Differential Equations jee mains questions | JEE Main Maths Differential Equations Previous Year Question

Q 1 :

The differential equation of the family of circles passing through the origin and having centre at the line y = x is [2024]

$(x^{2} + y^{2} + 2 x y) d x = (x^{2} + y^{2} - 2 x y) d y$
$(x^{2} - y^{2} + 2 x y) d x = (x^{2} - y^{2} - 2 x y) d y$
$(x^{2} + y^{2} - 2 x y) d x = (x^{2} + y^{2} + 2 x y) d y$
$(x^{2} - y^{2} + 2 x y) d x = (x^{2} - y^{2} + 2 x y) d y$

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(2)

Let (k, k) be the centre of circle, so equation of circle with radius r is given by

${(x - k)}^{2} + {(y - k)}^{2} = r^{2}$

Now, circle is passing through origin

$\Rightarrow r^{2} = 2 k^{2}$

So, equation of circle becomes

$x^{2} + y^{2} - 2 k x - 2 k y = 0$ ... (i)

On differentiating (i) w.r.t. x, we get

2x + 2yy' – 2k – 2ky' = 0

$\Rightarrow k = \frac{x + y y'}{(1 + y')}$

Substituting the value of k in (i) we get

$x^{2} + y^{2} - 2 x (\frac{x + y y'}{1 + y'}) - 2 y (\frac{x + y y'}{1 + y'}) = 0$

$\Rightarrow (x^{2} + y^{2}) (1 + y') - 2 x^{2} - 2 x y y' - 2 x y - 2 y^{2} y' = 0$

$\Rightarrow - x^{2} + y^{2} + x^{2} y' - y^{2} y' - 2 x y y' - 2 x y = 0$

$\Rightarrow (x^{2} - y^{2} - 2 x y) y' = x^{2} - y^{2} + 2 x y$

$\Rightarrow (x^{2} - y^{2} - 2 x y) d y = (x^{2} - y^{2} + 2 x y) d x$

is the required differential equation.

Q 2 :

Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is $e^{- a} + 4 a^{2} + a - 1$ . Then the differential equation, whose general solution is $y = c_{1} f (x) + c_{2}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is [2024]

$(8 e^{x} - 1) \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0$
$(8 e^{x} + 1) \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} = 0$
$(8 e^{x} + 1) \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0$
$(8 e^{x} - 1) \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} = 0$

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(3)

Given $\int_{0}^{a} f (x) d x = e^{- a} + 4 a^{2} + a - 1$

On differentiating both sides, we get $f (a) = - e^{- a} + 8 a + 1$

i.e., $f (x) = - e^{- x} + 8 x + 1$

Now, $y = c_{1} f (x) + c_{2}$

$\Rightarrow y' = c_{1} f' (x) = c_{1} (8 + e^{- x})$ ... (i)

Again, on differentiating, we get $c_{1} = - e^{x} y''$ ... (ii)

From (i) and (ii), we get

$y' = y'' e^{x} (- 8 - e^{- x}) \Rightarrow (8 e^{x} + 1) y'' + y' = 0$

Q 3 :

$Let y = f (x) = \sin^{3} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{\frac{3}{2}}))) .$ $Then, at x = 1,$ [2023]

$2 y' + \sqrt{3} π^{2} y = 0$
$\sqrt{2} y' - 3 π^{2} y = 0$
$2 y' + 3 π^{2} y = 0$
$y' + 3 π^{2} y = 0$

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(3)

$Let - 4 x^{3} + 5 x^{2} + 1 = t, then$

$y = \sin^{3} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} t^{3 / 2})))$

$\frac{d t}{d x} = - 12 x^{2} + 10 x$

$∴$ $\frac{d y}{d x} = 3 \sin^{2} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} t^{3 / 2}))) \cdot \cos (\frac{π}{3} \cos (\frac{π}{3 \sqrt{2}} t^{3 / 2})) \cdot \frac{π}{3} (- \sin (\frac{π}{3 \sqrt{2}} t^{3 / 2}))$

$\frac{π}{3 \sqrt{2}} \cdot \frac{3}{2} t^{1 / 2} \cdot \frac{d t}{d x}$

$\Rightarrow \frac{d y}{d x} = - 3 \sin^{2} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{3 / 2}))$ $\cdot \cos (\frac{π}{3} \cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{3 / 2})) \cdot \frac{π}{3} \sin (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{3 / 2})$ $\cdot \frac{π}{2 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{1 / 2} (- 12 x^{2} + 10 x)$

