NDA Differential Equations previous year questions with solutions

Answer: (a) 3

Method. rationalise: clear the fractional powers

Square both sides: (d²ydx²)³ = (dydx)⁵.
The highest derivative is d²ydx², now with power 3, so the degree is 3.

Answer: (c) Both I and II

Method. substitute each function into its equation

I. y = xe^2x: dydx = e^2x + 2xe^2x = e^2x(1 + 2x).
Right side: y(2 + 1x) = xe^2x·2x + 1x = e^2x(2x + 1). Equal, so I is correct.
II. y = x ln|x| + cx: dydx = ln|x| + 1 + c.
Right side: x + yx = 1 + ln|x| + c. Equal, so II is correct.
Both are correct.

Answer: (a) 1 and 1

Method. read off order and degree

Write it as dxdy + xy = y² (dividing by y dy), or y + (x − y³)dydx = 0 (dividing by dx).
Only first derivatives occur, to the first power: order 1, degree 1.

Answer: (d) 4xy − y⁴ = c

Method. exact differential

y dx + x dy = y³dy, i.e. d(xy) = y³dy.
Integrate: xy = y⁴4 + C, so 4xy − y⁴ = c.

Shortcut. y dx + x dy is d(xy): integrate at once.

Answer: (d) 2, 5

Method. rationalise the fractional power, then read off order and degree

0.6 = 35. Raise both sides to the 5th power: (2 − (dydx)²)³ = (d²ydx²)⁵.
Highest derivative is d²ydx², so the order is 2; its power is 5, so the degree is 5.
Order, degree = 2, 5.

Answer: (d) 3, 2

Method. read off order and degree

The highest derivative is d³ydx³, so the order is 3.
It occurs with power 2, and the equation is a polynomial in the derivatives (x² and sin x are only coefficients), so the degree is 2.

Answer: (b) 2 only

Method. order is the highest derivative; degree needs a polynomial in the derivatives

Statement 1: cos(dydx) is not a polynomial in dydx, so the degree of the equation is not defined. The claim 'degree is 1' is false.
Statement 2: the highest derivative present is d²ydx², so the order is 2. The cosine does not affect the order. The statement is true.
Only 2 is correct.

Answer: (d) 4

Method. degree = power of the highest derivative after clearing fractional powers

The highest derivative is d²ydx², raised to 43. Cube both sides to remove the fraction: (1 + (dydx)²)³ = (d²ydx²)⁴.
Now the equation is a polynomial in the derivatives and the highest derivative has power 4, so the degree is 4 (not 43).

Answer: (a) 1

Method. degree after removing the radical

x = ⁠√(1⁠ + ⁠d²ydx²⁠). Squaring: x² = 1 + d²ydx², i.e. d²ydx² = x² - 1.
This is a polynomial in the derivatives; the highest derivative d²ydx² has power 1. Degree 1.

Answer: (b) d²ydx² − y = 0

Method. eliminate a and b by differentiating twice

y = ae^x + be^-x gives y' = ae^x - be^-x and y'' = ae^x + be^-x = y.
Hence d²ydx² − y = 0.

Answer: (b) 2 and 3

Method. differentiate to remove the integral, then clear the fractional power

Differentiating kdydx = ∫[1 + (dydx)²]^2/⁠3 dx gives kd²ydx² = [1 + (dydx)²]^2/⁠3.
Cube both sides: k³(d²ydx²)³ = [1 + (dydx)²]².
Highest derivative d²ydx², power 3: order 2, degree 3.

Answer: (a) 1

Method. degree = power of the highest-order derivative

The highest-order derivative is d⁴ydx⁴, and it occurs to the first power (the factor x² is a coefficient, and dydx squared is a lower-order derivative).
Degree 1.

Answer: (c) Both 1 and 2

Method. read off order and degree

Highest derivative is d²ydx², appearing to the first power (the squared term is dydx, a lower-order derivative): degree 1, so statement 1 is correct.
The order is 2, so statement 2 is correct. Both are correct.

Answer: (a) d²xdt² + μx = 0

Method. a periodic solution needs a sine/⁠cosine solution

(a) x'' + μx = 0 with μ > 0 has x = A cos(⁠√μ⁠ t) + B sin(⁠√μ⁠ t), periodic.
(b) x'' - μx = 0 gives exponentials e^{±⁠√μ⁠ t}, not periodic.
(c) x dx = -μt dt gives x² = c − μt², which is not periodic (and exists only on a bounded interval).
(d) dxx = -μt dt gives x = ce^-μt2/⁠2, not periodic.
Only (a).

