Find the general solution of the following differential equations (complementary function + particular solution). Find the particular solution by inspection. Also find a computer solution and reconcile differences if necessary, noticing especially whether the particular solution is in simplest form.

(D22D+1)y=2cosx

Short Answer

Expert verified

The solution of differential equation isy(x)=(C1x+C2)ex+sinx

Step by step solution

01

Given information 

Given equation is

(D22D+1)y=2cosx

02

Definition of differential equation

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself to its derivatives of various orders.

03

Solve the given differential equation 

Substitute the values in given equation

(D22D+1)y=2cosxaem22m+1=0m2mm+1=0m(m1)1(m1)=0m=1,1

C.F=(C1x+C2)ezP.I=1w23+12cosx112mb12cosx(D2=a2)=1π2cosx    ((1πx=x)

(As integration is reciprocal of differentiation)

sinx

P.I=sinx

HenceCS=(C1x+C2)e2+sinx

The solution of the differential equation is

y(x)=(C1x+C2)ex+sinx

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Most popular questions from this chapter

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

y¨+4y˙+5y=2e-2tcost,y0=0,y˙0=3

a) Show that(er/r2)=0forr>0.

(b) Show that(1/r)=-er/r2.

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Several Terms on the Right-Hand Side: Principle of Superposition So far we have brushed over a question which may have occurred to you: What do we do if there are several terms on the right-hand side of the equation involving different exponentials?

In Problem 33 to 38 , solve the given differential equations by using the principle of superposition [see the solution of equation (6.29) . For example, in Problem 33 , solve three differential equations with right-hand sides equal to the three different brackets. Note that terms with the same exponential factor are kept together; thus a polynomial of any degree is kept together in one bracket.

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Find the orthogonal trajectories of each of the following families of curves. In each case, sketch or computer plot several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from y'for the original curves; this constant takes different values for different curves of the original family, and you want an expression for y'which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. See equations (2.10)to (2.12)

y=kx2

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