$At x = 1, t = - 4 + 5 + 1 = 2$

${\frac{d y}{d x}}_{(x = 1)} = - 3 \sin^{2} (\frac{π}{3} \cos (\frac{π}{3 \sqrt{2}} 2^{3 / 2})) \cdot \cos (\frac{π}{3} \cos (\frac{π}{3 \sqrt{2}} 2^{3 / 2}))$ $\cdot \frac{π}{3} \sin (\frac{π}{3 \sqrt{2}} 2^{3 / 2}) \cdot \frac{π}{2 \sqrt{2}} 2^{1 / 2} \cdot (- 2)$

$= π^{2} \sin^{2} (\frac{π}{3} (\cos \frac{2 π}{3})) \cdot \cos (\frac{π}{3} \cos \frac{2 π}{3}) \cdot \sin (\frac{2 π}{3})$

$= π^{2} \sin^{2} (- \frac{π}{3} \cdot \frac{1}{2}) \cdot \cos (- \frac{π}{3} \cdot \frac{1}{2}) \frac{\sqrt{3}}{2}$

$= \frac{\sqrt{3}}{2} π^{2} \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{3}{16} π^{2}$

$at x = 1, y = \sin^{3} (\frac{π}{3} (\cos \frac{π}{3 \sqrt{2}} {(2)}^{3 / 2})) = \sin^{3} (\frac{π}{3} (\cos \frac{2 π}{3}))$

$= \sin^{3} (- \frac{π}{3} \times \frac{1}{2}) = - \frac{1}{8}$

$\Rightarrow 2 y' + 3 π^{2} y = 2 \times \frac{3}{16} π^{2} + 3 π^{2} \times (- \frac{1}{8}) = 0$

Q 4 :

$Let a curve y = f (x), x \in (0, \infty) pass through the points P (1, \frac{3}{2}) and Q (a, \frac{1}{2}) .$

$If the tangent at any point R (b, f (b)) to the given curve cuts the y-axis at the point S (0, c)$ $such that b c = 3, then {(P Q)}^{2} is equal to$ ______. [2023]

(5)

$Equation of tangent at R (b, f (b)) is$

$y - f (b) = f' (b) (x - b)$

$Now, it passes through S (0, c)$

$∴$ $c - f (b) = f' (b) (0 - b)$

$\Rightarrow \frac{3}{b} - f (b) = - b f' (b) \Rightarrow b f' (b) - f (b) = - \frac{3}{b}$

$\Rightarrow \frac{b f' (b) - f (b)}{b^{2}} = - \frac{3}{b^{3}}$ $\Rightarrow$ $d (\frac{f (b)}{b}) = - \frac{3}{b^{3}}$

$\Rightarrow \frac{f (b)}{b} = \frac{3}{2 b^{2}} + M$

$Now, the curve passes through P (1, \frac{3}{2}) .$

$∴$ $\frac{3}{2} = \frac{3}{2} + M So, f (b) = \frac{3}{2 b}$

$Also, it passes through Q (a, \frac{1}{2})$

$\Rightarrow \frac{1}{2} = \frac{3}{2 a} \Rightarrow a = 3 \Rightarrow Q \neq (3, \frac{1}{2})$

$∴$ ${(P Q)}^{2} = 2^{2} + 1^{2} = 5$

Q 5 :

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x \sin (\frac{d y}{d x})$ is:

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3
not defined

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(4)

Ans. not defined

Explanation :

The degree of the above differential equation is not defined because when we expand $\sin (\frac{d y}{d x}),$ we get an infinite series in the increasing powers of $\frac{d y}{d x} .$ Therefore, its degree is not defined.

Q 6 :

The degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}$ $= \frac{d^{2} y}{d x^{2}}$ is

$4$
$\frac{3}{2}$
$not defined$
$2$

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(4)

Ans. 2

Explanation:

Given that,

${[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}$ $= \frac{d^{2} y}{d x^{2}}$

On squaring both sides, we get

${[1 + {(\frac{d y}{d x})}^{2}]}^{3}$ $=$ ${(\frac{d^{2} y}{d x^{2}})}^{2}$

So, the degree of differential equation is 2.

Q 7 :

Which of the following is a second order differential equation?

