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]

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(5)

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

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

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

$∴$ $c - f (b) = f' (b) (0 - b)$

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

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

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

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

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

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

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

$∴$ ${(P Q)}^{2} = 2^{2} + 1^{2} = 5$

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]

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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]

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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]