# Solve the following IVP using Laplace Transform y'-2y =1-t , y(0)=4

Solve the following IVP using Laplace Transform
${y}^{\prime }-2y=1-t,y\left(0\right)=4$
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berggansS
Step 1
The Laplace transform is given as:
$y\prime -2y=1-t,y\left(0\right)=4$
Taking the laplace trasform on both sides:
$L\left\{y\prime \right\}-2L\left\{y\right\}=L\left(1\right)-L\left(t\right)$
$sL\left(y\right)-y\left(0\right)-2L\left(y\right)=\frac{1}{s}-\frac{1}{{s}^{2}}$
$sL\left(y\right)-4-2L\left(y\right)=\frac{s-1}{{s}^{2}}$
$L\left(y\right)\left(s-2\right)=\frac{s-1}{{s}^{2}}+4$
$L\left(y\right)=\frac{4{s}^{2}+s-1}{{s}^{2}\left(s-2\right)}$
Step 2
Taking the inverse Laplace transform and make into the fraction form:
$\left(y\right)={L}^{-1}\left\{\frac{4{s}^{2}+s-1}{{s}^{2}\left(s-2\right)}\right\}\dots \left(i\right)$
$\frac{4{s}^{2}+s-1}{{s}^{2}\left(s-2\right)}=\frac{A}{s}+\frac{B}{{s}^{2}}+\frac{C}{\left(s-2\right)}$
$4{s}^{2}+s-1=As\left(s-2\right)+B\left(s-2\right)+Cs2$
$4{s}^{2}+s-1=\left(A+C\right){s}^{2}+\left(-2A+B\right)s-2B$
Step 3
Compare the coefficient and find out the values of A, B and C:

$-2A+B=1⇒-2A=1-\frac{1}{2}⇒A=-\frac{1}{4}$
$A+C=4⇒C=4-\left(-\frac{1}{4}\right)⇒C=\frac{17}{4}$
$\frac{4{s}^{2}+s-1}{{s}^{2}\left(s-2\right)}=\frac{-\frac{1}{4}}{s}+\frac{\frac{1}{2}}{{s}^{2}}+\frac{\frac{17}{4}}{\left(s-2\right)}$
Step 4
Now, form equation (i):
$\left(y\right)={L}^{-1}\left\{\frac{4{s}^{2}+s-1}{{s}^{2}\left(s-2\right)}\right\}={L}^{-1}\left\{\frac{-\frac{1}{4}}{s}+\frac{\frac{1}{2}}{{s}^{2}}+\frac{\frac{17}{4}}{s-2}\right\}$
$y=-\frac{1}{4}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{1}{2}{L}^{-1}\left\{\frac{1}{{s}^{2}}\right\}+\frac{17}{4}{L}^{-1}\left\{\frac{1}{s-2}\right\}$

$y=-\frac{1}{4}+\frac{t}{2}+\frac{17}{4}{e}^{2t}$