Answer to Question #243673 in Calculus for JaytheCreator

Question

With the use of Laplace transform, solve the following I.V.Ps (i) 𝑦 ′′ + 2𝑦 ′ + 𝑦 = 𝑒 −𝑡 , 𝑦(0) = 1, 𝑦 ′ (0) = 3, (ii) 𝑦 ′′ + 𝑦 = 𝑡 2 sin 𝑡, 𝑦(0) = 𝑦 ′ (0) = 0’ (iii) 𝑦 ′′ + 3𝑦 ′ + 7𝑦 = cos 𝑡, 𝑦(0) = 0, 𝑦 ′ (0) = 2, (iv) 𝑦 ′′ + 𝑦 ′ + 𝑦 = 𝑡 3 , 𝑦(0) = 2, 𝑦 ′ (0) = 0.

Answer 1

(i)

$$L(y''+2y'+y)=L(e^{-t})$$

$$s^2L(y)-sy(0)-y'(0)+2(sL(y)-y(0))+L(y)$$

$$=\dfrac{1}{s+1}$$

$$s^2L(y)-s-3+2sL(y)-2+L(y)=\dfrac{1}{s+1}$$

$$(s^2+2s+1)L(y)=\dfrac{s^2+5s+s+5+1}{s+1}$$

$$L(y)=\dfrac{s^2+6s+6}{(s+1)^3}$$

$$\dfrac{s^2+6s+6}{(s+1)^3}=\dfrac{s^2+2s+1+4s+4+1}{(s+1)^3}$$

$$=\dfrac{1}{(s+1)}+\dfrac{4}{(s+1)^2}+\dfrac{1}{(s+1)^3}$$

Taking the inverse transform then gives

$$y(t)=e^{-t}+4te^{-t}+\dfrac{1}{2}t^2e^{-t}$$

(ii)

$$L(y''+y)=L(t^2\sin t)$$

$$s^2L(y)-sy(0)-y'(0)+L(y)=(-1)^2\dfrac{d^2}{ds^2}(\dfrac{1}{s^2+1})$$

$$\dfrac{d}{ds}(\dfrac{1}{s^2+1})=-\dfrac{2s}{(s^2+1)^2}$$

$$\dfrac{d^2}{ds^2}(\dfrac{1}{s^2+1})=-\dfrac{2(s^2+1)^2-2s(2(s^2+1)(2s))}{(s^2+1)^4}$$

$$=-\dfrac{2(s^2+1-4s^2)}{(s^2+1)^3}=\dfrac{2(3s^2-1)}{(s^2+1)^3}$$

$$s^2L(y)-0-0+L(y)=\dfrac{2(3s^2-1)}{(s^2+1)^3}$$

$$(s^2+1)L(y)=\dfrac{2(3s^2-1)}{(s^2+1)^3}$$

$$L(y)=\dfrac{2(3s^2-1)}{(s^2+1)^4}$$

$$\dfrac{2(3s^2-1)}{(s^2+1)^4}=\dfrac{2(3s^2+3-4)}{(s^2+1)^4}$$

$$=\dfrac{6}{(s^2+1)^3}-\dfrac{8}{(s^2+1)^4}$$

Taking the inverse transform then gives

$$y(t)=6(-\dfrac{1}{8}t^2\sin t-\dfrac{3}{8}t\cos t+\dfrac{3}{8}\sin t)$$

$$-8(\dfrac{1}{48}t^3\cos t-\dfrac{1}{8}t^2\sin t-\dfrac{5}{16}t\cos t+\dfrac{5}{16}\sin t)$$

$$y(t)=-\dfrac{1}{6}t^3\cos t+t^2\sin t+\dfrac{5}{2}t\cos t-\dfrac{5}{2}\sin t$$

$$-\dfrac{3}{4}t^2\sin t-\dfrac{9}{4}t\cos t+\dfrac{9}{4}\sin t$$

$$y(t)=-\dfrac{1}{6}t^3\cos t+\dfrac{1}{4}t^2\sin t+\dfrac{1}{4}t\cos t-\dfrac{1}{4}\sin t$$