${(y')}^{2} + x = y^{2}$
$y' y'' + y = \sin x$
$y''' + {(y')}^{2} + y = 0$
$y' = y^{2}$

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(2)

Ans. $y' y'' + y = \sin x$

Explanation:

The second order differential equation is $y' y'' + y = \sin x .$

Q 8 :

The order and degree of differential equation ${(\frac{d^{3} y}{d x^{3}})}^{2} - 3 \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{4} = y^{4}$ are:

1, 4
3, 4
2, 4
3, 2

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(4)

Ans. 3, 2

Explanation:

Given that,

${(\frac{d^{3} y}{d x^{3}})}^{2} - 3 \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{4} = y^{4}$

$∴$ Order = 3

and Degree = 2

Q 9 :

Order and degree of the differential equation $r \frac{d r}{d θ} + \cos θ = 5$ are:

order 2, degree 1
order 2, degree 2
order 1 and degree 1
order 2, degree not define

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(3)

Ans. order 1 and degree 1

Explanation:

$In r \frac{d r}{d θ} + \cos θ = 5, order 1 and degree = 1 .$

Q 10 :

Order and degree of the differential equation $e^{\frac{d y}{d x}} + \frac{d y}{d x} = x$ are:

order 2, degree 1
order 2, degree 2
order 1, degree 2
order 1, degree not defined

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(4)

Ans. order 1, degree not defined

Explanation:

Its order is 1, but the equation cannot be expressed as a polynomial differential equation.

$∴$ The degree is not defined.

Q 11 :

The order and the degree of differential equation $\frac{d^{4} y}{d x^{4}} - 4 \frac{d^{3} y}{d x^{3}} + 8 \frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} + 4 y = 0$ respectively are:

order 4, degree 1
order 1, degree 4
order 1, degree 1
none of the above

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(1)

Ans. order 4, degree 1

Explanation:

In the given equation, highest differential power is 4 and degree is one.

Q 12 :

If $m$ and $n$ are the order and degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = x^{2} - 1$ , then:

$m = 3$ and $n = 5$
$m = 3$ and $n = 1$
$m = 3$ and $n = 3$
$m = 3$ and $n = 2$

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(4)

Ans. $m = 3$ and $n = 2$

Explanation:

The highest order $(m)$ of the given equation is $\frac{d^{3} y}{d x^{3}} = 3$

and degree $(n)$ of the given equation is ${(\frac{d^{3} y}{d x^{3}})}^{2} = 2 .$

Therefore, $m = 3$ and $n = 2$ .

Q 13 :

The degree of the differential equation ${(1 + \frac{d y}{d x})}^{3} = {(\frac{d^{2} y}{d x^{2}})}^{2}$ is:

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4

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(2)

Ans. 2

Explanation:

We have,

${(1 + \frac{d y}{d x})}^{3} = {(\frac{d^{2} y}{d x^{2}})}^{2}$

So, order = 2

and degree = 2.

Q 14 :

The order and degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{2}$ $=$ $\frac{d^{2} y}{d x^{2}}$ respectively, are:

1, 2
2, 2
2, 1
4, 2

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(3)

Ans. 2, 1

Explanation:

We have,

$[{(1 + \frac{d y}{d x})}^{2}] = \frac{d^{2} y}{d x^{2}}$

So, order = 2

and degree = 1

Q 15 :

Number of arbitrary constants in general solution of differential equation of fourth order is:

0
2
4
3

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4

(3)

Ans. 4

Explanation:

Number of arbitrary constants in the general solution of differential equation is equal to the order of differential equation.

Q 16 :

The solution of differential equation $x d y - y d x = 0$ represents:

a rectangular hyperbola
parabola whose vertex is at origin
straight line passing through origin
a circle whose centre is at origin

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(3)

Ans. straight line passing through origin

Explanation:

Given that,

$x d y - y d x = 0$

$\Rightarrow$ $x d y = y d x$

$\Rightarrow$ $\frac{d y}{y} = \frac{d x}{x}$

On integrating both sides, we get

$\log y = \log x + \log C$

$\Rightarrow$ $\log y = \log C x$

$\Rightarrow$ $y = C x$

which is a straight line passing through origin.

