JEE Main Differential Equations Practice Questions With Solutions

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JEE Main Mathematics Differential Equations Practice Questions

Question 1.

Let $\int_{0}^{x} \sqrt{1 - {(y^{'} (t))}^{2}} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

\sqrt{2}

1/2

Question 2.

The solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0$ , is :

{| x^{2} - 4 y^{2} |}^{5} = x^{2}

{| x^{2} - 2 y^{2} |}^{6} = x

{| x^{2} - 2 y^{2} |}^{5} = x^{2}

{| x^{2} - 4 y^{2} |}^{6} = x

Question 3.

The solution curve, of the differential equation $2 y \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$ , passing through the point $(0, 1)$ is a conic, whose vertex lies on the line :

2 x + 3 y = - 9

2 x + 3 y = - 6

2 x + 3 y = 9

2 x + 3 y = 6

Question 4.

Let $y = y (x)$ be the solution curve of the differential equation $\sec y \frac{d y}{d x} + 2 x \sin y = x^{3} \cos y, y (1) = 0$ . Then $y (\sqrt{3})$ is equal to:

\frac{π}{6}

\frac{π}{12}

\frac{π}{3}

\frac{π}{4}

Question 5.

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

(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

Question 6.

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{\tan x} d x + \cos^{2} x (1 + e^{2 \tan x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

\frac{1}{e^{2}}

\frac{2}{e^{2}}

\frac{2}{e}

\frac{1}{e}

Question 7.

Suppose the solution of the differential equation $\frac{d y}{d x} = \frac{(2 + α) x - β y + 2}{β x - 2 α y - (β γ - 4 α)}$ represents a circle passing through origin. Then the radius of this circle is :

\sqrt{17}

\frac{\sqrt{17}}{2}

\frac{1}{2}

Question 8.

Let $y = y (x)$ be the solution of the differential equation $(2 x \log_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} \log_{e} x, x > 0$ and $y (e^{- 1}) = 0$ . Then, $y (e)$ is equal to

- \frac{3}{e}

- \frac{3}{2 e}

- \frac{2}{3 e}

- \frac{2}{e}

Question 9.

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$ , $y (1) = 0$ . Then $y (0)$ is

\frac{1}{4} (e^{π / 2} - 1)

\frac{1}{2} (1 - e^{π / 2})

\frac{1}{4} (1 - e^{π / 2})

\frac{1}{2} (e^{π / 2} - 1)

Question 10.

The differential equation of the family of circles passing through the origin and having centre at the line $y = x$ is :

(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

Question 1.

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + 2 y = \sin (2 x), y (0) = \frac{3}{4}$ , then $y (\frac{π}{8})$ is equal to :

e^{- π / 8}

e^{π / 4}

e^{- π / 4}

e^{π / 8}

Question 2.

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

2 π

\frac{π}{8}

\frac{π}{16}

\frac{π}{32}

Question 3.

If the solution $y = y (x)$ of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then $y (0)$ is equal to :

- \frac{π}{12}

\frac{π}{2}

\frac{π}{4}

Question 4.

Let

α

be a non-zero real number. Suppose

f : R \to R

is a differentiable function such that

f (0) = 2

and

lim_{x \to - \infty} f (x) = 1

. If

f^{'} (x) = α f (x) + 3

, for all

x \in R

, then

f (- \log_{e} 2)

is equal to :

Question 5.

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} = 2 x (x + y)^{3} - x (x + y) - 1, y (0) = 1

.

Then,

{(\frac{1}{\sqrt{2}} + y (\frac{1}{\sqrt{2}}))}^{2}

equals :

\frac{4}{4 + \sqrt{e}}

\frac{3}{3 - \sqrt{e}}

\frac{2}{1 + \sqrt{e}}

\frac{1}{2 - \sqrt{e}}

Question 6.

The temperature $T (t)$ of a body at time $t = 0$ is $160^{\circ} F$ and it decreases continuously as per the differential equation $\frac{d T}{d t} = - K (T - 80)$ , where $K$ is a positive constant. If $T (15) = 120^{\circ} F$ , then $T (45)$ is equal to

^{\circ}

^{\circ}

^{\circ}

^{\circ}

Question 7.

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

\sqrt{3} (2 + \log_{e} 3)

\sqrt{3} (1 + 2 \log_{e} 3)

\sqrt{3} (2 + \log_{e} \sqrt{3})

\frac{\sqrt{3}}{2} (2 + \log_{e} 3)

Question 8.

The solution curve of the differential equation $y \frac{d x}{d y} = x (\log_{e} x - \log_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

| \log_{e} \frac{y}{x} | = y^{2}

| \log_{e} \frac{y}{x} | = x

| \log_{e} \frac{x}{y} | = y

2 | \log_{e} \frac{x}{y} | = y + 1

Question 9.

Let $y = y (x)$ be the solution of the differential equation $\sec x d y + {2 (1 - x) \tan x + x (2 - x)} d x = 0$ such that $y (0) = 2$ . Then $y (2)$ is equal to:

2 {\sin (2) + 1}

2 {1 - \sin (2)}

Question 10.

If $\sin (\frac{y}{x}) = \log_{e} | x | + \frac{α}{2}$ is the solution of the differential equation $x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

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JEE Main Differential Equations Practice Questions With Solutions

JEE Main Mathematics Differential Equations Practice Questions

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

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