Explore topic-wise InterviewSolutions in .

Correct option  (a) y² = sin x

Explanation:

(2ydy/dx) sinx + y²  cosx = sin 2x

d/dx(y² sinx) = sin 2x

Integrating both sides,

y² sin x = -1/2 soc2x + C

Now yπ/2 = 1

1.sinπ/2 = -1/2.(-1) + C

C = 1/2

y² sin x = 1/2(1 - cos 2x)

y² sin x = 1/2.2 sin²x

y² = sin x

y = e^x(a cos x + b sin x) ….(1) 

differentiate w.r.t x 

y₁ = e^x(a cosx + b sinx) + e^x(-a sin x + b cos x) ….(2) 

putting (1) in (2); we get 

⇒ y₁ = y + e^x(b cos x – a sin x) 

⇒ y₁ – y = e^x(b cos x – a sinx) ….(3) 

again differentiating w.r.t x, we get 

⇒ y₂ – y₁ = e^x (b cos x – a sin x) + e^x (-b sin x – a cos x) ….(4) 

putting (3) and (1) in (4), we get 

⇒ y₂ – y₁= (y₁ – y) +(-y) 

⇒ y₂ – 2y₁+ 2y = 0 is the required differential equation.

y = e^2x (a + bx) …(1) 

differentiating w.r.t x, we get 

⇒ y₁= e^2x (0 + b) + (a + bx) e^2x 2 

⇒ y₁ = e^2xxb + (a + bx) e^2x 2 ….(2) 

putting (1) in (2), 

y₁ = be^2x + 2y 

⇒ y₁ – 2y = b e^2x ….(3) 

again differentiating w.r.t x, we get 

y₂ – 2y₁ = b.e^2x.2 ….(4) 

putting (3) in (4) we get 

y₂ – 2y₁ = 2y₁ – 4y 

⇒ y₂ – 4y₁ + 4y = 0 is the required differential equation.

y = a e^3x + b e^-2x ….(1) 

Differentiating w.r.t x , y, = ae^3x.3 + be^2x (-2) ….(2) 

multiply (1) by 3 

⇒ 3y = 3ae^3x + 3be^-2x 

⇒ 3ae^3x = 3y -3be^-2x …(3) 

put (3) in (2), we get y, = 3y – 3be^-2x – 2be^-2x 

⇒ y – 3y = -5 be^-2x …….. (4) 

differentiating again ⇒ y₂ – 3y₁ = -5be^-2x (-2) ….(5) 

multiply (4) by 2, we get 

⇒ 2y,- 6y = -10be^2x ….(6) 

putting (6) in (5), we get 

⇒ -(2y₁ – 6y) = y₂ -3y₁ 

⇒ y₂ – 3y₁ + 2(y₁ 3y) = 0 

⇒ y₂ -y₁ – 6y = 0 is the required differential equation.

y² = a (b² – x²) 

Differentiating w.r.t x, we get 

= yy, = -ax ………(1) 

differentiating again 

yy₂ + yy₁ = -a …(2) 

putting value of (-a) from (2) in (1) 

yy₁= (yy₂ + y₁y₁) x 

⇒ yy, = x (y₁² + yy₂) which is required differential equation. 

Correct answer is A.

2x²(d²y/dx²) - 3(dy/dx) + y = 0

Order = Highest order derivative present in the differential equation which is \(\frac{d^2y}{dx^2}\)

∴ Order = 2

Answer is (D)

In particular solution of differential equation, there is no arbitrary constant 0.

Correct answer is D.

We know,

The number of arbitrary constants of an ordinary differential equation (ODE) is given by the order of the highest derivative.

∵ differential equation is of fourth order then it will have 4 arbitrary constants in the general solution.

Correct answer is D.

We know,

The number of arbitrary constants of an ordinary differential equation (ODE) is given by the order of the highest derivative and if we give particular values to those arbitrary constants, we get particular solution in which we have 0 arbitrary constants

∴ The number of arbitrary constants in the particular solution of a differential equation of third order is 0

we have:
\(\frac{dy}{dx}\) = y² tan2x

Given that, y = 2 when x = 0

⇒ \(\frac{dy}{y^2}\) = tan2x dx

⇒ \(\int\frac{dy}{y^2}\) = ∫tan2x dx

integrating both sides

⇒ - \(\frac{1}y\)= \(\frac{log(sec2x)}2\)

⇒ - \(\frac{1}2\) = 0 + c

⇒ c = - \(\frac{1}2\)

⇒ y(1 + log cos2x) = 2

is the particular solution

we have

\(\frac{dy}{dx}\) = y tan x

given that: y=1 when x = 0

⇒ \(\frac{dy}{dx}\) = y tan x

⇒ \(\frac{dy}{y}\) = tan y dx

⇒ log y = cosec x + c

⇒ 0 = 0 + c

⇒ y cos x = 1 is the particular solution

The differential equation is \(\frac{d^2y}{dx^2}-2\frac{dy}{dx} + 2y = 0\) and the function to be proven as the solution is

y = e^x(A cosx + B sinx), we need to find the value of \(\frac{dy}{dx}.\)

