

By question, dv/dt = 3t^2 – 5t

Or, dv = 3t^2 dt – 5tdt

Or, ∫dv = 3∫t^2 dt – 5∫t dt

Or, v = t^3 – (5/2)t^2 + c……….(1)

Given, v = 5, when t = 0; HENCE PUTTING these values in equation (1) we GET, c = 5

Thus v = t^3 – (5/2)t^2 + 5

Or, dx/dt = t^3 – (5/2)t^2 + 5 ………..(2)

Thus, the velocity of the particle at the end of 4 seconds,

= [v]t = 4 = (4^3 – (5/2)4^2 + 5 ) cm/sec [putting t = 4 in (2)]

= 29 cm/sec

The CORRECT choice is (a) ORDER -3, Degree-1

Easy EXPLANATION: In the D.E y”’-(4y’)^3=0, the highest order derivative is y”’. Therefore, the order of the D.E is 3. Since it is a polynomial equation, the degree will be the power RAISED to y”’. Therefore, the degree is 1.

Correct CHOICE is (C) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]

BEST explanation: Put x = X + h, Y = Y + k,

We have, dY/dX = (X + 2Y +(h + 2k + 3))/ 2X + 3Y + (2h + 3k + 4)

So, (a – b)x = (a – b)

To determine h and k we set,

2h + 3k + 4 = 0 and h + 2k + 3 = 0

=> h = 1 and k = – 2

Therefore, dY/dX = (X + 2Y) / (2X + 3Y)

Putting Y = VX, we get,

V + X dV/dX = (1 + 2V)/(2 + 3V)

= (1 + 2V)/(3V^2 – 1)*dV = -dX/X

=> [(2 + √3)/(2(√3V – 1)) – (2 – √3)/(2(√3V – 1))] dV = -dX/X

Simplifying it further, we get;

[(2 + √3)/2√3 * (log (√3Y – X)) – (2 – √3)/2√3 * (log (√3Y – X))]

Where, X = x – 1 and Y = y + 2

The correct answer is (c) ORDER – 3, Degree – 1

The EXPLANATION is: In the differential equation, the highest order derivative is \(\frac{d^3 y}{dx^3}\). THEREFORE, the order of the differential equation is 3. The given is a POLYNOMIAL equation in \(\frac{d^3 y}{dx^3}\). Hence, the degree will be the power raised to \(\frac{d^3 y}{dx^3}\) i.e. 1

Right answer is (c) X \(\frac{d^2 y}{dx^2}\)–\(\frac{dy}{dx}\)=0

The EXPLANATION is: Consider the function y=3x^2

Differentiating w.r.t x, we get

\(\frac{dy}{dx}\)=6x –(1)

Differentiating (1) w.r.t x, we get

\(\frac{d^2 y}{dx^2}\)=6

∴\(\frac{xd^2 y}{dx^2}-\frac{6dy}{dx}\)=6x-6x=0

Hence, the function y=3x^2 is a solution for the differential EQUATION x \(\frac{d^2 y}{dx^2}\)-6 \(\frac{dy}{dx}\)=0.

Right answer is (c) d^3y/dx^3 = 0

To EXPLAIN I would say: The equation of family of parabolas is Ax^2 + Bx + C = 0 where, A, B, C are arbitrary constant.

By differentiating the equation with respect to x TILL all the CONSTANTS get eliminated,

Hence, d^3y/dx^3 = 0

The correct option is (d) Circle

The explanation is: Equation of the normal at a point P(x, y) is given by

Y – y = -1/(dy/dx)(X – x) ….(1)

Let the point Q at the x-axis be (x1 , 0).

From (1), we get

y(dy/dx) = x1 – x….(2)

Now, giving that PQ^2 = k^2

Or, x1 – x + y^2 = k^2

=>y(dy/dx) = ± √(k^2 – y^2)….(3)

(3) is the required differential equation for such curves,

Now solving (3) we get,

∫-dy/√(k^2 – y^2) = ∫-dx

Or, x^2 + y^2 = k^2 passes through (0, k)

Thus, it is a circle.