Q 17 :

The solution of $\frac{d y}{d x} - y = 1,$ $y (0) = 1$ is given by:

$x y = - e^{x}$
$x y = - e^{- x}$
$x y = - 1$
$y = 2 e^{x} - 1$

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(4)

Ans. $y = 2 e^{x} - 1$

Explanation :

Given that,

$\frac{d y}{d x} - y = 1$

$\Rightarrow$ $\frac{d y}{d x} = 1 + y$

$\Rightarrow$ $\frac{d y}{1 + y} = d x$

On integrating both sides, we get

$\log (1 + y) = x + C$ ...(i)

When $x = 0$ and $y = 1$ , then

$\log 2 = 0 + C$

$\Rightarrow$ $C = \log 2$

The required solution is

$\log (1 + y) = x + \log 2$

$\Rightarrow$ $\log (\frac{1 + y}{2}) = x$

$\Rightarrow$ $\frac{1 + y}{2} = e^{x}$

$\Rightarrow$ $1 + y = 2 e^{x}$

$\Rightarrow$ $y = 2 e^{x} - 1$

Q 18 :

The differential equation $y \frac{d y}{d x} + x = C$ represents:

family of hyperbolas
family of parabolas
family of ellipses
family of circles

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(4)

Ans. family of circles

Explanation :

Given that,

$y \frac{d y}{d x} + x = C$

$\Rightarrow$ $y \frac{d y}{d x} = C - x$

$\Rightarrow$ $y d y = (C - x) d x$

On integrating both sides, we get

$\frac{y^{2}}{2} = C x - \frac{x^{2}}{2} + K$ (where $K$ is an integration constant)

$\Rightarrow$ $\frac{x^{2}}{2} + \frac{y^{2}}{2} = C x + K$

$\Rightarrow$ $\frac{x^{2}}{2} + \frac{y^{2}}{2} - C x = K$

which represents a family of circles.

Q 19 :

The solution of differential equation $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$ is:

$y = \tan^{- 1} x$
$y - x = k (1 + x y)$
$x = \tan^{- 1} y$
$\tan (x y) = k$

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(2)

Ans. $y - x = k (1 + x y)$

Explanation:

Given that,

$\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$

$\Rightarrow$ $\frac{d y}{1 + y^{2}} = \frac{d x}{1 + x^{2}}$

On integrating both sides, we get

$\tan^{- 1} y = \tan^{- 1} x + C$

$\Rightarrow$ $\tan^{- 1} y - \tan^{- 1} x = C$

$\Rightarrow$ $\tan^{- 1} (\frac{y - x}{1 + x y}) = C$

$\Rightarrow$ $\frac{y - x}{1 + x y} = \tan C$

$\Rightarrow$ $y - x = \tan C (1 + x y)$

$\Rightarrow$ $y - x = k (1 + x y)$

where, $k = \tan C$

Q 20 :

The general solution of $\frac{d y}{d x} = 2 x e^{x^{2} - y}$ is:

$e^{x^{2} - y} = C$
$e^{- y} + e^{x^{2}} = C$
$e^{y} = e^{x^{2}} + C$
$e^{x^{2} + y} = C$

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(3)

Ans. $e^{y} = e^{x^{2}} + C$

Explanation:

Given that,

$\frac{d y}{d x} = 2 x e^{x^{2} - y} = 2 x e^{x^{2}} \cdot e^{- y}$

$\Rightarrow$ $e^{y} \frac{d y}{d x} = 2 x e^{x^{2}}$

$\Rightarrow$ $e^{y} d y = 2 x e^{x^{2}} d x$

On integrating both sides, we get

$\int e^{y} d y = 2 \int x e^{x^{2}} d x$

Put $x^{2} = t$ in RHS integral, we get

$2 x d x = d t$

$∴$ $\int e^{y} d y = \int e^{t} d t$

$\Rightarrow$ $e^{y} = e^{t} + C$

$\Rightarrow$ $e^{y} = e^{x^{2}} + C$

Q 21 :

The solution of equation $(2 y - 1) d x - (2 x + 3) d y = 0$ is:

$\frac{2 x - 1}{2 y + 3} = k$
$\frac{2 y + 1}{2 x - 3} = k$
$\frac{2 x + 3}{2 y - 1} = k$
$\frac{2 x - 1}{2 y - 1} = k$

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(3)

Ans. $\frac{2 x + 3}{2 y - 1} = k$

Explanation:

Given that,

$(2 y - 1) d x - (2 x + 3) d y = 0$

$\Rightarrow$ $(2 y - 1) d x = (2 x + 3) d y$

$\Rightarrow$ $\frac{d x}{2 x + 3} = \frac{d y}{2 y - 1}$

On integrating both sides, we get

$\frac{1}{2} \log (2 x + 3) = \frac{1}{2} \log (2 y - 1) + \log C$

$\Rightarrow$ $\frac{1}{2} [\log (2 x + 3) - \log (2 y - 1)] = \log C$

$\Rightarrow$ $\frac{1}{2} \log (\frac{2 x + 3}{2 y - 1}) = \log C$

$\Rightarrow$ ${(\frac{2 x + 3}{2 y - 1})}^{1 / 2} = C$

$\Rightarrow$ $\frac{2 x + 3}{2 y - 1} = C^{2}$

$\Rightarrow$ $\frac{2 x + 3}{2 y - 1} = k,$ $where k = C^{2}$

Q 22 :

