Solveeit Logo
Login

Question

Question: Solve the shifted data IVPs by the Laplace transform. Show the details. \( y'' + 3y' - 4y = 6{e^{...

Solve the shifted data IVPs by the Laplace transform. Show the details.
y′′+3y′−4y=6e2t−3,    y(1.5)=4,    y′(1.5)=5y'' + 3y' - 4y = 6{e^{2t - 3}},\;\;y\left( {1.5} \right) = 4,\;\;y'\left( {1.5} \right) = 5

Explanation

Solution

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- 1.5 . The Laplace Transform Formula is given as,
F(s)∫Lf(x)e−sxdxF\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 xx into a function of a complex variable ss .

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=6e2t−3,    y(1.5)=4,    y′(1.5)=5            ...(1)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 1.51.5 so we need a shift of −1.5- 1.5 .
We can substitute τ=t−1.5⇒t=τ+1.5\tau = t - 1.5 \Rightarrow t = \tau + 1.5
And y(t)=u(τ)=u(t−1.5)y\left( t \right) = u\left( \tau \right) = u\left( {t - 1.5} \right)
Substituting this in equation (1)\left( 1 \right) we get,
u′′+3u′−4u=6e2(τ+1.5)−3 ⇒u′′+3u′−4u=6e2τ,    u(0)=4,    u′(0)=5            ...(2)  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 (2)\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,
(4s−7)(s−1)(s−2)=a(s−1)+b(s−2)=a(s−2)+b(s−1)(s−1)(s−2) ⇒(4s−7)=a(s−2)+b(s−1)   \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⇒−3=−a⇒a=3s = 1 \Rightarrow - 3 = - a \Rightarrow a = 3
And putting s=2⇒1=b⇒b=1s = 2 \Rightarrow 1 = b \Rightarrow b = 1
Thus,
U(s)=3s−1+1s−2U\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(τ)=3eτ+e2τu\left( \tau \right) = 3{e^\tau } + {e^{2\tau }}
Further we substitute back τ=t−1.5\tau = t - 1.5 to get,

y(t)=u(t−1.5)=3e(t−1.5)+e2(t−1.5) ⇒y(t)=3e(t−1.5)+e2t−3   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(t)=3e(t−1.5)+e2t−3y\left( t \right) = 3{e^{\left( {t - 1.5} \right)}} + {e^{2t - 3}}.
So, the correct answer is “y(t)=3e(t−1.5)+e2t−3y\left( t \right) = 3{e^{\left( {t - 1.5} \right)}} + {e^{2t - 3}}”.

Note : Since the initial values were not given at t=0t = 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. 1.51.5 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 tt after substituting back all the values.