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A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation.θ''-θ'-2θ=1-2t, â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰Î¸p(t)=t-1

Short Answer

Expert verified

The general solution of the given differential equation isθ=c1e2t+c2e-t+t-1.

Step by step solution

01

Write the auxiliary equation of the given differential equation

The differential equation is:

θ''-θ'-2θ=1-2t â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€¦(1)

Write the homogeneous differential equation of the equation (1),

θ''-θ'-2θ=0

The auxiliary equation for the above equation,

m2-m-2=0

02

Now find the complementary solution of the given equation is

Solve the auxiliary equation,

m2-m-2=0m2-2m+m-2=0m(m-2)+1(m-2)=0(m-2)(m+1)=0

The roots of the auxiliary equation are,

m1=2, â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰m2=-1

The complementary solution of the given equation is,

θc=c1e2t+c2e-t

03

Use the given particular solution to find a general solution for the equation. 

The given particular solution,

θp(t)=t-1

Therefore, the general solution is,

θ=θc(t)+θp(t)θ=c1e2t+c2e-t+t-1

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Most popular questions from this chapter

Using the mass-spring analogy, predict the behavior as t→∞of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem.

(a). y''+16y=0;y(0)=2,y'(0)=0(b). y''+100y'+y=0;y(0)=1,y'(0)=0(c). y''-6y'+8y=0;y(0)=1,y'(0)=0(d). y''+2y'-3y=0;y(0)=-2,y'(0)=0(e). y''-y'-6y=0;y(0)=1,y'(0)=1

Discontinuous Forcing Term. In certain physical models, the nonhomogeneous term, or forcing term, g(t) in the equation

ay''+by'+cy=g(t)

may not be continuous but may have a jump discontinuity. If this occurs, we can still obtain a reasonable solution using the following procedure. Consider the initial value problem;

y''+2y'+5y=g(t); â¶Ä‰â¶Ä‰â¶Ä‰y(0)=0, â¶Ä‰â¶Ä‰â¶Ä‰y'(0)=0

Where,

g(t)=10, â¶Ä‰if 0≤t≤3Ï€20, â¶Ä‰â¶Ä‰â¶Ä‰â€‰if t>3Ï€2

  1. Find a solution to the initial value problem for 0≤t≤3π2 .
  2. Find a general solution fort>3Ï€2.
  3. Now choose the constants in the general solution from part (b) so that the solution from part (a) and the solution from part (b) agree, together with their first derivatives, att=3Ï€2 . This gives us a continuously differentiable function that satisfies the differential equation except at t=3Ï€2.

Find a particular solution to the differential equation.

θ''(t)-θ(t)=tsint

Find a general solution to the differential equation.

y''-2y'-3y=3t2-5

Solve the given initial value problem. y''-2y'+y=0;y(0)=1,y'(0)=-2

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