Free Differential Equations 01 Practice Test - 12th Grade - Commerce

Question 1

The solution of $(y (1 + x^{- 1}) + s i n y) d x + (x + l o g x + x c o s y) d y = 0$ is

(1 + y^{- 1} s i n y) + x^{- 1} l o g x = C

(y + s i n y) + x y l o g x = C

x y + y l o g x + x s i n y = C

N o n e o f t h e s e

SOLUTION

Solution : C

The given equation can be written as $y (1 + x^{- 1}) d x + (x + l o g x) d y + s i n y d x + x c o s y d y = 0$
$\Rightarrow d (y (x + l o g x)) + d (x s i n y) = 0 \to y (x + l o g x) + x s i n y = C$

Question 2

The solution of $y_{2} - 7 y_{1} + 12 y = 0$ is

y = C_{1} e^{3 x} + C_{2} e^{4 x}

y = C_{1} x e^{3 x} + C_{2} e^{4 x}

y = C_{1} e^{3 x} + C_{2} x e^{4 x}

N o n e o f t h e s e

SOLUTION

Solution : A

The given equation can be written as $(\frac{d}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0 . . . . (i)$
If $\frac{d y}{d x} - 4 y = u$ then (I) reduces to $\frac{d u}{d x} - 3 u = 0$
$\Rightarrow \frac{d u}{u} = 3 d x \Rightarrow u = C_{1} e^{3 x}$ . Therefore, we have $\frac{d y}{d x} - 4 y = C_{1} e^{3 x}$ which is a linear equation whose I.F.is $e^{- 4 x}$ . So $\frac{d}{d x} (y e^{- 4 x}) = C_{1} e^{- x}$
$\Rightarrow y e^{- 4 x} = - C_{1} e^{- x} + C_{2} \Rightarrow y = C_{1} e^{3 x} + C_{2} e^{4 x}$

Question 3

The solution lof $\frac{d y}{d x} - x t a n (y - x) = 1$

c e \frac{e^{x^{2}}}{2} = s i n (y - x)

c e \frac{e^{x^{2}}}{2} = s i n (y + x)

c e \frac{e^{x^{2}}}{2} = s i n (y - x)^{2}

c e \frac{e^{x^{2}}}{2} = c o s (y - x)

SOLUTION

Solution : A

Put y-x = z. Then $\frac{d y}{d x} - 1 = \frac{d z}{d x} \Rightarrow \frac{d y}{d x} = 1 + \frac{d z}{d x}$
Given $\frac{d y}{d x} - x t a n (y - x) = 1 \Rightarrow 1 + \frac{d z}{d x} - x t a n z = 1 \Rightarrow \frac{d z}{d x} = x t a n z \Rightarrow \frac{1}{t a n z} d z = x d x \Rightarrow \int c o t z d z = \int x d x$
$\Rightarrow l o g | s i n z | - l o g c = \frac{x^{2}}{2} \Rightarrow l o g | \frac{s i n z}{c} | = \frac{x^{2}}{2} = \frac{s i n z}{c} = \frac{e^{x^{2}}}{2} \Rightarrow s i n (y - x) = c \frac{e^{x^{2}}}{2}$
$∴$ The solution is $s i n (y - x) = c \frac{e^{x^{2}}}{2}$ , where c is arbitrary constant.

Question 4

Solution of the differential equation : $\frac{d y}{d x} = \frac{3 x^{2} y^{4} + 2 x y}{x^{2} - 2 x^{3} y^{3}}$ is

x^{2} y^{2} + \frac{x^{2}}{y} = c

x^{3} y^{2} + \frac{x^{2}}{y} = c

x^{3} y^{2} + \frac{y^{2}}{x} = c

x^{2} y^{3} + \frac{x^{2}}{y} = c

SOLUTION

Solution : B

$x^{2} d y - 2 x^{3} y^{3} d y = 3 x^{2} y^{4} d x + 2 x y d x \Rightarrow x^{2} d y - 2 x y d x = 3 x^{2} y^{4} d x + 2 x^{3} y^{3} d y \Rightarrow \frac{2 x y d x - x^{2} d y}{y^{2}} + 3 x^{2} y^{2} d x + 2 x^{3} y d y = 0 \Rightarrow d (\frac{x^{2}}{y}) + d (x^{3} y^{2}) = 0 \Rightarrow \frac{x^{2}}{y} + x^{3} y^{2} = C$

Question 5

The degree of the differential equation satisfying the relation $\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}} - y \sqrt{1 + x^{2}})$ is