The solution of differential equation $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ is:

$y = e^{x - y} - x^{2} e^{- y} + C$
$e^{y} - e^{x} = \frac{x^{3}}{3} + C$
$e^{x} + e^{y} = \frac{x^{3}}{3} + C$
$e^{x} - e^{y} = \frac{x^{3}}{3} + C$

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(2)

Ans. $e^{y} - e^{x} = \frac{x^{3}}{3} + C$

Explanation:

Given that, $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$

$\Rightarrow$ $\frac{d y}{d x} = e^{x} e^{- y} + x^{2} e^{- y}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{e^{x} + x^{2}}{e^{y}}$

$\Rightarrow$ $e^{y} d y = (e^{x} + x^{2}) d x$

On integrating both sides, we get

$\int e^{y} d y = \int (e^{x} + x^{2}) d x$

$\Rightarrow$ $e^{y} = e^{x} + \frac{x^{3}}{3} + C$

$\Rightarrow$ $e^{y} - e^{x} = \frac{x^{3}}{3} + C$

Q 23 :

The solution of the equation $(1 + x^{2}) \frac{d y}{d x} = 1$ is:

$y = \log (1 + x^{2}) + C$
$y + \log (1 + x^{2}) + C = 0$
$y - \log (1 + x) + C$
$y = \tan^{- 1} x + C$

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(4)

Ans. $y = \tan^{- 1} x + C$

Explanation:

$(1 + x^{2}) \frac{d y}{d x} = 1$

$\Rightarrow$ $\frac{d y}{d x} = \frac{1}{1 + x^{2}}$

$\Rightarrow$ $d y = \frac{d x}{1 + x^{2}}$

On integrating, $y = \tan^{- 1} x + C$

Q 24 :

The general solution of differential equation $\frac{d y}{d x} = e^{x + y}$ is:

$e^{x} + e^{- y} = C$
$e^{- x} + e^{y} = C$
$e^{x} + e^{y} = C$
$e^{- x} + e^{- y} = C$

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(1)

Ans. $e^{x} + e^{- y} = C$

Explanation:

$\frac{d y}{d x} = e^{x} \times e^{y}$

$\Rightarrow$ $e^{- y} d y = e^{x} d x$

On integrating both sides,

$\Rightarrow$ $\int e^{- y} d y = \int e^{x} d x$

$\Rightarrow$ $- e^{- y} = e^{x} + A$

$\Rightarrow$ $- e^{- y} - e^{x} = A$

$\Rightarrow$ $e^{- y} + e^{x} = - A$

$e^{- y} + e^{x} = C,$ $where C = - A$

Q 25 :

The general solution of the differential equation $\frac{d y}{d x} = \cot x \cot y$ is:

$\cos x = C c o s e c y$
$\sin x = C \sec y$
$\sin x = C \cos y$
$\cos x = C \sin y$

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(2)

Ans. $\sin x = C \sec y$

Explanation:

$\frac{d y}{d x} = \cot x \cot y$

$\Rightarrow$ $\cot x d x - \tan y d y = 0$

Integrating on both sides, we get

$\Rightarrow$ $\int \cot x d x - \int \tan y d y = 0$

$\log \sin x - \log \sec y = \log C$

$\Rightarrow$ $\log (\frac{\sin x}{\sec y}) = \log C$

$\Rightarrow$ $\sin x = C \sec y$

Q 26 :

Which of the following is a homogeneous differential equation?

$(4 x + 6 y + 5) d y - (3 y + 2 x + 4) d x = 0$
$x y d x - (x^{3} + y^{3}) d x = 0$
$(x^{3} + 2 y^{2}) d x + 2 x y d y = 0$
$y^{2} d x + (x^{2} - x y - y^{2}) d y = 0$

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(4)

Ans. $y^{2} d x + (x^{2} - x y - y^{2}) d y = 0$

Explanation:

By equation, $y^{2} d x + (x^{2} - x y - y^{2}) d y = 0$

$∴$ $\frac{d y}{d x} = \frac{- y^{2}}{x^{2} - x y - y^{2}}$

Power of numerator and denominator are of same order. Hence, it is a homogeneous differential equation.

