# 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

Solve the third-order initial value problem below using the method of Laplace transforms
${y}^{‴}+3{y}^{″}-18{y}^{\prime }-40y=-120$
$y\left(0\right)=6$
${y}^{\prime }\left(0\right)=45$
$y"\left(0\right)=-21$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Delorenzoz
$\text{Step 1}$
$\text{Given equation,}$
${y}^{‴}+3{y}^{″}-18{y}^{\prime }-40y=-120$
$y\left(0\right)=6$
${y}^{\prime }\left(0\right)=45$
$y"\left(0\right)=-21$
$\text{take Laplace transform of both sides of the equation (1)}\phantom{\rule{0ex}{0ex}}$
Step 2$L\left\{{y}^{‴}+3{y}^{″}-18{y}^{\prime }-40y\right\}=L\left\{-120\right\}$
${s}^{3}L\left\{y\right\}-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-{y}^{″}\left(0\right)+3\left({s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)$
$-18\left(sL\left\{y\right\}-y\left(0\right)\right)-40L\left\{y\right\}=\frac{-120}{s}$

${s}^{3}L\left\{y\right\}-{s}^{2}×6-s×45-\left(-21\right)+3\left({s}^{2}L\left\{y\right\}-s×6-45\right)-18\left(sL\left\{y\right\}-6\right)-40L\left\{y\right\}=\frac{-120}{s}$
${s}^{3}L\left\{y\right\}-6{s}^{2}-45s+21+3\left({s}^{2}L\left\{y\right\}6s-45\right)-18\left(sL\left\{y\right\}-6\right)-40L\left\{y\right\}=\frac{-120}{s}$
${s}^{3}L\left\{y\right\}+3{s}^{2}L\left\{y\right\}-18sL\left\{y\right\}-40L\left\{y\right\}-6{s}^{2}-63s-6=\frac{-120}{s}$
$L\left\{y\right\}\left({s}^{3}+3{s}^{2}-18s-40\right)=\frac{-120}{s}+6{s}^{2}+63s+6$
$L\left\{y\right\}\left({s}^{3}+3{s}^{2}-18s-40\right)=\frac{-120+6{s}^{3}+63{s}^{2}+6s}{s}$
$L\left\{y\right\}=\frac{-120+6{s}^{3}+63{s}^{2}+6s}{s\left({s}^{3}+3{s}^{2}-18s-40\right)}$
$\text{Step 3}$
$\text{Resolve into partial fractions,}$
$\frac{6{s}^{3}+63{s}^{2}+6s-120}{s\left({s}^{3}+3{s}^{2}-18s-40\right)}=\frac{6{s}^{3}+63{s}^{2}+6s-120}{s\left(s+2\right)\left(s-4\right)\left(s+5\right)}$
$=\frac{A}{s}\frac{B}{s+2}\frac{C}{s-4}\frac{D}{s+5}$