Consider the non-homogeneous differential equation y^\'\'+ 1y = - cos(t) + 3 sin

Question

Consider the non-homogeneous differential equation y^''+ 1y = - cos(t) + 3 sin(t). a. (1/10) Find the roots of the characteristic polynomial, r_+ and r_-, respectively, b. (1/10) Find the real-valued fundamental solutions to the differential equation associated to the roots in part (a) y_+(t) = y_(t) = c. (1/10) Use the functions in part (b) to find the general solution to the homogeneous equation, (use the names c and d for any arbitrary constants) yh(t) = d. (4/10) Find a particular solution y_p to the non-homogeneous equation, y_p(t) = e. (3/10) Find the solution of the initial value problem y^'' + 1 y = - cos(t) + 3 sin(t), y(0) = 1, y^?(0) = -3. y(t)=

Explanation / Answer

a. Characteristic equation is given by: r2+1=0 => r = +sqr(-1) and r = -sqr(-1) or r+ = i and r- = -i

b. y+(t) = cost, y-(t) = sin(t) (because the roots are imaginary)

c. yh(t) = csin(t) + dcos(t)

d. Let the guess of solution be : (A+Bt)cos(t)+ (C+Dt)sin(t) = y

Then y' = Bcos(t)-(A+Bt)sin(t)+Dsin(t)+(C+Dt)cos(t)

y" = -Bsin(t)-Bsin(t)-(A+Bt)cos(t)+Dcos(t)+Dcos(t)-(C+Dt)sin(t) = -2Bsin(t)-(A+Bt)cos(t)+2Dcos(t)-(C+Dt)sin(t)

So y"+y = -cos(t)+3sin(t) => -2Bsin(t)-(A+Bt)cos(t)+2Dcos(t)-(C+Dt)sin(t) = -cos(t)+3sin(t)

Comparing the coefficients, we get:

-2B-C = 3, -A+2D = -1, -B=0, -D =0

=> C = -3, A =1, B =0, D =0

And so particular solution is given by: yp(t) = cos(t)-3sin(t)

e. y(t) = csin(t)+dcos(t)+cos(t)-3sin(t)

y(0)=1 => 1 = 0+d+1-0 = d+1 => d=0

y'(t) = ccos(t)-dsin(t)-sin(t)-3cos(t)

y'(0) = c-0-0-3 =-3 => c=0

Hence solution of initial value problem is y(t) = cos(t)-3sin(t)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.