A. 1

B. 2

C. 3

D. none of these

SOLUTION

Solution : A

On Putting $x = t a n A, y = t a n B$ we get
$s e c A + s e c B = λ (t a n A s e c B - t a n B s e c A) c o s A + c o s B = λ (s i n A - s i n B) t a n (\frac{A - B}{2}) = \frac{1}{λ} t a n^{- 1} x - t a n^{- 1} y = 2 t a n^{- 1} \frac{1}{λ}$
On differentiating $\frac{1}{1 + x^{2}} - \frac{1}{1 + y^{2}} \frac{d y}{d x} = 0$

Question 6

The order of the differential equation of all tangent lines to the parabola $y = x^{2}$ is

A. 1

B. 2

C. 3

D. 4

SOLUTION

Solution : A

The parametric form of the given equation is $x = t, y = t^{2}$ .The equation of any tangent at t is $2 x t = y + t^{2}$ , Differentiating we get $2 t = y_{1} (= \frac{d y}{d x})$ putting this value in the above equation, we have $2 x \frac{y_{1}}{2} = y + {(\frac{y_{1}}{2})}^{2} \Rightarrow 4 x y_{1} = 4 y + y_{1}^{2}$
The order of this equation is 1
Hence (A) is the correct answer

Question 7

A function y =f(x) has a second order derivative f"=6(x-1). If its graph passes through the point(2,1) and at that point the tangent to the graph is y =3x -5, then the function is

(x - 1)^{2}

(x - 1)^{3}

(x + 1)^{2}

(x + 1)^{3}

SOLUTION

Solution : B

Since f"(x)=6(x-1)
$\Rightarrow f^{'} (x) = 3 (x - 1)^{2} + c$ (integrating) ----(i)
Also, at the point (2,1), the tangent to the graph is y =3x-5and slope of thetangent = 3
$\Rightarrow f^{'} (2) = 3$
$3 (2 - 1)^{3} + c = 3$ [from eq(i)
$\Rightarrow 3 + c = 3 \Rightarrow c = 0$
From Eq (i) we have
$f^{'} (x) = 3 (x - 1)^{2}$
$\Rightarrow f (x) = (x - 1)^{3} + k$ (Integrating)----(ii)
$∴ 1 = (2 - 1)^{3} + k \Rightarrow k = 0$
Hence the equation of the function is $f (x) = (x - 1)^{3}$ .

Question 8

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

2,2

3,3

2,3

3,2

SOLUTION

Solution : C

Here power on the differential coefficient is fractional, therefore change it into positive integer, so
${[4 + {(\frac{d y}{d x})}^{2}]}^{2 / 3} = \frac{d^{2} y}{d x^{2}} \Rightarrow {[4 + {(\frac{d y}{d x})}^{2}]}^{2} = {[\frac{d^{2} y}{d x^{2}}]}^{3}$
Hence order is 2 and degree is 3.

Question 9

The order and degree of the differential equation $y = x \frac{d y}{d x} + \sqrt{a^{2} {(\frac{d y}{d x})}^{2} + b^{2}}$ are

1,2

2,1

1,1

2,2

SOLUTION

Solution : A

Given differential equation can be written as
$y^{2} = x^{2} {(\frac{d y}{d x})}^{2} - 2 x y . \frac{d y}{d x} = a^{2} {(\frac{d y}{d x})}^{2} + b^{2} .$
Hence it is of 1^st order and 2^nddegree differential equation.

Question 10

The general solution of $y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$ is

[EAMCET 2003]

$t a n^{- 1} (\frac{x}{y}) + l o g y + c = 0$

$2 t a n^{- 1} (\frac{x}{y}) + l o g x + c = 0$

$l o g (y + \sqrt{x^{2} + y^{2}}) + l o g y + c = 0$

$s i n h^{- 1} (\frac{x}{y}) + l o g y + c = 0$

SOLUTION

Solution : A

$\frac{d x}{d y} + \frac{x^{2} - x y + y^{2}}{y^{2}} = 0 \frac{d x}{d y} + {(\frac{x}{y})}^{2} - (\frac{x}{y}) + 1 = 0$
Put $v = x / y \Rightarrow x = v y \Rightarrow \frac{d x}{d y} = v + y \frac{d v}{d y} v + y \frac{d v}{d y} + v^{2} - v + 1 = 0 \Rightarrow \frac{d v}{v^{2} + 1} + \frac{d y}{y} = 0 \Rightarrow \int \frac{d v}{v^{2} + 1} + \int \frac{d y}{y} = 0 \Rightarrow t a n^{- 1} (v) + l o g y + C = 0 \Rightarrow t a n^{- 1} (x / y) + l o g y + c = 0.$

Free Differential Equations 01 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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Inverse Trigonometric Functions 02
Probability 01
Application of Derivatives 02