\(\frac{dy}{dx}=\) e^x(A cos x + B sin x) + e^x(–A sin x + B cos x)

\(\frac{d^2y}{dx^2}=\) e^x(A cos x + B sin x) + e^x(–A sin x + B cos x) + e^x(–A sin x + B cos x) + e^x(–A cos x – B sin x)

= 2e^x(–A sin x + B cos x)

Putting the values in equation,

2e^x(–A sin x + B cos x) – 2e^x(A cos x + B sin x) – 2e^x(–A sin x + B cos x) + 2e^x(A cos x + B sin x) = 0

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, hence this function is the solution of the differential equation.

As the given equation has 3 different arbitrary constants so we can differentiate it thrice with respect to x

So, differentiating once with respect to x,

\(\frac{dy}{dx} = 2ax + b\)

Differentiating twice with respect to x,

\(\left(\frac{d^2y}{dx^2}\right)=2a\)

Now, differentiating thrice with respect to x we get,

\(\frac{d^3y}{dx^3}=0\)

Hence, \(\frac{d^3y}{dx^3}=0\) is the differential equation corresponding to y = ax² + bx + c.

The differential equation is \(\frac{dy}{dx}=\frac{y^2-x^2}{y-x}=y+x\) and the function to be proven as the solution is

y = – x – 1, now we need to find the value of \(\frac{dy}{dx}.\)

\(\frac{dy}{dx}=-1\)

Putting the values in equation,

–1 = –x –1 + x

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

The differential equation is \(2\left(\frac{dy}{dx}\right)^2 + x\frac{dy}{dx}-y=0\) and the function to be proven as the solution is y = cx + 2c², now we need to find the value of \(\frac{dy}{dx}.\)

\(\frac{dy}{dx}=\) c + 0

Putting the values,

2c² + xc – cx – 2c² = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Given curve is

y = e^-x +ax + b

Differentiating with respect to x, we get

\(\frac{dy}{dx}=e^{-x}+a\)

Differentiating again with respect to x, we get

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

\(y=x \left(\frac{dy}{dx}\right)^3+\frac{d^2y}{dx^2}.\)

Order = Highest order derivative present in the differential equation.

∴ Order = 2

Degree = Highest power of highest order derivative which is \(\frac{d^2y}{dx^2}\)

∴ Degree = 1

∴ Sum of the order and degree = 2 + 1 = 3

Given y  = cx + c²

^{⇒ \(\frac{dy}{dx}= c + 0\)}

[Differentiating with respect to x]

⇒ \(\frac{dy}{dx}=c\)

(b) xe^2x 

PI = 1/(3D² + D - 14)13e^2x

Replace D by 2.   3D² + D - 14 = 0

∴ PI = x(1/6D + 1)13e^2x

Replace D by 2.   PI = xe^2x

The auxiliary equations A.E is m² – 4m + 4 = 0 

(m – 2)² = 0 

m = 2, 2 

Roots are real and equal

The complementary function (C.F) is (Ax + B) e^2x 

The general solution is y = (Ax + B) e^2x

Given (D – 6D + 8) y = 0, D = d/dx

The auxiliary equations is 

m² – 6m + 8 = 0 

(m – 4)(m – 2) = 0 

m = 4, 2 

Roots are real and different 

The complementary function (C.F) is (Ae^4x + Be^2x ) 

The general solution is y = Ae^4x + Be^2x

(a) xdy/dx = y

y = mx; dy/dx = m ⇒ y = xdy/dx

(a) y = x/2 + c/x

(b) 1/x

(c) d²y/dx² + y = 0

(d) linear 

(a) – (iii) 

(b) – (i) 

(c) – (iv) 

(d) – (ii)

Writing the equation as 

x² dx + y² dy = a (x dy + y dx) 

x² dx + y² dy = a d(xy) 

∫x² dx + ∫y² dy = a ∫d(xy) + c 

x³/3 + y³/3 = axy + c 

Hence the general solution is x³ + y³ = 3axy + c

(a) A + B e^x 

m² – m = 0 

m(m – 1) = 0 

CF = Ae^0x + Be^x = A + B e^x

(a) (x²/2)e^2x

The highest order derivative d²y/dx² and its power is 1.