(iii)

$$L(y''+3y'+7y)=L(\cos t)$$

$$s^2L(y)-sy(0)-y'(0)+3(sL(y)-y(0))+7L(y)$$

$$=\dfrac{s}{s^2+1}$$

$$s^2L(y)-0-2+3sL(y)-0+7L(y)=\dfrac{s}{s^2+1}$$

$$(s^2+3s+7)L(y)=\dfrac{2s^2+2+s}{s^2+1}$$

$$L(y)=\dfrac{2s^2+s+2}{(s^2+1)(s^2+3s+7)}$$

$$\dfrac{2s^2+s+2}{(s^2+1)(s^2+3s+7)}=\dfrac{As+B}{s^2+1}+\dfrac{Cs+D}{s^2+3s+7}$$

$$=\dfrac{As^3+3As^2+7As+Bs^2+3Bs+7B}{(s^2+1)(s^2+3s+7)}$$

$$+\dfrac{Cs^3+Cs+Ds^2+D}{(s^2+1)(s^2+3s+7)}$$

$s^3: A+C=0$

$s^2: 3A+B+D=2$

$s^1: 7A+3B+C=1$

$s^0: 7B+D=2$

$$A=2B, C=-2B, D=2-7B$$

$$14B+3B-2B=1=>B=\dfrac{1}{15}$$

$$A=\dfrac{2}{15}, C=-\dfrac{2}{15}, D=\dfrac{23}{15}$$

$$L(y)=\dfrac{1}{15}(\dfrac{2s+1}{s^2+1})+\dfrac{1}{15}(\dfrac{-2s+23}{s^2+3s+7})$$

Taking the inverse transform then gives

$$y(t)=\dfrac{1}{15}(\sin t+2\cos t)$$

$$+\dfrac{e^{-3t/2}}{15}(\dfrac{52\sqrt{19}\sin(\dfrac{\sqrt{19}t}{2})-38\cos(\dfrac{\sqrt{19}t}{2})}{19})$$

(iv)

$$L(y''+y'+y)=L(t^3)$$

$$s^2L(y)-sy(0)-y'(0)+(sL(y)-y(0))+L(y)$$

$$=\dfrac{6}{s^4}$$

$$s^2L(y)-2s-0+sL(y)-2+L(y)=\dfrac{6}{s^4}$$

$$(s^2+s+1)L(y)=\dfrac{2s^5+2s^4+6}{s^4}$$

$$L(y)=\dfrac{2s^5+2s^4+6}{s^4(s^2+s+1)}$$

$$\dfrac{2s^5+2s^4+6}{s^4(s^2+s+1)}=\dfrac{As+B}{s^2+s+1}$$

$$+\dfrac{C}{s}+\dfrac{D}{s^2}+\dfrac{E}{s^3}+\dfrac{F}{s^4}$$

$$2s^5+2s^4+6=As^5+Bs^4+Cs^5+Cs^4+Cs^3$$

$$+Ds^4+Ds^3+Ds^2+Es^3+Es^2+Es+Fs^2+Fs+F$$

$s^5: A+C=2$

$s^4: B+C+D=2$

$s^3: C+D+E=0$

$s^2: D+E+F=0$

$s^1: E+F=0$

$s^0: F=6$

$$F=6, E=-6, D=0, C=6, B=-4, A=-4$$

$$L(y)=-4\dfrac{s+1}{s^2+s+1}+\dfrac{6}{s}-\dfrac{6}{s^3}+\dfrac{6}{s^4}$$

Taking the inverse transform then gives

$$y(t)=-\dfrac{4}{3}e^{-t/2}(\sqrt{3}\sin(\dfrac{\sqrt{3}t}{2})+3\cos(\dfrac{\sqrt{3}t}{2}))$$

$$+6-3t^2+t^3$$