Q 27 :

The integrating factor of differential equation $\cos x \frac{d y}{d x} + y \sin x = 1$ is:

$\cos x$
$\tan x$
$\sec x$
$\sin x$

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(3)

Ans. $\sec x$

Explanation:

Given that,

$\cos x \frac{d y}{d x} + y \sin x = 1$

$\Rightarrow$ $\frac{d y}{d x} + y \tan x = \sec x$

Here, $P = \tan x$ and $Q = \sec x$

$IF = e^{\int P d x} = e^{\int \tan x d x} = e^{\log \sec x}$

$\Rightarrow$ $= \sec x$

Q 28 :

The integrating factor of differential equation $(1 - x^{2}) \frac{d y}{d x} - x y = 1$ is:

$- x$
$\frac{x}{1 + x^{2}}$
$\sqrt{1 - x^{2}}$
$\frac{1}{2} \log (1 - x^{2})$

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(3)

Ans. $\sqrt{1 - x^{2}}$

Explanation:

Given that,

$(1 - x^{2}) \frac{d y}{d x} - x y = 1$

$\Rightarrow$ $\frac{d y}{d x} - \frac{x}{1 - x^{2}} y = \frac{1}{1 - x^{2}}$

which is a linear differential equation.

$∴$ $IF = e^{\int \frac{- x}{1 - x^{2}} d x}$

Put $1 - x^{2} = t$

$\Rightarrow$ $- 2 x d x = d t$

$\Rightarrow$ $x d x = - \frac{d t}{2}$

Now, $IF = e^{\frac{1}{2} \int \frac{d t}{t}} = e^{\frac{1}{2} \log t}$

$= e^{\frac{1}{2} \log (1 - x^{2})}$

$= \sqrt{1 - x^{2}}$

Q 29 :

The integrating factor of differential equation $\frac{d y}{d x} + y \tan x - \sec x = 0$ is:

$\cos x$
$\sec x$
$e^{\cos x}$
$e^{\sec x}$

1

2

3

4

(2)

Ans. $\sec x$

Explanation:

Given that,

$\frac{d y}{d x} + y \tan x - \sec x = 0$

Here, $P = \tan x,$ $Q = \sec x$

$IF = e^{\int P d x} = e^{\int \tan x d x}$

$= e^{(\log \sec x)}$

$= \sec x$

Q 30 :

The solution of $x \frac{d y}{d x} + y = e^{x}$ is:

$y = \frac{e^{x}}{x} + \frac{k}{x}$
$y = x e^{x} + C x$
$y = x e^{x} + k$
$x = \frac{e^{y}}{y} + \frac{k}{y}$

1

2

3

4

(1)

Ans. $y = \frac{e^{x}}{x} + \frac{k}{x}$

Explanation:

Given that,

$x \frac{d y}{d x} + y = e^{x}$

$\Rightarrow$ $\frac{d y}{d x} + \frac{y}{x} = \frac{e^{x}}{x}$

which is a linear differential equation.

$∴$ $IF = e^{\int \frac{1}{x} d x} = e^{\log x} = x$

The general solution is

$\Rightarrow$ $y \cdot x = \int (\frac{e^{x}}{x} \cdot x) d x$

$\Rightarrow$ $y \cdot x = \int e^{x} d x$

$\Rightarrow$ $y \cdot x = e^{x} + k$

$\Rightarrow$ $y = \frac{e^{x}}{x} + \frac{k}{x}$

Topic Question Set

Q 1 :

The differential equation of the family of circles passing through the origin and having centre at the line y = x is [2024]

Q 2 :

Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is $e^{- a} + 4 a^{2} + a - 1$ . Then the differential equation, whose general solution is $y = c_{1} f (x) + c_{2}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is [2024]

Q 3 :

$Let y = f (x) = \sin^{3} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{\frac{3}{2}}))) .$ $Then, at x = 1,$ [2023]

Q 4 :

$Let a curve y = f (x), x \in (0, \infty) pass through the points P (1, \frac{3}{2}) and Q (a, \frac{1}{2}) .$

$If the tangent at any point R (b, f (b)) to the given curve cuts the y-axis at the point S (0, c)$ $such that b c = 3, then {(P Q)}^{2} is equal to$ ______. [2023]

Q 5 :

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x \sin (\frac{d y}{d x})$ is:

Q 6 :

The degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}$ $= \frac{d^{2} y}{d x^{2}}$ is

Q 7 :

Which of the following is a second order differential equation?