Answer: (c) 3, 2

Method. order = highest derivative; degree = its power

(d³ydx³)² = y⁴ + (dydx)⁵ is polynomial in the derivatives.
Highest derivative d³ydx³: order 3, power 2: degree 2.
Order 3, degree 2, which the option lists as '3, 2'.

Answer: (b) 2 and 2

Method. order = highest derivative; degree = its power once the equation is a polynomial in the derivatives

The equation [1 + (dydx)²]³ = ρ²(d²ydx²)² is already a polynomial in the derivatives (ρ is a constant).
Highest derivative: d²ydx², so order 2; its power is 2, so degree 2.

Answer: (d) 4, 1

Method. degree = power of the highest-order derivative after the equation is made a polynomial in the derivatives

Put p = dydx, so dxdy = 1p. The equation is y = xp² + 1p².
Multiply by p²: yp² = xp⁴ + 1, a polynomial in p.
Highest derivative is dydx (order 1) and its highest power is 4 (degree 4).
Degree and order respectively: 4, 1.

Answer: (c) x dydx − y = 0

Method. y = mx, eliminate m

Lines through the origin: y = mx, m arbitrary.
dydx = m = yx, so xdydx − y = 0.

Answer: (a) 1

Method. eliminate c; one derivative to the first power

y² = ⁠√c⁠ x. Differentiate: 2yy₁ = ⁠√c.⁠
Substitute: y² = 2yy₁x, so y = 2xdydx (y ≠ 0).
It is first order and first power in dydx, so the degree is 1.

Answer: (c) d²ydx² − 2dydx + 2y = 0

Method. differentiate twice and eliminate a and b

y = e^x(a cos x + b sin x).
y₁ = y + e^x(-a sin x + b cos x).
y₂ = y₁ + (y₁ - y) + e^x(-a cos x − b sin x) = 2y₁ - y − y = 2y₁ - 2y.
So d²ydx² − 2dydx + 2y = 0.

Answer: (d) y d²ydx² + (dydx)² = 0

Method. differentiate twice and eliminate a

y² = 4a(x − b), a and b arbitrary.
Differentiate: 2yy₁ = 4a.
Differentiate again: 2y₁² + 2yy₂ = 0, so yd²ydx² + (dydx)² = 0.

Answer: (a) 1

Method. count independent constants

Only c is arbitrary in y² = 2cx + 2c⁠√c⁠ (the family has one parameter), so the differential equation has order 1.

Answer: (b) 3

Method. eliminate c, rationalise, read off the degree

Differentiate: 2ydydx = 2c, so c = yy₁.
Substitute: y² = 2xyy₁ + 2(yy₁)^3/⁠2, so y² - 2xyy₁ = 2(yy₁)^3/⁠2.
Square to remove the fractional power: (y² - 2xyy₁)² = 4y³y₁³.
The highest power of y₁ is 3 (the left side has degree 2 only), so the degree is 3.

Answer: (b) xdydx − 2y = 0

Method. form the family, differentiate, eliminate the constant

Vertex at the origin, axis along positive y: x² = 4ay.
Differentiate: 2x = 4adydx, so 4a = 2xy₁.
Put back: x² = 2xy₁· y, so xdydx = 2y, i.e. xdydx − 2y = 0.
(Option (d) reduces to 2xdydx = y, which is a different equation.)

Answer: (b) 2

Method. count independent arbitrary constants

All such ellipses are x²a² + y²b² = 1 with two independent constants a and b.
The differential equation of a family with n independent constants has order n, so the order is 2.

Shortcut. Two constants, so order 2.

Answer: (b) 2

Method. form the equation of the family, eliminate the parameter, rationalise, read off the degree

A circle in the first quadrant touching both axes has centre (a, a) and radius a: (x − a)² + (y − a)² = a², one constant.
Differentiate: (x − a) + (y − a)y₁ = 0, so a = x + yy₁1 + y₁.
Then x − a = y₁(x − y)1 + y₁ and y − a = y − x1 + y₁.
Put in the circle equation: (x − y)²(1 + y₁²) = (x + yy₁)².
Expand: (x² - 2xy)y₁² - 2xy y₁ + (y² - 2xy) = 0, a polynomial of degree 2 in y₁, order 1.
Degree = 2.

Answer: (b) xy₂ + 2y₁ = 0

Method. differentiate twice and eliminate the constants

y = A − Bx: y₁ = Bx², y₂ = -2Bx³.
So y₂ = -2xy₁, i.e. xy₂ + 2y₁ = 0.

Answer: (a) 1

Method. order = number of arbitrary constants

The solution y² + 2cy − cx + c² = 0 contains one arbitrary constant c.
Eliminating it needs one differentiation, so the order is 1.

Answer: (d) dydx = -4xy²

Method. differentiate and express in terms of y

y = (2x² - c)^-1 gives dydx = -(2x² - c)^-2 · 4x = -4x y².
Hence dydx = -4xy².

Answer: (b) e^x tan y dx + (1 − e^x)sec²y dy = 0

Method. differentiate once and eliminate c

tan y = c(1 − e^x) gives sec²y dydx = -ce^x.
From the given relation c = tan y1 − e^x, so (1 − e^x) sec²y dydx = -e^x tan y.
Hence e^x tan y dx + (1 − e^x) sec²y dy = 0.

Answer: (c) x² - y² + 2xy dydx = 0

Method. write the family, differentiate once, eliminate the parameter

A circle touching the y-axis at the origin has its centre on the x-axis: x² + y² - 2ax = 0.
Differentiate: 2x + 2yy' - 2a = 0, so a = x + yy'.
Substitute: x² + y² - 2x(x + yy') = 0, i.e. y² - x² - 2xyy' = 0.
Hence x² - y² + 2xydydx = 0.

Answer: (d) d²ydx² + a²y = 0

Method. differentiate twice

y = p cos(ax) + q sin(ax); y' = -ap sin(ax) + aq cos(ax); y'' = -a²p cos(ax) − a²q sin(ax) = -a²y.
Hence d²ydx² + a²y = 0.

Answer: (d) 4

Method. order = number of independent arbitrary constants

cos x, sin x, e^-x and 1 are linearly independent, so a, b, c, d are four genuinely independent constants.
The equation is (D² + 1)(D + 1)D y = 0, i.e. y'''' + y''' + y'' + y' = 0, of order 4.

Answer: (a) 1, 2

Method. eliminate a, then read order and degree

Differentiating y² = 4a(x − a): 2yy' = 4a, so a = yy'2.
Substituting: y² = 2yy'(x − yy'2) = 2xyy' - y²y'².
So y²(y')² - 2xyy' + y² = 0: highest derivative y' (order 1), power 2 (degree 2).
Order 1, degree 2.

Answer: (d) y'' + y = 0

Method. reduce to a single sinusoid, which has only two essential constants

sin(x + C) + cos(x + C) = ⁠√2⁠ sin(x + C + π4), so y = A⁠√2⁠ sin(x + C + π4) = B sin(x + D): only two independent constants, hence order 2.
y' = B cos(x + D), y'' = -B sin(x + D) = -y.
Hence y'' + y = 0.

Answer: (a) (x² - 2y²)p² - 4pxy − x² = 0

Method. differentiate once and eliminate a

Differentiating y² - 2ay + x² = a²: 2yp − 2ap + 2x = 0, so a = yp + xp.
The given relation is y² + x² = a(a + 2y), and a + 2y = 3yp + xp.
So (x² + y²)p² = (yp + x)(3yp + x) = 3y²p² + 4xyp + x².
Hence (x² - 2y²)p² - 4xyp − x² = 0.

Answer: (a) sec(y − x) = ce^x2/⁠2

Method. substitute v = y − x

Let v = y − x, so dvdx = dydx − 1 = x cot v.
Separate: tan v dv = x dx. Integrate: -ln|cos v| = x²2 + C.
So ln|sec v| = x²2 + C, i.e. sec(y − x) = ce^x2/⁠2.

Answer: (a) cos(y − qx) = p

Method. dydx is a constant

cos(dydx) = p with p, q constants means dydx = m, a constant with cos m = p.
Then y = mx + C and y(0) = q gives C = q, so m = y − qx.
Hence cos(y − qx) = p.

Answer: (a) -2

Method. differentiate dydx = e^-2y once

e^x+ydydx = e^x-y gives dydx = e^-2y.
Differentiate: d²ydx² = -2e^-2ydydx = -2(dydx)².
So d²ydx²(dxdy)² = -2(dydx)²·1(dy/⁠dx)² = -2.

Answer: (c) 2y = ln (2x + 1)

Method. separate variables and use y(0) = 0

dydx = e^-2y gives e^2ydy = dx.
Integrate: 12e^2y = x + C. At x = 0, y = 0: C = 12.
e^2y = 2x + 1, so 2y = ln(2x + 1).

