IIT JEE Main Maths -Unit 9 -Ordinary differential equations- Exam Style Questions- New Syllabus

Question

Let $x = x(y)$ be the solution of the differential equation $\left(\dfrac{y^2+1}{y}\right)dx – x\,dy = 0$. If $x(1) = 1$, then $\dfrac{x(1/2)}{2}$ is:

$(1)\ 1+\dfrac{e}{2}$

$(2)\ 3+\dfrac{e}{2}$

$(3)\ 3 – e$

$(4)\ 3 + e$

▶️ Answer/Explanation

$\frac{dx}{dy} + \left(\frac{1}{y^2}\right)x = \frac{1}{y^3}$

$\text{I.F.} = e^{\int \frac{1}{y^2}dy} = e^{-\frac{1}{y}}$

$\Rightarrow x \cdot e^{-\frac{1}{y}} = \int \left(e^{-\frac{1}{y}}\right)\frac{1}{y^3}dy$

Put $-\frac{1}{y} = t$

$+\frac{1}{y^2}dy = dt$

$x \cdot e^{-\frac{1}{y}} = – \int t e^t dt$

$x \cdot e^{-\frac{1}{y}} = -te^t + e^t + C$

$x \cdot e^{-\frac{1}{y}} = \frac{1}{e} + \frac{1}{y} e^{-\frac{1}{y}} + C$

For $x=1, \, y=1$:

$\frac{1}{e} = \frac{1}{e} + \frac{1}{e} + C$

$\Rightarrow C = -\frac{1}{e}$

Put $y = \frac{1}{2}$:

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

$\therefore x = 3 – e$

Question

If $x = f(y)$ is the solution of the differential equation $(1 + y^2) + (\tan^{-1} y) \frac{dy}{dx} (x – 2e^{-1}) = 0$, $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$ with $f(0) = 1$, then $f(\frac{1}{\sqrt{3}})$ is equal to :
(1) $e^{\pi/4}$
(2) $e^{\pi/12}$
(3) $e^{\pi/3}$
(4) $e^{\pi/6}$

▶ Answer/Explanation

Ans. (4)

$ \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{2e^{\tan^{-1}y}}{1 + y^2} $

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

$ x e^{\tan^{-1}y} = \int \frac{2(e^{\tan^{-1}y})^2}{1 + y^2} \, dy $

$\text{Put } \tan^{-1}y = t, \, \frac{dy}{1 + y^2} = dt $

$ x e^{\tan^{-1}y} = \int 2e^{2t} \, dt $

$ x e^{\tan^{-1}y} = e^{2t} + c $

$ x = e^{t – \tan^{-1}y} + c e^{-\tan^{-1}y} $

$\text{When } y = 0, \, x = 1: \, 1 = 1 + c \Rightarrow c = 0 $

$ y = \frac{1}{\sqrt{3}}, \, x = e^{\pi/6} $

Question

Let $y = f(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{xy^2 + x}{x^2 – 1} = \frac{4x}{1 – x^2}$, $-1 < x < 1$ such that $f(0) = 0$. If $6 \int_{-1/2}^{1/2} f(x) dx = \alpha \pi$, then $\alpha^2$ is equal to __________.

▶ Answer/Explanation

Ans. (27)

$ \text{I.F.} = e^{-\frac{1}{2} \int \frac{2x}{1 – x^2} \, dx} = e^{-\frac{1}{2} \ln(1 – x^2)} = \sqrt{1 – x^2} $

$ y \sqrt{1 – x^2} = \int (x^6 + 4x) \, dx = \frac{x^7}{7} + 2x^2 + c $

$\text{Given } y(0) = 0 \Rightarrow c = 0 $

$ y = \frac{\frac{x^7}{7} + 2x^2}{\sqrt{1 – x^2}} $

$\text{Now, } 6 \int_{0}^{\frac{1}{2}} \frac{\frac{x^7}{7} + 2x^2}{\sqrt{1 – x^2}} \, dx = 6 \int_{0}^{\frac{1}{2}} \frac{2x^2}{\sqrt{1 – x^2}} \, dx $

$ = 24 \int_{0}^{\frac{1}{2}} \frac{x^2}{\sqrt{1 – x^2}} \, dx $

\text{Put } $ x = \sin \theta $

$ dx = \cos \theta \, d\theta $

$ = 24 \int_{0}^{\frac{\pi}{6}} \frac{\sin^2 \theta}{\cos \theta} \cos \theta \, d\theta $

$ = 24 \int_{0}^{\frac{\pi}{6}} \frac{1 – \cos 2\theta}{2} \, d\theta = 12 \left[ \theta – \frac{\sin 2\theta}{2} \right]_{0}^{\frac{\pi}{6}} $

$ = 12 \left( \frac{\pi}{6} – \frac{\sqrt{3}}{4} \right) $

$ = 2\pi – 3\sqrt{3} $

$ \alpha^2 = (3\sqrt{3})^2 = 27 $

Question

5. Let a curve $ y = f(x) $ pass through the points $ (0, 5) $ and $ (\log_e 2, k) $. If the curve satisfies the differential equation
$ 2(3 + y)e^{2x}dx – (7 + e^{2x})dy = 0 $, then $ k $ is equal to

(1) $16 $

(2) $8$

(3) $32 $

(4) $4$

▶️ Answer/Explanation

Ans. (2)

Sol.
$ 2(3 + y)e^{2x}dx = (7 + e^{2x})dy $

$ \Rightarrow \frac{dy}{dx} = \frac{2(3 + y)e^{2x}}{7 + e^{2x}} $

$ \Rightarrow \frac{dy}{dx} – \frac{2e^{2x}}{7 + e^{2x}}y = \frac{6e^{2x}}{7 + e^{2x}} $

Integrating factor $ I.F. = e^{-\int \frac{2e^{2x}}{7 + e^{2x}}dx} = e^{-\ln(7 + e^{2x})} = \frac{1}{7 + e^{2x}} $

$ \therefore y \cdot \frac{1}{7 + e^{2x}} = \int \frac{6e^{2x}}{(7 + e^{2x})^2} dx $

Put $ 7 + e^{2x} = t \Rightarrow dt = 2e^{2x} dx $

$ \Rightarrow \int \frac{6e^{2x}}{(7 + e^{2x})^2} dx = 3 \int \frac{dt}{t^2} = -\frac{3}{t} + C $

$ \therefore y = -3(7 + e^{2x}) + C(7 + e^{2x}) $

Using $ (0, 5) \Rightarrow 5 = -3(8) + 8C \Rightarrow 5 = -24 + 8C \Rightarrow C = \frac{29}{8} $

$ \therefore y = -3 + 7 + e^{2x} = e^{2x} + 4 $

$ \therefore k = 8 $

IIT JEE Main Maths -Unit 9 -Ordinary differential equations- Exam Style Questions- New Syllabus

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