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

Q 81 :

If the solution curve of the differential equation $(y - 2 \log_{e} x) d x + (x \log_{e} x^{2}) d y = 0, x > 1$ passes through the points $(e, \frac{4}{3})$ and $(e^{4}, α),$ then $α$ is equal to ____ . [2023]

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$Given differential equation is$ $(y - 2 \log_{e} x) d x + (x \log_{e} x^{2}) d y = 0$

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

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

This is a linear differential equation of the type $\frac{d y}{d x} + P y = Q$

$Here, P = \frac{1}{x \log_{e} x^{2}}, Q = \frac{1}{x}$

$I.F. = e^{\int P d x} = e^{\int \frac{1}{2 x \log_{e} x} d x} = \sqrt{\log_{e} x}$

Solution of the differential equation is

$y \sqrt{\log_{e} x} = \int \frac{1}{x} \cdot \sqrt{\log_{e} x} d x$

$\Rightarrow y \sqrt{\log_{e} x} = \frac{2}{3} {(\log_{e} x)}^{3 / 2} + c \dots (i)$

$Now, put (e, \frac{4}{3}) in (i), we get$ $\frac{4}{3} = \frac{2}{3} + c \Rightarrow c = \frac{2}{3}$

$Now, put x = e^{4}, y = α, c = \frac{2}{3} in (i), we get$

$α \sqrt{\log_{e} e^{4}} = \frac{2}{3} {(\log_{e} e^{4})}^{3 / 2} + \frac{2}{3}$

$\Rightarrow α \times 2 = \frac{2}{3} \times 8 + \frac{2}{3} \Rightarrow α = \frac{8}{3} + \frac{1}{3} = 3$

Q 82 :

Let the solution curve $x = x (y), 0 < y < \frac{π}{2}$ , of the differential equation $(\log_{e} {(\cos y))}^{2} \cos y d x - (1 + 3 x \log_{e} (\cos y)) \sin y d y = 0$ satisfy $x (\frac{π}{3}) = \frac{1}{2 \log_{e} 2} .$ If $x (\frac{π}{6}) = \frac{1}{\log_{e} m - \log_{e} n},$ where $m$ and $n$ are coprime, then $m n$ is equal to _______ . [2023]

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Given differential equation is, $\cos y {(logcos y)}^{2} d x = (1 + 3 x (logcos y)) \sin y d y$

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

$\Rightarrow \frac{d x}{d y} - (\frac{3 \tan y}{logcos y}) x = \frac{\tan y}{{(logcos y)}^{2}}$

$I.F. = e^{\int - \frac{3 \tan y}{logcos y} d y} = (\log {(\cos y))}^{3}$

Solution is given by,

$x \cdot {(logcos y)}^{3} = \int \frac{\tan y}{{(logcos y)}^{2}} \times {(logcos y)}^{3} d y + C$

$= - \frac{{(logcos y)}^{2}}{2} + C$

$Given, x (\frac{π}{3}) = \frac{1}{2 \log 2}$

So, $\frac{1}{2 \log 2} (\log {(\frac{1}{2}))}^{3} = - \frac{(\log {(\frac{1}{2}))}^{2}}{2} + C \Rightarrow C = 0$

For $y = \frac{π}{6}$ , we have $x {(\log \frac{\sqrt{3}}{2})}^{3} = - \frac{1}{2} {(\log \frac{\sqrt{3}}{2})}^{2} + 0$

$\Rightarrow x = - \frac{1}{2 \log (\frac{\sqrt{3}}{2})} = \frac{1}{\log (\frac{4}{3})}$

$= \frac{1}{\log 4 - \log 3} = \frac{1}{\log_{e} m - \log_{e} n} \Rightarrow m = 4, n = 3$

$∴$ $m n = 4 \times 3 = 12$

Q 83 :

Let the tangent at any point P on a curve passing through the points (1,1) and $(\frac{1}{10}, 100),$ intersect the positive $x$ -axis and $y$ -axis at the points A and B respectively. If $P A : P B = 1 : k$ and $y = y (x)$ is the solution of the differential equation $e^{\frac{d y}{d x}} = k x + \frac{k}{2},$ $y (0) = k$ , then $4 y (1) - 5 \log_{e} 3$ is equal to ________ . [2023]

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Equation of tangent at $P (x, y)$

$Y - y = \frac{d y}{d x} (X - x)$

$Y = 0 at point A, then X = \frac{- y d x}{d y} + x$

Point $P (x, y)$ divides $A B$ in $k : 1$ ratio,

$∴$ $\frac{k α + 0}{k + 1} = x \Rightarrow α = \frac{k + 1}{k} x$

$x + \frac{x}{k} = - y \frac{d x}{d y} + x$

$\Rightarrow \frac{d y}{d x} + \frac{k}{x} y = 0$

$\Rightarrow y \cdot x^{k} = C$

$\Rightarrow C = 1 [∵ Tangent passing through (1, 1)]$

From point $(\frac{1}{10}, 100),$

$100 {(\frac{1}{10})}^{k} = 1 \Rightarrow k = 2$

$\frac{d y}{d x} = \ln (2 x + 1) [Given, e^{\frac{d y}{d x}} = k x + \frac{k}{2}]$

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

Now, $2 = \frac{1}{2} (0 - 1) + C [∵ y (0) = k and k = 2]$

$C = 2 + \frac{1}{2} = \frac{5}{2}$

Now, $y (1) = \frac{3}{2} (\ln 3 - 1) + \frac{5}{2} = \frac{3}{2} \ln 3 + 1 \Rightarrow 4 y (1) = 6 \ln 3 + 4$

$∴$ $4 y (1) - 5 \ln 3 = 6 \ln 3 + 4 - 5 \ln 3 = \ln 3 + 4 \approx 5.09 \approx 5$ (approx.)

Q 84 :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{4 x}{(x^{2} - 1)} y = \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}}, x > 1$ such that $y (2) = \frac{2}{9} \log_{e} (2 + \sqrt{3})$ and $y (\sqrt{2}) = α \log_{e} (\sqrt{α} + β) + β - \sqrt{γ}, α, β, γ \in ℕ,$ then $α β γ$ is equal to ______ . [2023]

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$\frac{d y}{d x} + \frac{4 x}{(x^{2} - 1)} y = \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}}, x > 1$

$I.F. = e^{2 \ln (x^{2} - 1)} = {(x^{2} - 1)}^{2}$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \int \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}} \times {(x^{2} - 1)}^{2} d x$

$\Rightarrow$ $y \times {(x^{2} - 1)}^{2} = \int \frac{x + 2}{\sqrt{x^{2} - 1}} d x$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \int \frac{x d x}{\sqrt{x^{2} - 1}} + 2 \int \frac{d x}{\sqrt{x^{2} - 1}}$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \sqrt{x^{2} - 1} + 2 \ln$ $| x + \sqrt{x^{2} - 1} |$ $+ C$

$\Rightarrow y (2) = {(3)}^{- 3 / 2} + \frac{2 \ln | 2 + \sqrt{3} |}{9} + \frac{C}{9}$

$\Rightarrow \frac{2}{9} \log_{e} (2 + \sqrt{3}) = {(3)}^{- 3 / 2} + \frac{2}{9} \log_{e}$ $| 2 + \sqrt{3} |$ $+ \frac{C}{9}$

$\Rightarrow \frac{C}{9} = - {(3)}^{- 3 / 2} \Rightarrow C = - \sqrt{3}$

$∴$ $y = {(x^{2} - 1)}^{- \frac{3}{2}} + \frac{2 \log | x + \sqrt{x^{2} - 1} |}{{(x^{2} - 1)}^{2}} - \frac{\sqrt{3}}{{(x^{2} - 1)}^{2}}$

$Now, y (\sqrt{2}) = {(2 - 1)}^{- \frac{3}{2}} + \frac{2 \log | \sqrt{2} + \sqrt{1} |}{1} - \frac{\sqrt{3}}{1}$

$\Rightarrow α \log_{e} (\sqrt{α} + β) + β - \sqrt{γ} = 1 + \log (\sqrt{2} + 1) - \sqrt{3}$

$∴$ $α = 2, β = 1, γ = 3 \Rightarrow α β γ = 6$

Q 85 :

If $f : R \to R$ is a differentiable function such that $f' (x) + f (x) = \int_{0}^{2} f (t) d t .$ If $f (0) = e^{- 2},$ then $2 f (0) - f (2)$ is equal to _____ . [2023]

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$f' (x) + f (x) = \int_{0}^{2} f (t) d t and f (0) = e^{- 2}$