Correct choice is (c) [2x/3 *(dy/dx) – y][ 2x/3 *dy/dx]^2 = x^3

The best I can explain: The given FAMILY is c(y + c)^2 = x^3

Differentiating once, we get

c[2(y + c)]dy/dx = 3x^2

=> 2x^3 (y + c)/(y + c)^2 * dy/dx = 3x^2

=> 2x^3/(y + c) * dy/dx = 3x^2

Or, 2x (y + c)/(y + c)^2 * dy/dx = 3

=> 2x/3 *[dy/dx] = (y + c)

 => c = 2x/3 *[dy/dx] – y

Substituting c back into equation (1), we get

[2x/3 *(dy/dx) – y][ 2x/3 *dy/dx]^2 = x^3

which is the REQUIRED differential equation

Correct CHOICE is (a) 1

The explanation is: In the polynomial differential equation \(3(\FRAC{d^2 y}{dx^2})-(\frac{dy}{dx})^2\)+sin⁡y=0, the POWER raised to the highest DERIVATIVE \(\frac{d^2 y}{dx^2}\) is 1. Therefore, the degree of the differential equation is 1.

Right choice is (B) 2

Explanation: The HIGHEST ORDER derivative in the given D.E is y”. Therefore, the order of the D.E is 2.

Right answer is (d) 1

To explain I would say: The highest order derivative in the given D.E is \(\FRAC{dy}{DX}\). Hence, the order of the given differential EQUATION is 1.

Right choice is (b) 30(2/3)

Easiest EXPLANATION: Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the END of t SECONDS. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.

By question, dv/dt = 3t^2 – 5t

Or, dv = 3t^2 dt – 5t dt

Or, ∫dv = 3∫t^2 dt – 5∫t dt

Or, v = t^3 – (5/2)t^2 + c……….(1)

Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5

Thus v = t^3 – (5/2)t^2 + 5

Or, dx/dt = t^3 – (5/2)t^2 + 5………..(2)

Or, dx = t^3 dt – (5/2)t^2 dt + 5 dt

Integrating this we get,

x = (1/4)t^4 – (5/2)t^3/3 + 5t + k ……….(3)

By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.

Thus, x = (1/4)t^4 – (5/6)t^3 + 5t……….(4)

Thus, the velocity of the particle at the end of 4 seconds,

= [x]t = 4 = (1/4)4^4 – (5/6)4^3 + 5(4) [putting t = 4 in (4)]

= 30(2/3) cm

Correct choice is (c) y=5x-x^2+1

Best explanation: GIVEN that, \(\FRAC{DY}{dx}+2x=5\)

\(\frac{dy}{dx}=5-2x\)

Separating the variables, we get

dy=(5-2x)dx

Integrating both sides, we get

\(\int dy=\int 5-2x \,dx\)

y=5x-x^2+C –(1)

Given that, y=5, when x=1

⇒5=5(1)-(1)^2+C

∴C=1

Substituting value of C to equation (1), we get

y=5x-x^2+1 which is the particular solution of the given differential equation.

The correct OPTION is (b) y = e^2x sin3x + Ax + B

The explanation is: Given, d^2y/dx^2 = e^2x(12 cos3x – 5 sin3x) ………(1)

Integrating (1) we get,

dy/dx = 12∫ e^2xcos3x dx – 5∫ e^2x sin3x dx

= 12 * (e^2x/2^2 + 3^2)[2cos3x + 3sin3x] – 5 * (e^2x/2^2 + 3^2)[2sin3x – 3cos3x] + A (A is integrationconstant)

So dy/dx = e^2x/13 [24 cos3x + 36 sin3x – 10 sin3x + 15 cos3x] + A

= e^2x/13(39 cos3x + 26 sin3x) + A

=> dy/dx = e^2x(3 cos3x + 2 sin3x) + A ……….(2)

Again integrating (2) we get,

y = 3*∫ e^2x cos3xdx + 2∫ e^2x sin3xdx + A ∫dx

y = 3*(e^2x/2^2 + 3^2)[2cos3x + 3sin3x] + 2*(e^2x/2^2 + 3^2)[2sin3x – 3cos3x] + Ax + B(B is integration constant)

y = e^2x/13(6 cos3x + 9 sin3x + 4 sin3x – 6 cos3x) + Ax + B

or, y = e^2x sin3x + Ax + B