Q 8 :

The order and degree of differential equation ${(\frac{d^{3} y}{d x^{3}})}^{2} - 3 \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{4} = y^{4}$ are:

Q 9 :

Order and degree of the differential equation $r \frac{d r}{d θ} + \cos θ = 5$ are:

Q 10 :

Order and degree of the differential equation $e^{\frac{d y}{d x}} + \frac{d y}{d x} = x$ are:

Q 11 :

The order and the degree of differential equation $\frac{d^{4} y}{d x^{4}} - 4 \frac{d^{3} y}{d x^{3}} + 8 \frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} + 4 y = 0$ respectively are:

Q 12 :

If $m$ and $n$ are the order and degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = x^{2} - 1$ , then:

Q 13 :

The degree of the differential equation ${(1 + \frac{d y}{d x})}^{3} = {(\frac{d^{2} y}{d x^{2}})}^{2}$ is:

Q 14 :

The order and degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{2}$ $=$ $\frac{d^{2} y}{d x^{2}}$ respectively, are:

Q 15 :

Number of arbitrary constants in general solution of differential equation of fourth order is:

Q 16 :

The solution of differential equation $x d y - y d x = 0$ represents:

Q 17 :

The solution of $\frac{d y}{d x} - y = 1,$ $y (0) = 1$ is given by:

Q 18 :

The differential equation $y \frac{d y}{d x} + x = C$ represents:

Q 19 :

The solution of differential equation $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$ is:

Q 20 :

The general solution of $\frac{d y}{d x} = 2 x e^{x^{2} - y}$ is:

Q 21 :

The solution of equation $(2 y - 1) d x - (2 x + 3) d y = 0$ is:

Q 22 :

The solution of differential equation $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ is:

Q 23 :

The solution of the equation $(1 + x^{2}) \frac{d y}{d x} = 1$ is:

Q 24 :

The general solution of differential equation $\frac{d y}{d x} = e^{x + y}$ is:

Q 25 :

The general solution of the differential equation $\frac{d y}{d x} = \cot x \cot y$ is:

Q 26 :

Which of the following is a homogeneous differential equation?

Q 27 :

The integrating factor of differential equation $\cos x \frac{d y}{d x} + y \sin x = 1$ is:

Q 28 :

The integrating factor of differential equation $(1 - x^{2}) \frac{d y}{d x} - x y = 1$ is:

Q 29 :

The integrating factor of differential equation $\frac{d y}{d x} + y \tan x - \sec x = 0$ is:

Q 30 :

The solution of $x \frac{d y}{d x} + y = e^{x}$ is:

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Topic Question Set

Q 1 : The differential equation of the family of circles passing through the origin and having centre at the line y = x is [2024]

Q 2 : Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e-a + 4a2 + a - 1. Then the differential equation, whose general solution is y = c1f(x) + c2, where c1 and c2 are arbitrary constants, is [2024]

Q 3 : Let y=f(x)=sin3(π3(cos(π32(-4x3+5x2+1)32))). Then, at x=1, [2023]

Q 4 : Let a curve y=f(x),x∈(0,∞) pass through the points P(1,32) and Q(a,12). If the tangent at any point R(b,f(b)) to the given curve cuts the y-axis at the point S(0,c) such that bc=3,then (PQ)2 is equal to ______. [2023]

Q 5 : The degree of the differential equation (d2ydx2)2+(dydx)2=xsin(dydx) is:

Q 6 : The degree of the differential equation [1+(dydx)2]3/2=d2ydx2 is

Q 7 : Which of the following is a second order differential equation?

Q 8 : The order and degree of differential equation (d3ydx3)2-3d2ydx2+2(dydx)4=y4 are:

Q 9 : Order and degree of the differential equation rdrdθ+cosθ=5 are:

Q 10 : Order and degree of the differential equation edydx+dydx=x are:

Q 11 : The order and the degree of differential equation d4ydx4-4d3ydx3+8d2ydx2-8dydx+4y=0 respectively are:

Q 12 : If m and n are the order and degree of the differential equation (d2ydx2)5+4(d2ydx2)3(d3ydx3)+d3ydx3=x2-1, then:

Q 13 : The degree of the differential equation (1+dydx)3=(d2ydx2)2 is:

Q 14 : The order and degree of the differential equation [1+(dydx)2]2=d2ydx2 respectively, are:

Q 15 : Number of arbitrary constants in general solution of differential equation of fourth order is:

Q 16 : The solution of differential equation xdy-ydx=0 represents:

Q 17 : The solution of dydx-y=1, y(0)=1 is given by:

Q 18 : The differential equation ydydx+x=C represents:

Q 19 : The solution of differential equation dydx=1+y21+x2 is:

Q 20 : The general solution of dydx=2xex2-y is:

Q 21 : The solution of equation (2y-1) dx-(2x+3) dy=0 is:

Q 22 : The solution of differential equation dydx=ex-y+x2e-y is:

Q 23 : The solution of the equation (1+x2)dydx=1 is:

Q 24 : The general solution of differential equation dydx=ex+y is:

Q 25 : The general solution of the differential equation dydx=cotx coty is:

Q 26 : Which of the following is a homogeneous differential equation?

