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Question 1


Question 2

The path of a train A is given by the equation 3x + 4y – 12 = 0 and the path of another train B is given by the equation 6x + 8y – 48 = 0. Represent this situation graphically. Linear Equation: In reference to a linear line, the basic representation of it is y = a x + b . The coefficient of x in the equation is termed as the slope of the line. The slope is rate of change of the dependent variable with the change in the independent variable. The y-intercept is b .
Question 3

The homogeneous equation y ( 4 ) + 8 y ′′′ + 28 y ′′ + 48 y ′ + 48 y = 0 has the general solution y n ( t ) = c 1 e − 2 t + c 2 t − 2 t + c 3 e − 2 t c o s ( 2 t ) + c 4 e − 2 t s i n ( 2 t ) . For each question below, suppose that one plans to use the Method of Undetermined Coefficients in order to solve the given nonhomogeneous equation. For each equation, write the appropriate attempt for the particular solution. Do NOT solve the ODEs. (a) y ( 4 ) + 8 y ′′′ + 28 y ′′ + 48 y ′ + 48 y = t 2 (b) y ( 4 ) + 8 y ′′′ + 28 y ′′ + 48 y ′ + 48 y = e − 2 t (c) y ( 4 ) + 8 y ′′′ + 28 y ′′ + 48 y ′ + 48 y = s i n ( 2 t ) (d) y ( 4 ) + 8 y ′′′ + 28 y ′′ + 48 y ′ + 48 y = ( e 2 t + e − 2 t ) s i n ( 2 t ) (e) y ( 4 ) + 8 y ′′′ + 28 y ′′ + 48 y ′ + 48 y = u ( t ) Procedure to find the complementary function: In a differential equation F ( D ) y = 0 , assume D as an algebraic quantity and put F ( D ) = 0 , by solving it we get roots. Based on the nature of these roots we write the complementary function as follows: If roots are real and distinct that is D = m 1 , m 2 , m 3 , then the complementary function of the differential equation is C . F . = c 1 e m 1 x + c 2 e m 2 x + c 3 e m 3 x If roots are real and repeated that is D = m 1 , m 1 , m 3 , then the complementary function of the differential equation is C . F . = ( c 1 + c 2 x ) e m 1 x + c 3 e m 3 x If roots are complex and distinct that is D = a ± i b , m 3 , then the complementary function of the differential equation is C . F . = e a x ( c 1 cos b x + c 2 sin b x ) + c 3 e m 3 x If roots are complex and repeated that is D = a ± i b , a ± i b , then the complementary function of the differential equation is C . F . = e a x [ ( c 1 + c 2 x ) cos b x + ( c 3 + c 4 x ) sin b x ]
Question 4

What is the length of the major axis of an ellipse whose equation is 12 x 2 + 8 y 2 = 48 ? The ellipse:- In a plane, we can say that an ellipse is a set of all points, the sum of the whose distance from two fixed point points in the plane is a constant, and this constant is greater than the distance between the two fixed points. We call these two points as the foci of the ellipse. And the distance between the two foci is labeled as 2 c . The major axis of the ellipse:- we define a line segment passing through the two foci and terminating at the ellipse as the major axis of the ellipse. The length of the semi-major axis of the ellipse is denoted by a The minor axis of the ellipse:- we define a line segment perpendicular to the major axis and passing through the center 0f the ellipse and terminating at the ellipse is called the called minor axis of the ellipse. The length of the semi-minor axis of the ellipse is denoted by b Standard equation of an ellipse:- x 2 a 2 + y 2 b 2 = 1 ( b > a ) Here a is the semi – minor axis and b is the semi major axis.
Answer to question 1


Answer to question 2

Given Data

  • The path of train A is: A=3x+4y−12=0A=3x+4y−12=0.
  • The path of train B is: B=6x+8y−48=0B=6x+8y−48=0.

The path of train A can be further simplified as:

34x+y−3=0y=−34x+334x+y−3=0y=−34x+3

The slope of the path of train A is mA=−34mA=−34 and y-intercept is 33.


Answer to question 3

The homogeneous differential equation is

y(4)+8y′′′+28y′′+48y′+48y=0y(4)+8y‴+28y″+48y′+48y=0

whose general solution is given by

yn(t)=c1e−2t+c2te−2t+c3e−2tcos2t+c4e−2tsin2tyn(t)=c1e−2t+c2te−2t+c3e−2tcos⁡2t+c4e−2tsin⁡2t

(a).

The non-homogeneous differential equation is

y(4)+8y′′′+28y′′+48y′+48y=t2y(4)+8y‴+28y″+48y′+48y=t2

Therefore by using method of undetermined coefficient the particular solution of the above non-homogeneous differential equation is of the form

yp=At2+Bt+Cyp=At2+Bt+C

(b).

The non-homogeneous differential equation is

y(4)+8y′′′+28y′′+48y′+48y=e−2ty(4)+8y‴+28y″+48y′+48y=e−2t

Therefore by using method of undetermined coefficient the particular solution of the above non-homogeneous differential equation is of the form

yp=At2e−2typ=At2e−2t

(c).

The non-homogeneous differential equation is

y(4)+8y′′′+28y′′+48y′+48y=sin2ty(4)+8y‴+28y″+48y′+48y=sin⁡2t

Therefore by using method of undetermined coefficient the particular solution of the above non-homogeneous differential equation is of the form

yp=Asin2t+Bcos2typ=Asin⁡2t+Bcos⁡2t

(d).

The non-homogeneous differential equation is

y(4)+8y′′′+28y′′+48y′+48y=e2tsin2t+e−2tsin2ty(4)+8y‴+28y″+48y′+48y=e2tsin⁡2t+e−2tsin⁡2t

Therefore by using method of undetermined coefficient the particular solution of the above non-homogeneous differential equation is of the form

yp=Ae2tsin2t+e2tcos2t+Cte−2tsin2t+Dte−2tcos2typ=Ae2tsin⁡2t+e2tcos⁡2t+Cte−2tsin⁡2t+Dte−2tcos⁡2t

(e).

The non-homogeneous differential equation is

y(4)+8y′′′+28y′′+48y′+48y=u(t)y(4)+8y‴+28y″+48y′+48y=u(t)

Here u(t)u(t) is the unit step function. So we cannot find the particular solution of this using method of undetermined coefficient because unit step function is a piece-wise discontinuous function.


Answer to question 4

Given:-

12×2+8y2=4812×2+8y2=48

Now converting given equation in the standard form of the ellipse:-

12×2+8y2=4812×248+8y248=1×24+y26=1×222+y2(√6)2=112×2+8y2=4812×248+8y248=1×24+y26=1×222+y2(6)2=1

So by comparing from the standard equation of the ellipse, we have

Semi minor axesa=2a=2

and

Semi major axisb=√6b=6

So the major axis =2×√6=2√6=2×6=26 is the required answer.

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