Question

Use the Laplace transform to solve the following initial-value problem. y" +5y' +4y = 20 sin 2t, y(0)=-1, y'(0) = 2

Note: Use partial fractions when solving

Answer 1

Let us assume the Laplace transform of y(t), L(y(t))=Y(s) then, we have L(y'(t))= sY(s)-y(0) and L(y"(t))= s​​​​​​2​​​​​Y(s)-sy(0)-y'(0) . By putting the initial values y(0)=-1 and y'(0)=2 and then taking the laplace transform both sides of the given differential equation, we will get the expression of Y(s).

Then, using the Method of Partial Fraction, we will get expression of Y(s) in partial fraction form. By taking the inverse Laplace transform on both sides of Y(s), and using the rule L​​​​-1​( s/(s​​​​​​2​​​​​ + b​2))= Cos (bt) and L​-1​(1/(s+a))= e​​​​​​-at

The step by step explanatory solution is provided below.

following gritial value problem. Y" + sy'tly 20 cin at 22. Lly HJ = Y() solution use the Laplace transform to Lolve the 'ylo): -1, y'=)= 2 Let the laplace transform of yet be then lly')) - 640) -Yeo) - syce)-(1) 44/45+! and LCy"+)) $2 Y62) - byto)- y'ro) 42Y(s)- "S(-1) - 2 = 2 y(s) + 4-2 "Taking Laplace transform on both sides of the given differential equation, we get Lly" + sy' +"y) = Lly") + 5 Lly') + Y L(y) = 2 (2o linat) > (sayas) + 5-2) +5(5468+1) + 4405) 20 LC sinat) (32+ 58+4) 46) +(5-2+5) = 20 ( 224 225 2) (uring LC sin (bt)) = (8² +58 +4) Y(s) + (+3) V b 370) s2 +62 yo s2+4 Y(s) = [line) - ceny] (assen yo (8+3)(8²+4) S²+55+ 4 (8²+4) 40-23-45-35² - 12 ***") (osten +38 ) (**5547) s²+4 -83-382_45+28 ²+ 4 (-0.35 Y(s)

Now , & Method of partial Fraction, Y(s) = (-83 38²- Us + 28) (x²+4) s²+55+ 4 33-362 us+ 28 8² + 4 4) (4+1) 4 + Ast 2²ty sty 4+1 pultiply by (2²+4) (4+4) Cs+1) both sides, we get (23-342_45+28) (As+B) (6+4)(4+1) + C ( 8 + 4) (4+1) + D (22+4) (4+4) (ab. 342 - 48 + 28) = (AS+B)( 12+5 644) + C (as 2 + 1²+ 45+4)* + C 44 4++ / 16 ) = CA53+ 5A 8+ 4 As+B53++48) + Ces340821405+49) + CD33+ 4D527405 +10) - ² - 38²43 728 = (AAC+D) 3 + (SATB + C + y DJs² 4+ (4A+58+4C HUD) s + C8+46+ By comparing the coefficients both sides, we get ATC+D = 1 ; 5A+B+C+YD YA+5+40 + 4D = - 4 Y B+ 4C +16D = 28 4x (A+C+D) = 4x4 YAA 46 +48 = -4. YA + 5 Btyc+ y = -4 + -3 7 - 5B B=0 Lola ( u Bt uc + 16 D- 28 2 fuct 16 D=28 4(0) + 40 +16D - 28 .. C + 4D = 7 c= 7-4D

Putting we B=0 and CF 7-UD in 5A+B+C+ 4D = -3 and A+C+D=-1, get 5A to +(7-400 +40.2-3. 3) 5A + +=-3 5 5A =-3-7 = 10. A=-2 OH U ATC+D = -24 (7-4D) +D = -) wa 5- 3D - - | - 3D = – 1-5 = – 6 D=2 a. TV Thus, ©= 7-4D = 7-4 (2) = 7-8 = [ Putting these values, we get w Yls) = 3-38² – us +28 -) (+&+1) 12 tu no 2 + + (-2) s t (0) 6² +4 (1) sty S+ (1 w 2 ; y(s) 25 + s2+4 sty st + 463) - Lly 4) = -25 2 sti 1²44 sty we get Inverse both sides, Taking Laplace Inverce juos. (d) -1) ( + 2 LY sty sti - 2 Lt & 62 +4

- t § yut) = -2 Cos (2+) - cos (bt) ; (: LY " ( PC -yt - 2 cos(at) - e +2 e s2+62 28 -at e LY ( _yt +(sta The solution is given by Lyct) = yu) - - 2 Coscat) -e +2e -t