Q 27 : The integrating factor of differential equation cosxdydx+ysinx=1 is:

Q 28 : The integrating factor of differential equation (1-x2)dydx-xy=1 is:

Q 29 : The integrating factor of differential equation dydx+ytanx-secx=0 is:

Q 30 : The solution of xdydx+y=ex is:

Want Us to Email you About Special Offers & Updates?

Q 1 :

The differential equation of the family of circles passing through the origin and having centre at the line y = x is [2024]

Q 2 :

Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is $e^{- a} + 4 a^{2} + a - 1$ . Then the differential equation, whose general solution is $y = c_{1} f (x) + c_{2}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is [2024]

Q 3 :

$Let y = f (x) = \sin^{3} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{\frac{3}{2}}))) .$ $Then, at x = 1,$ [2023]

Q 4 :

$Let a curve y = f (x), x \in (0, \infty) pass through the points P (1, \frac{3}{2}) and Q (a, \frac{1}{2}) .$

$If the tangent at any point R (b, f (b)) to the given curve cuts the y-axis at the point S (0, c)$ $such that b c = 3, then {(P Q)}^{2} is equal to$ ______. [2023]

Q 5 :

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x \sin (\frac{d y}{d x})$ is:

Q 6 :

The degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}$ $= \frac{d^{2} y}{d x^{2}}$ is

Q 7 :

Which of the following is a second order differential equation?

Q 8 :

The order and degree of differential equation ${(\frac{d^{3} y}{d x^{3}})}^{2} - 3 \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{4} = y^{4}$ are:

Q 9 :

Order and degree of the differential equation $r \frac{d r}{d θ} + \cos θ = 5$ are:

Q 10 :

Order and degree of the differential equation $e^{\frac{d y}{d x}} + \frac{d y}{d x} = x$ are:

Q 11 :

The order and the degree of differential equation $\frac{d^{4} y}{d x^{4}} - 4 \frac{d^{3} y}{d x^{3}} + 8 \frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} + 4 y = 0$ respectively are:

Q 12 :

If $m$ and $n$ are the order and degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = x^{2} - 1$ , then:

Q 13 :

The degree of the differential equation ${(1 + \frac{d y}{d x})}^{3} = {(\frac{d^{2} y}{d x^{2}})}^{2}$ is:

Q 14 :

The order and degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{2}$ $=$ $\frac{d^{2} y}{d x^{2}}$ respectively, are:

Q 15 :

Number of arbitrary constants in general solution of differential equation of fourth order is:

Q 16 :

The solution of differential equation $x d y - y d x = 0$ represents:

Q 17 :

The solution of $\frac{d y}{d x} - y = 1,$ $y (0) = 1$ is given by:

Q 18 :

The differential equation $y \frac{d y}{d x} + x = C$ represents:

Q 19 :

The solution of differential equation $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$ is:

Q 20 :

The general solution of $\frac{d y}{d x} = 2 x e^{x^{2} - y}$ is:

Q 21 :

The solution of equation $(2 y - 1) d x - (2 x + 3) d y = 0$ is:

Q 22 :

The solution of differential equation $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ is:

Q 23 :

The solution of the equation $(1 + x^{2}) \frac{d y}{d x} = 1$ is:

Q 24 :

The general solution of differential equation $\frac{d y}{d x} = e^{x + y}$ is:

Q 25 :

The general solution of the differential equation $\frac{d y}{d x} = \cot x \cot y$ is:

Q 26 :

Which of the following is a homogeneous differential equation?

Q 27 :

The integrating factor of differential equation $\cos x \frac{d y}{d x} + y \sin x = 1$ is:

Q 28 :

The integrating factor of differential equation $(1 - x^{2}) \frac{d y}{d x} - x y = 1$ is:

Q 29 :

The integrating factor of differential equation $\frac{d y}{d x} + y \tan x - \sec x = 0$ is:

Q 30 :

The solution of $x \frac{d y}{d x} + y = e^{x}$ is: