Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

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

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

$\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} = 0$
$\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$
$\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y = 0$
$\frac{d^{2} y}{d x^{2}} + 2 y = 0$

hint Hint:

In this question we have to differentiate the function twice to get the required result.

The correct answer is: $fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction plus 2 y equals 0$

$I f space y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis d i f f e r e n t i a t i n g space b o t h space s i d e s comma space w e space g e t fraction numerator d y over denominator d x end fraction equals fraction numerator d over denominator d x end fraction e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis rightwards double arrow fraction numerator d y over denominator d x end fraction equals negative e to the power of negative x end exponent open parentheses A cos x plus B sin x close parentheses plus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses rightwards double arrow fraction numerator d y over denominator d x end fraction equals negative y plus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses A g a i n comma space d i f f e r e n t i a t i n g space b o t h space s i d e s comma space w e space g e t rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator negative d y over denominator d x end fraction plus e to the power of negative x end exponent open parentheses negative A cos x minus B sin x close parentheses minus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals negative e to the power of negative x end exponent open parentheses A cos x plus B sin x close parentheses minus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals negative y minus y minus fraction numerator d y over denominator d x end fraction rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction plus 2 y equals 0$

Related Questions to study

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

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

parallel

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

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

parallel

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

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

parallel

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

The figure shows electric potential V as a function of $x$ . Rank the four regions according to the magnitude of $x$ -component of the electric field E within them, greatest first

The figure shows electric potential V as a function of $x$ . Rank the four regions according to the magnitude of $x$ -component of the electric field E within them, greatest first

physics-General

parallel

The general solution of the equation $fraction numerator d y over denominator d x end fraction equals 1 plus x y$ is

The general solution of the equation $fraction numerator d y over denominator d x end fraction equals 1 plus x y$ is

Solution of $2 y sin space x fraction numerator d y over denominator d x end fraction equals 2 sin space x cos space x minus y squared cos space x comma x equals pi over 2 comma y equals 1$ is given by

Solution of $2 y sin space x fraction numerator d y over denominator d x end fraction equals 2 sin space x cos space x minus y squared cos space x comma x equals pi over 2 comma y equals 1$ is given by

Equation of the curve passing through (3, 9) which satisfies the differential equation $fraction numerator d y over denominator d x end fraction equals x plus 1 over x squared$ is

Equation of the curve passing through (3, 9) which satisfies the differential equation $fraction numerator d y over denominator d x end fraction equals x plus 1 over x squared$ is

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.