Answer: (b) x + y = k tan(y − ck)

Method. substitute v = x + y, separate variables

Let v = x + y, so dvdx = 1 + dydx = 1 + k²v² = v² + k²v².
Separate: v²v² + k²dv = dx, i.e. (1 − k²v² + k²)dv = dx.
Integrate: v − k tan^-1(vk) = x + C. With v = x + y: y − C = k tan^-1(x + yk).
So x + y = k tan(y − ck).

Answer: (a) 1

Method. separate variables and apply y(0) = 12

dyy³ = 2e^xdx, so -12y² = 2e^x + C.
At x = 0, y = 12: -2 = 2 + C, so C = -4.
-12y² = 2e^x - 4, so 1y² = 8 − 4e^x = 4(2 − e^x).
Hence 4y²(2 − e^x) = 1.

Answer: (d) y = (x² - 2)2

Method. factor, then test each option

(dydx)² - xdydx = 0 gives dydx(dydx − x) = 0, so dydx = 0 or dydx = x.
The slopes of the options are 2, 2, 2x and x respectively.
Only y = x² - 22 has dydx = x for all x. (y = 2x and y = 2x + 4 have the constant slope 2, which equals neither 0 nor x.)

Answer: (c) y = 2tan(x2) − x + c

Method. separate variables with half-angle identity

(dy − dx) + cos x(dy + dx) = 0 gives dy(1 + cos x) = dx(1 − cos x).
dydx = 1 − cos x1 + cos x = tan²(x2) = sec²(x2) − 1.
Integrate: y = 2tan(x2) − x + c.

Answer: (d) 5 ln 5

Method. separate variables, use the initial value

dyy = ln 5 dx, so ln y = x ln 5 + C and y = A· 5^x.
y(0) = A = ln 5.
y(1) = 5 ln 5.

Answer: (c) e^x - e^y = c

Method. exponentiate, then separate variables

ln(dydx) = x − y, so dydx = e^x-y = e^xe^-y.
Separate: e^y dy = e^x dx.
Integrate: e^y = e^x + k, i.e. e^x - e^y = c (c = -k).
Checking: y = ln(e^x - c) gives ln(dydx) + y = x.

Answer: (b) y = tan (x + c)

Method. separate variables

dy1 + y² = dx gives tan^-1y = x + c.
Hence y = tan(x + c).

Answer: (b) x² + y² = 2xy + c

Method. factorise

x(dx − dy) + y(dy − dx) = (x − y)(dx − dy) = (x − y) d(x − y) = 0.
Integrating: (x − y)²2 = const, i.e. x² - 2xy + y² = c.
Hence x² + y² = 2xy + c.

Answer: (a) y = e^x + c

Method. remove the logarithm and integrate

ln(dydx) = x gives dydx = e^x.
Hence y = e^x + c.

Answer: (a) e^x[sec (y − x) − tan (y − x)] = c

Method. substitute v = y − x

Let v = y − x. Then dvdx = dydx − 1 = cos v.
So sec v dv = dx and ln|sec v + tan v| = x + C, i.e. sec v + tan v = Ke^x.
Since (sec v + tan v)(sec v − tan v) = 1, sec v − tan v = e^-xK.
Hence e^x[sec(y − x) − tan(y − x)] = c.

Answer: (a) x² + y² = c

Method. separate variables

dydx = -xy gives y dy = -x dx.
Integrating: y²2 = -x²2 + C, so x² + y² = c.

Answer: (d) 1a e^ax + 1b e^-by = c

Method. separate variables

ln(dydx) = ax + by gives dydx = e^axe^by.
So e^-by dy = e^ax dx. Integrating: -1be^-by = 1ae^ax + C.
Hence 1ae^ax + 1be^-by = c.

Answer: (c) y − x + ln(x + y) = c

Method. substitute v = x + y

Let v = x + y. Then dvdy = 1 + dxdy = 1 + v + 1v − 1 = 2vv − 1.
Separate: v − 12v dv = dy, i.e. 12(1 − 1v) dv = dy.
Integrate: 12(v − ln v) = y + C, so x + y − ln(x + y) = 2y + 2C.
Hence y − x + ln(x + y) = c, option (c).

Options (b) and (d) are printed identically (both wrong); (c) is the only correct option, so the answer stands.

Answer: (b) y = cx

Method. separate variables

x dy = y dx gives dyy = dxx.
ln y = ln x + ln c, so y = cx.

