Question

Solve the shifted data IVPs by the Laplace transform. Show the details.
 $ y'' + 3y' - 4y = 6{e^{2t - 3}},\;\;y\left( {1.5} \right) = 4,\;\;y'\left( {1.5} \right) = 5 $

Answer 1

Hint: An Initial Value Problem (IVP) is a differential equation along with an appropriate number of initial conditions. The given equation has a shifted value IVP needing a shift of $ - 1.5 $ . The Laplace Transform Formula is given as,
 $ F\left( s \right)\int\limits_L {f\left( x \right){e^{ - sx}}dx} $
With the help of the Laplace formula we convert a function of a real variable $ x $ into a function of a complex variable $ s $ .

Complete step by step solution:
We have been given to solve an IVP with shifted data by Laplace transform. The differential equation is given as,
 $ y'' + 3y' - 4y = 6{e^{2t - 3}},\;\;y\left( {1.5} \right) = 4,\;\;y'\left( {1.5} \right) = 5\;\;\;\;\;\;...\left( 1 \right) $
The initial conditions are about $ 1.5 $ so we need a shift of $ - 1.5 $ .
We can substitute $ \tau = t - 1.5 \Rightarrow t = \tau + 1.5 $
And $ y\left( t \right) = u\left( \tau \right) = u\left( {t - 1.5} \right) $
Substituting this in equation $ \left( 1 \right) $ we get,
 $
  u'' + 3u' - 4u = 6{e^{2\left( {\tau + 1.5} \right) - 3}} \\
   \Rightarrow u'' + 3u' - 4u = 6{e^{2\tau }},\;\;u\left( 0 \right) = 4,\;\;u'\left( 0 \right) = 5\;\;\;\;\;\;...\left( 2 \right) \\
  $
Now we will use some of the following standard Laplace transform given as,
 $
  L\left\{ {f\left( t \right)} \right\} = F\left( s \right) \\
  L\left\{ {f'\left( t \right)} \right\} = sF\left( s \right) - f\left( 0 \right) \\
  L\left\{ {f''\left( t \right)} \right\} = {s^2}F\left( s \right) - sf\left( 0 \right) - f'\left( 0 \right) \\
  L\left\{ {{e^{at}}} \right\} = \dfrac{1}{{s - a}} \\
  $
We can use this Laplace transforms in equation $ \left( 2 \right) $ with $ U\left( s \right) = L\left\{ {u\left( \tau \right)} \right\} $ ,
 $
  L\left\{ {u''} \right\} + 3L\left\{ {u'} \right\} - 4L\left\{ u \right\} = 6L\left\{ {{e^{2\tau }}} \right\} \\
   \Rightarrow \left[ {{s^2}U - su\left( 0 \right) - u'\left( 0 \right)} \right] + 3\left[ {sU - u\left( 0 \right)} \right] - 4U = 6\left[ {\dfrac{1}{{s - 2}}} \right] \\
   \Rightarrow \left[ {{s^2}U - 4s - 5} \right] + 3\left[ {sU - 4} \right] - 4U = \dfrac{6}{{s - 2}} \\
   \Rightarrow {s^2}U - 4s - 5 + 3sU - 12 - 4U = \dfrac{6}{{s - 2}} \\
   \Rightarrow \left( {{s^2} + 3s - 4} \right)U - 4s - 17 = \dfrac{6}{{s - 2}} \\
   \Rightarrow \left( {{s^2} - s + 4s - 4} \right)U = \dfrac{6}{{s - 2}} + 4s + 17 \\
   \Rightarrow \left( {s\left( {s - 1} \right) + 4\left( {s - 1} \right)} \right)U = \dfrac{{6 + \left( {4s + 17} \right)\left( {s - 2} \right)}}{{s - 2}} \\
   \Rightarrow \left( {s + 4} \right)\left( {s - 1} \right)U = \dfrac{{6 + 4{s^2} - 8s + 17s - 34}}{{s - 2}} \\
   \Rightarrow \left( {s + 4} \right)\left( {s - 1} \right)U = \dfrac{{4{s^2} + 9s - 28}}{{s - 2}} = \dfrac{{4{s^2} + 16s - 7s - 28}}{{s - 2}} = \dfrac{{4s\left( {s + 4} \right) - 7\left( {s + 4} \right)}}{{s - 2}} \\
   \Rightarrow \left( {s + 4} \right)\left( {s - 1} \right)U = \dfrac{{\left( {4s - 7} \right)\left( {s + 4} \right)}}{{s - 2}} \\
   \Rightarrow U = \dfrac{{\left( {4s - 7} \right)}}{{\left( {s - 1} \right)\left( {s - 2} \right)}} \;
  $
We can write this function by converting into partial fractions as follows,
 $
  \dfrac{{\left( {4s - 7} \right)}}{{\left( {s - 1} \right)\left( {s - 2} \right)}} = \dfrac{a}{{\left( {s - 1} \right)}} + \dfrac{b}{{\left( {s - 2} \right)}} = \dfrac{{a\left( {s - 2} \right) + b\left( {s - 1} \right)}}{{\left( {s - 1} \right)\left( {s - 2} \right)}} \\
   \Rightarrow \left( {4s - 7} \right) = a\left( {s - 2} \right) + b\left( {s - 1} \right) \;
  $
Putting $ s = 1 \Rightarrow - 3 = - a \Rightarrow a = 3 $
And putting $ s = 2 \Rightarrow 1 = b \Rightarrow b = 1 $
Thus,
 $ U\left( s \right) = \dfrac{3}{{s - 1}} + \dfrac{1}{{s - 2}} $
We can use inverse Laplace transform $ {L^{ - 1}}\left\{ {\dfrac{1}{{s - a}}} \right\} = {e^{at}} $ for $ u\left( \tau \right) = {L^{ - 1}}\left\{ {U\left( s \right)} \right\} $ to get,
 $ u\left( \tau \right) = 3{e^\tau } + {e^{2\tau }} $
Further we substitute back $ \tau = t - 1.5 $ to get,
\[
  y\left( t \right) = u\left( {t - 1.5} \right) = 3{e^{\left( {t - 1.5} \right)}} + {e^{2\left( {t - 1.5} \right)}} \\
   \Rightarrow y\left( t \right) = 3{e^{\left( {t - 1.5} \right)}} + {e^{2t - 3}} \;
 \]
Hence, the final solution for the given shifted data IVP is \[y\left( t \right) = 3{e^{\left( {t - 1.5} \right)}} + {e^{2t - 3}}\].
So, the correct answer is “\[y\left( t \right) = 3{e^{\left( {t - 1.5} \right)}} + {e^{2t - 3}}\]”.

Note: Since the initial values were not given at $ t = 0 $ we concluded that the given data is shifted and the shift is equal to the time at which the conditions are given, i.e. $ 1.5 $ in this question. We first shifted the data and then used Laplace transform to solve the differential equation. The final result is written as a function of $ t $ after substituting back all the values.