$Let \int_{0}^{2} f (t) d t = K$

$\Rightarrow \frac{d y}{d x} + y = K [Where \frac{d y}{d x} = f' (x), y = f (x)]$

$\Rightarrow y e^{x} = K e^{x} + C;$ $Now f (0) = e^{- 2}$

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

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

$Now, K = \int_{0}^{2} f (x) d x = \int_{0}^{2} (K + (e^{- 2} - K) e^{- x}) d x$

$= 2 K + {[(K - e^{- 2}) e^{- x}]}_{0}^{2}$ $= 2 K + (K - e^{- 2}) (e^{- 2} - 1)$

$\Rightarrow K = 2 K + K e^{- 2} - K - e^{- 4} + e^{- 2}$ $\Rightarrow K = \frac{e^{- 4} - e^{- 2}}{e^{- 2}}$

$\Rightarrow K = e^{- 2} - 1$

$Now, 2 f (0) - f (2) = 2 e^{- 2} - 2 e^{- 2} + 1 = 1$

Q 86 :

Let $f$ be a differentiable function defined on $[0, \frac{π}{2}]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e, \forall x \in [0, \frac{π}{2}] .$ Then $(6 \log_{e} f {(\frac{π}{6}))}^{2}$ is equal to _____ . [2023]

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$Given, f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e$ ...(i)

$\Rightarrow f (0) = e$

$Differentiating (i) both sides, we get$

$f' (x) + f (x) \sqrt{1 - (\ln f {(x))}^{2}} = 0$

$Put f (x) = y$

$\Rightarrow \frac{d y}{d x} = - y \sqrt{1 - {(\ln y)}^{2}}$ $\Rightarrow \int \frac{d y}{y \sqrt{1 - {(\ln y)}^{2}}} = - \int d x$

$Put \ln y = t$

$\Rightarrow \int \frac{d t}{\sqrt{1 - t^{2}}} = - x + c \Rightarrow \sin^{- 1} t = - x + c$

$\Rightarrow \sin^{- 1} (\ln y) = - x + c$

$\Rightarrow \sin^{- 1} (\ln f (x)) = - x + c; f (0) = e$

$\Rightarrow \frac{π}{2} = C \Rightarrow \sin^{- 1} (\ln (f (x))) = - x + \frac{π}{2}$

$\Rightarrow \sin^{- 1} (\ln f (\frac{π}{6})) = \frac{- π}{6} + \frac{π}{2} = \frac{π}{3}$

$∴$ $\ln f (\frac{π}{6}) = \sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$

$Thus,$ $(6 \log_{e} f {(\frac{π}{6}))}^{2} = {(6 \times \frac{\sqrt{3}}{2})}^{2} = 27$

Q 87 :

Let $y = y (x)$ be the solution of the differential equation $x^{4} d y + (4 x^{3} y + 2 \sin x) d x = 0, x > 0,$ $y (\frac{π}{2}) = 0 .$ Then $π^{4} y (\frac{π}{3})$ is equal to: [2026]

81
72
92
64

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Q 88 :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - \tan^{- 1} x) d x = 0, y (0) = 1 .$ Then the value of $y (1)$ is: [2026]

$\frac{4}{e^{π / 4}} - \frac{π}{2} - 1$
$\frac{2}{e^{π / 4}} + \frac{π}{4} - 1$
$\frac{4}{e^{π / 4}} + \frac{π}{2} - 1$
$\frac{2}{e^{π / 4}} - \frac{π}{4} - 1$

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

$I.F. = e^{\tan^{- 1} x}$

$y \times e^{\tan^{- 1} x} = \int e^{\tan^{- 1} x} \cdot \frac{\tan^{- 1} x}{1 + x^{2}} d x$

$y \times e^{\tan^{- 1} x} = \tan^{- 1} x (e^{\tan^{- 1} x}) - e^{\tan^{- 1} x} + c$

$y (0) = 1 \Rightarrow c = 2$

$y (1) = \frac{2}{e^{π / 4}} + \frac{π}{4} - 1$

Q 89 :

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

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Q 90 :

Let y=y(x) be the solution of the differential equation $x \frac{d y}{d x} - y = x^{2} \cot x, x \in (0, π) .$ If $y (\frac{π}{2}) = \frac{π}{2},$ then $6 y (\frac{π}{6}) - 8 y (\frac{π}{4})$ is equal to: [2026]

$π$
$3 π$
$- π$
$- 3 π$

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