Answer: (a) x − y − 2xy = c

Method. separate variables

(1 + 2x) dy = (1 − 2y) dx gives dy1 − 2y = dx1 + 2x.
Integrating: -12ln|1 − 2y| = 12ln|1 + 2x| + C, so (1 − 2y)(1 + 2x) = K.
Expanding: 1 + 2x − 2y − 4xy = K, i.e. 2(x − y − 2xy) = K − 1.
Hence x − y − 2xy = c.

Answer: (a) y = xe^a + c

Method. separate variables after removing the logarithm

ln(dydx) = a gives dydx = e^a, a constant.
Integrating: y = xe^a + c.

Answer: (a) y = xc + sin x

Method. recognise d(xy)

y dx − x dy = y²cos x dx. Divide by y²: y dx − x dyy² = cos x dx.
The left side is d(xy). Integrate: xy = sin x + c.
So y = xc + sin x.

Answer: (d) y = φ(x)x + c

Method. substitute u = φ(x)y

Let u = φy. Then u' = φ'y − φy'y².
Using φy' = yφ' - y²: φ'y − φy' = y², so u' = 1.
Thus u = x + c, i.e. y = φ(x)x + c.

Answer: (a) 3

Method. linear equation in x as a function of y

x dy = y dx + y² dy gives (x − y²) dy = y dx, i.e. dxdy − xy = -y.
Integrating factor = 1y, so d(xy) = -dy and xy = -y + c, i.e. x = -y² + cy.
y(1) = 1 gives 1 = -1 + c, so c = 2 and y² - 2y + x = 0.
Hence y = 1 ± ⁠√(1⁠ −⁠ x). With y(x) > 0 at x = -3 take y = 1 + ⁠√4⁠ = 3 (the other root, -1, is negative).

Answer: (c) x = 2y² + cy

Method. linear equation in x as a function of y

y dx = (x + 2y²) dy gives dxdy − xy = 2y.
Integrating factor = 1y: d(xy) = 2 dy, so xy = 2y + c.
Hence x = 2y² + cy.

Answer: (a) A straight line passing through (1, 4)

Method. integrate dydx = 4 and use the origin

dydx = 4 gives y = 4x + C; the curve passes through (0, 0), so C = 0 and y = 4x.
This is a straight line, and (1, 4) lies on it (4.1 = 4), while (-1, 4) does not (4·(-1) = -4).

Answer: (c) 32 square units

Method. area of a triangle

The line y = 4x with the x-axis and x = 4 bounds a right triangle with vertices (0, 0), (4, 0), (4, 16).
Area = 12·4.16 = 32 square units. (Or ∫₀⁴ 4x dx = 2.16 = 32.)

Answer: (a) No point

Method. integrate and compare the two solutions

dydx = x gives y = x²2 + C.
y₁(0) = 0 gives y₁ = x²2; y₂(0) = 4 gives y₂ = x²2 + 4.
They differ by 4 for every x, so they never meet. No point of intersection.

Answer: (a) ln 2100

Method. rate law dNdt = -kN

dNdt = -kN gives N = N₀e^-kt.
Half is left at t = 100: 12 = e^-100k, so 100k = ln 2.
k = ln 2100.

Answer: (c) A family of parabolas

Method. separate variables and identify the curves

xdydx = 2y gives dyy = 2dxx, so ln|y| = 2 ln|x| + C and y = cx².
This is a family of parabolas (axis the y-axis, vertex at the origin).

Answer: (c) 2 ln 5

Method. separate variables with x(0) = 0

dxx + 1 = dt gives ln(x + 1) = t + C; x = 0 at t = 0 gives C = 0, so x = e^t - 1.
x = 24: e^t = 25, so t = ln 25 = 2 ln 5.

Answer: (b) 6x²y + 17x² + 2x⁵ - 3 = 0

Method. integrate and fit the point

y = ∫(-x² - x^-3) dx = -x³3 + 12x² + C.
At (-1, -2): -2 = 13 + 12 + C, so C = -176.
Multiply by 6x²: 6x²y = -2x⁵ + 3 − 17x².
Hence 6x²y + 17x² + 2x⁵ - 3 = 0.

Answer: (b) a = -b ≠ 0

Method. separate variables and compare with the general second-degree equation of a circle

(by + k) dy = (ax + h) dx gives b2y² + ky = a2x² + hx + C.
That is ax² - by² + 2hx − 2ky = constant.
A circle needs equal coefficients of x² and y², both non-zero: a = -b ≠ 0.