# Solve the third-order initial value problem below using the method of Laplace transforms y'''+3y''−18y'−40y=−120 y(0)=6 y'(0)=45 y"(0)=-21

Question
Laplace transform
Solve the third-order initial value problem below using the method of Laplace transforms
$$y'''+3y''−18y'−40y=−120$$
$$y(0)=6$$
$$y'(0)=45$$
$$y"(0)=-21$$

2021-03-03
$$\text{Step 1}$$
$$\text{Given equation,}$$
$$y'''+3y''-18y'-40y=-120$$
$$y(0)=6$$
$$y'(0)=45$$
$$y"(0)=-21$$
$$\text{take Laplace transform of both sides of the equation (1)}\\$$
Step 2 $$L\left\{y'''+3y''-18y'-40y\right\}=L\left\{-120\right\}$$
$$s^3L\left\{y\right\}-s^2y(0)-sy'(0)-y''(0)+3(s^2L\left\{y\right\}-sy(0)-y'(0))$$
$$-18(sL\left\{y\right\}-y(0))-40L\left\{y\right\}=\frac{-120}{s}$$
$$\text{Plug in the initial conditions: }y(0)=6 , y'(0)=45 , y"(0)=-21$$
$$s^3L\left\{y\right\}-s^2\times6-s\times45-(-21)+3(s^2L\left\{y\right\}-s\times6-45)-18(sL\left\{y\right\}-6)-40L\left\{y\right\}=\frac{-120}{s}$$
$$s^3L\left\{y\right\}-6s^2-45s+21+3(s^2L\left\{y\right\}6s-45)-18(sL\left\{y\right\}-6)-40L\left\{y\right\}=\frac{-120}{s}$$
$$s^3L\left\{y\right\}+3s^2L\left\{y\right\}-18sL\left\{y\right\}-40L\left\{y\right\}-6s^2-63s-6=\frac{-120}{s}$$
$$L\left\{y\right\}(s^3+3s^2-18s-40)=\frac{-120}{s}+6s^2+63s+6$$
$$L\left\{y\right\}(s^3+3s^2-18s-40)=\frac{-120+6s^3+63s^2+6s}{s}$$
$$L\left\{y\right\}=\frac{-120+6s^3+63s^2+6s}{s(s^3+3s^2-18s-40)}$$
$$\text{Step 3}$$
$$\text{Resolve into partial fractions,}$$
$$\frac{6s^3+63s^2+6s-120}{s(s^3+3s^2-18s-40)}=\frac{6s^3+63s^2+6s-120}{s(s+2)(s-4)(s+5)}$$
$$=\frac{A}{s}\frac{B}{s+2}\frac{C}{s-4}\frac{D}{s+5}$$
$$6s^3+63s^2+6s-120=A(s+2)(s-4)(s+5)+Bs(s-4)(s+5)+Cs(s+2)(s+5)+Ds(s+2)(s-4)$$
$$\text{put } s=0$$
$$-120=-40A$$
$$A=3$$
$$\text{put } s=-2$$
$$B=2$$
$$\text{put } s=4$$
$$C=6$$
$$\text{put } s=-5$$
$$D=-5$$
$$\text{Then, }$$
$$\frac{6s^3+63s^2+6s-120}{s(s+2)(s-4)(s+5)}=\frac{3}{s}+\frac{2}{s+2}+\frac{6}{s-4}-\frac{5}{s+5}$$
$$\text{Step 4}$$
$$\text{Take the inverse Laplace transform }$$
$$y=L^{-1}\left\{\frac{6s^3+63s^2+6s-120}{s(s^3+3s^2-18s-40)}\right\}$$
$$=L^{-1}\left\{\frac{3}{s}+\frac{2}{s+2}+\frac{6}{s-4}-\frac{5}{s+5}\right\}$$
$$y=3+2e^{-2t}+6e^{4t}-5e^{-5t} \text{ is the required solution }$$

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