Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

The solution of $x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x$ is

$\tan \frac{y}{2 x} = c - \frac{1}{2 x^{2}}$
$\tan \frac{y}{x} = c + \frac{1}{x}$
$\cos (\frac{y}{x}) = 1 + \frac{c}{x}$
$x^{2} = (c + x^{2}) \tan \frac{y}{x}$

hint Hint:

In the question we have to integrate the given function.

The correct answer is: $tan space fraction numerator y over denominator 2 x end fraction equals c minus fraction numerator 1 over denominator 2 x squared end fraction$

$x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x rightwards double arrow x squared fraction numerator d y over denominator d x end fraction minus x y equals 2 cos squared fraction numerator y over denominator 2 x end fraction rightwards double arrow fraction numerator 1 over denominator cos squared begin display style fraction numerator y over denominator 2 x end fraction end style end fraction open square brackets x squared. fraction numerator d y over denominator d x end fraction minus x y close square brackets equals 2 rightwards double arrow 1 half s e c squared fraction numerator y over denominator 2 x end fraction open square brackets fraction numerator x. begin display style fraction numerator d y over denominator d x end fraction end style minus y over denominator x squared end fraction close square brackets equals 1 over x cubed rightwards double arrow fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction equals 1 over x cubed O n space i n t e g r a t i n g space w e space g e comma integral fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction d x equals integral 1 over x cubed d x tan fraction numerator y over denominator 2 x end fraction equals C minus fraction numerator 1 over denominator 2 x squared end fraction$

Related Questions to study

For the given figure, calculate zero correction.

For the given figure, calculate zero correction.

physics-General

Three capacitors $C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript$ are connected as shown in the figure to a battery of $V$ volt. If the capacitor $C subscript 3 end subscript$ breaks down electrically the change in total charge on the combination of capacitors is

Three capacitors $C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript$ are connected as shown in the figure to a battery of $V$ volt. If the capacitor $C subscript 3 end subscript$ breaks down electrically the change in total charge on the combination of capacitors is

physics-General

The charge deposited on $4 mu F$ capacitor the circuit is

The charge deposited on $4 mu F$ capacitor the circuit is

physics-General

parallel

If $y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis$ then y satisfies

If $y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis$ then y satisfies

In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

physics-General

Solution of $fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent$ is

Solution of $fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent$ is

parallel

Two charges $q subscript 1 end subscript$ and $q subscript 2 end subscript$ are placed 30 cm apart, as shown in the figure. A third charge $q subscript 3 end subscript$ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is $fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma$ where k is

Two charges $q subscript 1 end subscript$ and $q subscript 2 end subscript$ are placed 30 cm apart, as shown in the figure. A third charge $q subscript 3 end subscript$ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is $fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma$ where k is

physics-General

Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential difference $V subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript$ . If $t subscript 1 end subscript$ and $t subscript 2 end subscript$ be the distance between them. Then

Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential difference $V subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript$ . If $t subscript 1 end subscript$ and $t subscript 2 end subscript$ be the distance between them. Then

physics-General

Charges +q and –q are placed at points A and B respectively which are a distance $2 L$ apart, C is the mid-point between A and B. The work done in moving a charge+ $Q$ along the semicircle CRD is

Charges +q and –q are placed at points A and B respectively which are a distance $2 L$ apart, C is the mid-point between A and B. The work done in moving a charge+ $Q$ along the semicircle CRD is

physics-General

parallel

Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure $left parenthesis a less than b less than c right parenthesis.$ Their surface charge densities are $sigma comma negative sigma a n d sigma$ respectively. Calculate the electric potential on the surface of shell A

Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure $left parenthesis a less than b less than c right parenthesis.$ Their surface charge densities are $sigma comma negative sigma a n d sigma$ respectively. Calculate the electric potential on the surface of shell A

physics-General

Work required to set up the four charge configuration (as shown in the figure) is

Work required to set up the four charge configuration (as shown in the figure) is

physics-General

The points resembling equal potentials are

The points resembling equal potentials are

physics-General

parallel

In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as $W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript$ then

In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as $W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript$ then

physics-General

Three charges $Q subscript 0 end subscript comma negative q$ and $– q$ are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if $Q subscript 0 end subscript$ is equal to

Three charges $Q subscript 0 end subscript comma negative q$ and $– q$ are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if $Q subscript 0 end subscript$ is equal to

physics-General

A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. $V subscript A end subscript comma blank V subscript B end subscript comma blank V subscript c end subscript$ be the potentials at points A, B and C respectively. Then

A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. $V subscript A end subscript comma blank V subscript B end subscript comma blank V subscript c end subscript$ be the potentials at points A, B and C respectively. Then

physics-General

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.