A curve satisfies the differential equation d y d x = x + 1 - x y 2 x 2 y - y and passes through ( 0 , 0 )

The equation of the curve is x 2 + y 2 + 2 x + 2 y = x 2 y 2

The differential equation d y d x = 1 - y 2 y determines a family of circles with

fixed radius 1 and variable centres along the x -axis

variable radii and a fixed centre at ( 0 , 1 )

variable radii and a fixed centre at ( 0 , - 1 )

fixed radius 1 and variable centres along the y -axis

Solve the differential equation ( x d y - y d x ) y sin y x = ( y d x + x d y ) x cos y x .

Given, ( x d y - y d x ) y sin y x = ( y d x + x d y ) x cos y x ⇒ x y sin y x d y - y 2 sin y x d x = x y cos y x d x + x 2 cos y x d y ⇒ x y sin y x - x 2 cos y x d y = x y cos y x + y 2 sin y x d x ⇒ d y d x = x y cos y x + y 2 sin y x x y sin y x - x 2 cos y x ⇒ d y d x = y x cos y x + y x 2 cos y x y x sin y x - cos y x ⇒ y = v x ⇒ d y d x = x d v d x + v ⇒ v + xdv dx = vx x cos vx x + v 2 x 2 x 2 sin vx x vx x sin vx x - cos vx x ⇒ x dv dx = vcosv + v 2 sinv vsinv - cosv - v ⇒ x dv dx = vcosv + v 2 sinv - v 2 sinv + vcosv vsinv - cosv ⇒ xdv dx = 2 vcosv vsinv - cosv ⇒ vsinv - cosv vcosv dv = 2 dx x ⇒ vsinv vcosv - cosv vcosv dv = 2 dx x ⇒ tanv - 1 v dv = 2 dx x ⇒ ∫ tanv - 1 v dv = 2 ∫ dx x ⇒ ∫ tanvdv - ∫ dv v = 2 ∫ dx x ⇒ log | secv | - log | v | = 2 log | x | + logC ⇒ log secv v = logx 2 + logC ⇒ log secv v = logx 2 c ⇒ secv v = x 2 c ⇒ sec y x y x = x 2 c ⇒ sec y x = y x x 2 c ⇒ sec y x = Cxy is the required differential equation.

Let T denote a curve y = y x which is in the first quadrant and let the point 1 , 0 lie on it. Let the tangent to T at a point P intersect the y -axis at Y P . If P Y P has length 1 for each point P on T , then which of the following options is/are correct?

y = - l o g e 1 + 1 - x 2 x + 1 - x 2

E M B I B E

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 13

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IMPORTANT

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A curve satisfies the differential equation $\frac{d y}{d x} = \frac{x + 1 - x y^{2}}{x^{2} y - y}$ and passes through $(0, 0)$

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Important Questions on Differential Equations

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 14

Tangent is drawn at any point

P

of a curve which passes through

(1, 1)

cutting

X

-axis and

Y

-axis at

A

and

B

respectively. If

B P : A P = 3 : 1,

then

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Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 15

y = c_{1} x + c_{2} \sin (2 x + c_{3}), (c_{1}, c_{2}, c_{3}

are arbitrary constants)

HARD

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 16

Which of the following statements is/are true?

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Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 17

A curve passes through $(1, 0)$ and satisfies the defferential equation $(2 x \cos y + 3 x^{2} y) d x +$

$(x^{3} - x^{2} \sin y - y) d y = 0$

HARD

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 1

The differential equation

\frac{d y}{d x} = \frac{\sqrt{1 - y^{2}}}{y}

determines a family of circles with

HARD

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 2

A curve passes through the point

(1, \frac{π}{6}) .

Let the slope of the curve at each point

(x, y)

be

\frac{y}{x} + \sec (\frac{y}{x}), x > 0 .

Then, the equation of the curve is

HARD

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Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment>Q 3

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt{1 - x^{2}}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then, $\int_{- \frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f (x) d x$ .