Solve IVPs by Laplace Transform y−4y=6e2t−3 y(1.5)=4 y'(1.5)=5

Step 1Given y3y′−4y=6e2t−3y(1.5)=4y′(1.5)=5Since initial values are not given at t=0 , so we need to change the variableLet t=z+1.5⇒z=t−1.5So, our original equation is yt)+3y′(t)−4y(t)=6e2t−3which now becomes yz+1.5)+3y′(z+1.5)−4y(z+1.5)=6e2(z+1.5)−3=(yz+1.5)+3y′(z+1.5)−4y(z+1.5)=6e2zLet u(z)=y(z+1.5)⇒u′(z)=dudz=dydtdtdz=y′(z+1.5)similarly u"(z)=y"(z+1.5)The initial conditions for u(z) are u(0)=y(1.5)=4u′(0)=y′(1.5)=5Step 2So, the differential equation is u(z)+3u′(z)−4u(z)=6e2z with u(0)=4,u′(0)=5Taking Laplace transform both sides, we gets2u―−su(0)−u′(0)+3su―−3u(0)−4u―=6s−2⇒s2u―−4s−5+3su―−12−4u―=6s−2⇒(s2+3s−4)u―=6s−2+4s+17⇒(s2+3s−4)u―=4s2+9s−28s−2

Resolved: "Solve IVPs by Laplace Transform y−4y=6e2t−3 y(1.5)=4 y'(1.5)=5"

Trent Carpenter

Answered question

2021-09-08

Solve IVPs by Laplace Transform
$y - 4 y = 6 e^{2 t - 3}$
y(1.5)=4
y'(1.5)=5

Answer & Explanation

rogreenhoxa8

Skilled2021-09-09Added 109 answers

Step 1
Given $y 3 y^{'} - 4 y = 6 e^{2 t - 3}$
$y (1.5) = 4$
$y^{'} (1.5) = 5$
Since initial values are not given at t=0 , so we need to change the variable
Let $t = z + 1.5 \Rightarrow z = t - 1.5$
So, our original equation is $y t) + 3 y^{'} (t) - 4 y (t) = 6 e^{2 t - 3}$
which now becomes $y z + 1.5) + 3 y^{'} (z + 1.5) - 4 y (z + 1.5) = 6 e^{2 (z + 1.5) - 3}$
$= (y z + 1.5) + 3 y^{'} (z + 1.5) - 4 y (z + 1.5) = 6 e^{2 z}$
Let $u (z) = y (z + 1.5)$
$\Rightarrow u^{'} (z) = \frac{d u}{d z} = \frac{d y}{d t} \frac{d t}{d z} = y^{'} (z + 1.5)$
similarly $u " (z) = y " (z + 1.5)$
The initial conditions for $u (z)$ are $u (0) = y (1.5) = 4$
$u^{'} (0) = y^{'} (1.5) = 5$
Step 2
So, the differential equation is $u (z) + 3 u^{'} (z) - 4 u (z) = 6 e^{2 z}$ with $u (0) = 4, u^{'} (0) = 5$
Taking Laplace transform both sides, we get
$s^{2} \overset{―}{u} - s u (0) - u^{'} (0) + 3 s \overset{―}{u} - 3 u (0) - 4 \overset{―}{u} = \frac{6}{s - 2}$
$\Rightarrow s^{2} \overset{―}{u} - 4 s - 5 + 3 s \overset{―}{u} - 12 - 4 \overset{―}{u} = \frac{6}{s - 2}$
$\Rightarrow (s^{2} + 3 s - 4) \overset{―}{u} = \frac{6}{s - 2} + 4 s + 17$
$\Rightarrow (s^{2} + 3 s - 4) \overset{―}{u} = \frac{4 s^{2} + 9 s - 28}{s - 2}$

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