 # Solve the initial value problem below using the method of Laplace transforms. y"-4y'+40y=225 e^{5t} y(0)=5 y'(0)=31 he298c 2021-01-31 Answered
Solve the initial value problem below using the method of Laplace transforms.
$y"-4{y}^{\prime }+40y=225{e}^{5t}$
$y\left(0\right)=5$
${y}^{\prime }\left(0\right)=31$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it SoosteethicU
Step 1
Taking Laplace transform in the both the direction:
$y"-4{y}^{\prime }+40y=225{e}^{5t}$
$y\left(0\right)=5$
${y}^{\prime }\left(0\right)=31$
$L\left\{y"-4{y}^{\prime }+40y\right\}=L\left\{Y\left(s\right)\right\}$
$L\left\{y"\right\}-4L\left\{{y}^{\prime }\right\}+40L\left\{y\right\}=225L\left\{Y\left(s\right)\right\}$
$L\left\{y"\right\}={s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left({y}^{\prime }\right)=sY\left(s\right)-y\left(0\right)$
$L\left(y\right)=Y\left(s\right),L\left\{{e}^{at}\right\}=\frac{1}{\left(s-a\right)}$
$\left\{{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right\}-4\left\{sY\left(s\right)-y\left(0\right)\right\}+40\left\{Y\left(s\right)\right\}=225\left\{\frac{1}{\left(s-5\right)}\right\}$
$\left\{{s}^{2}-4s+40\right\}Y\left(s\right)+\left\{-5s-11\right\}=225\left\{\frac{1}{\left(s-5\right)}\right\}$
$\left\{{s}^{2}-4s+40\right\}Y\left(s\right)+\left\{-5s-11\right\}=225\left\{\frac{1}{\left(s-5\right)}\right\}$
$\left\{{s}^{2}-4s+40\right\}Y\left(s\right)=225\left\{\frac{1}{\left(s-5\right)}\right\}+5s+11$
$Y\left(s\right)=\frac{225\left\{\frac{1}{\left(s-5\right)}\right\}+5s+11}{{s}^{2}-4s+40}$
Step 2
Find the factor of the function:
$Y\left(s\right)=\frac{225\left\{\frac{1}{\left(s-5\right)}\right\}+5s+11}{{s}^{2}-4s+40}=$
$=\frac{225}{\left(s-5\right)\left({s}^{2}-4s+40\right)}+\frac{5s}{\left({s}^{2}-4s+40\right)}+\frac{11}{\left({s}^{2}-4s+40\right)}$
$=225\left\{\frac{-5s-5}{{s}^{2}-4s+40}+\frac{5}{\left(s-5\right)}\right\}+5\cdot \frac{\left(s-2\right)}{\left(s-2{\right)}^{2}+36}+10\cdot \frac{1}{\left(s-2{\right)}^{2}+36}+\left\{11\cdot \frac{1}{\left(s-2{\right)}^{2}+36}\right\}$
Step 3Taking inverse Laplace transform in both direction:
${L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{\left\{\frac{\left(-5s-5\right)}{\left({s}^{2}-4s+40\right)}+\frac{5}{\left(s-5\right)}\right\}+5\cdot \frac{\left(s-2\right)}{\left(s-2{\right)}^{2}+36}+10\cdot \frac{1}{\left(s-2{\right)}^{2}+36}+\left\{11\cdot \frac{1}{\left(s-2{\right)}^{2}+36}\right\}\right\}$