So the degree of differential equation is 1.

e^xdy - ye^xdx = e^3xdx

⇒ e^xdy = (e^3x + ye^x)dx

⇒ e^x dy  = e^x(e^2x + y)dx

⇒ \(\frac{dy}{dx} \) = y + e^2x

⇒ \(\frac{dy}{dx} \) - y = e^2x

\(\therefore\) P = -1 and Q = e^2x

\(\therefore\) I. F. = \(e^{\int pdx}=e^{\int-1dx}=e^{-x}\)

\(\therefore\) Complete solution is y x I. F. = \(\int\)(I.F.) x Q dx

⇒ y .e^-x = \(\int\)e^-x.e^2xdx = \(\int\)e^xdx = e^x + c

⇒ y = e^2x+ ce^x

i. Order = 3, degree = 1

ii. Order = 2, degree = 2

Given xdy + ydx = 0

xdy = -ydx

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

On integrating on both sides we get,

-log y = log x + c

log x + log y = c

log xy = c

xy = C

Conclusion: Therefore xy = c is the solution of xdy + ydx = 0

x \(\frac{dy}{dx} \) = cot y

Separating the variables, we get,

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

tan y dy = \(\frac{dx}{x}\)

Integrating both sides, we get,

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

log sec y = log x + log c

x cos y = c

Hence, A is the correct answer.

The order of a differential equation is the order of the highest derivative involved in the equation. So the

order comes out to be 2 as we have \(\cfrac{d^2y}{dx^2}\) and the degree is the highest power to which a derivative is raised. So the power at this order is 1. 

So the answer is 2, 1.

dy = \(\frac{x}{x^2+1}\)dx

Multiply and divide 2 in numerator and denominator of RHS,

y = \(\frac{1}2\).\((\frac{2x}{x^2+1}dx)\)

Integrating on both sides

y = \(\frac{1}2\).log (x² + 1) + c

The order of a differential equation is the order of the highest derivative involved in the equation. So the 

order comes out to be 4 as we have \(\cfrac{d^4y}{dx^4}\) and the degree is the highest power to which a derivative is raised. 

But when we open the Cos x series, we get \(1-\cfrac{x^2}{2!}+\cfrac{x^2}{2!}-\cfrac{x^6}{6!}+-----\) ,This leads to an undefined power on the highest derivative. Therefore the deg9ee of this function becomes undefined. 

So the answer is 4, not defined

\(\frac{1}{1+y^2}dy\) = (1 + x)dx

Integrating on both sides

\(\int\frac{1}{1+y^2}dy\) = \(\int(1+x)dx\)

⇒ tan^-1y = x + \(\frac{x^2}2\) + c

Given that the product of slope of tangent and y coordinate equals the x-coordinate i.e.,y \(\frac{dy}{dx}\) = x

We have ydy = xdx

⇒ ∫ydy = ∫xdx

⇒ y²/2 = x²/2 + c

For the curve passes through (0, -2), we get c = 2, 

Thus, the required particular solution is:-

∴ y² = x² + 4

(a) d²y/dx² = 0

5x(dy/dx)²-d²y/dx²-6y=logx

Degree of differential equation is the highest power of the highest derivative. In above d²y/dx² is the highest order of derivative.​

Its degree = 1.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

So, in this question the order of the differential equation is 2 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term \(\frac{dy}{dx}\) is multiplied by itself so the given equation is non-linear.

\(\frac{dy}{dx}\) = 1 - y

\(\frac{1}{1-y}dy\) = dx

Integrating on both sides

\(\int\frac{1}{1-y}dy\) = \(\int{dx}\)

⇒ log |1 - y| = x + c

The order of a differential equation is the order of the highest derivative involved in the equation. So the

order comes out to be 1 as we have \(\cfrac{dy}{dx}\) and the degree is the highest power to which a derivative is raised. 

So the power at this order is 3. Because the Sine function is not at any derivative, so it doesn’t destroy the polynomial .Hence the degree is 3. 

So the answer is 1, 3.

e^y = e^x + c from given equation, we have e^ydy = e^xdx.

The equation of the circle passing through the point (O,k)and radius 3 is of the form

(x - 0)² + (y - k)² = 9

⇒ x² + (y + k)² = 9 .........(i)

Differentiatitng w.r.t to x we get;

2x + 2(y - k) y₁ = 0 ⇒ x + (y - k)y₁ = 0

⇒ x + (y - k)y₁ = \(\frac{x}{y_1}\)

Substituting in (1) We get ;

⇒ x² + (\(-\frac{x}{y_1}\))² = 9 ⇒ x² + \(\frac{x^2}{{y_1}^2}\) = 9

⇒ x²y₁² + x^{2 =} 9y₁²

(y’) + y = e^x 

Here the order is 1 and hence the degree is 1.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

Here in this question the dependent variable is t, and thus the order of the equation is 2, and the degree of the equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable t and its derivative is multiplied together \(st\frac{dt}{ds}\)so this equation is non-linear differential equation.

\(\Big(\frac{d^2y}{dx^2}\Big)^1\) = cos3x + sin3x

Here order is 2 and hence degree is 1.

The order of a differential equation is the order of the highest derivative involved in the equation. So the 

order comes out to be 2 as we have \(\cfrac{d^2s}{dt^2}\) and the degree is the highest power to which a derivative is raised. 

So the power at this order is 2. 

So the answer is 2, 2.

Answer is (C) x = vy