# How to solve for third order differential equation of y"'-7y'+6y =2 sin (t) using Method of Laplace Transform when y(0)=0, y'(0)=0, y"(0)=0? Step by step

bobbie71G 2021-03-09 Answered
How to solve for third order differential equation of $y{"}^{\prime }-7{y}^{\prime }+6y=2\mathrm{sin}\left(t\right)$ using Method of Laplace Transform when $y\left(0\right)=0,{y}^{\prime }\left(0\right)=0,y"\left(0\right)=0$?
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## Expert Answer

lamanocornudaW
Answered 2021-03-10 Author has 85 answers

Step 1
Given third order differential equation is: $y{"}^{\prime }-7{y}^{\prime }+6y=2\mathrm{sin}\left(t\right)$
Apply Laplace transform on each term on both sides of the differential equation.
$L\left\{{y}^{‴}\right\}-7L\left\{{y}^{\prime }\right\}+6L\left\{y\right\}=2L\left\{\mathrm{sin}t\right\}$
Use the standard appropriate Laplace transforms for the higher order differentials:
$L\left\{{y}^{‴}\right\}={s}^{3}Y\left(s\right)-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-{y}^{″}\left(0\right)$
$L\left\{{y}^{\prime }\right\}=sY\left(s\right)-y\left(0\right)$
$L\left\{y\right\}=Y\left(s\right)$ The Laplace transform of $\mathrm{sin}\left(t\right)$ is: $L\left\{\mathrm{sin}t\right\}=\frac{1}{\left(1+{s}^{2}\right)}$
Substitute the formula in the given equation:
$\left[{s}^{3}Y\left(s\right)-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-{y}^{″}\left(0\right)\right]-7\left[sY\left(s\right)-y\left(0\right)\right]+6\left[Y\left(s\right)\right]=\frac{2}{\left(1+{s}^{2}\right)}$
Now, substitute the given boundary conditions in the above equation.
$⇒\left({s}^{3}-7s+6\right)Y\left(s\right)=\frac{2}{\left(1+{s}^{2}\right)}$
$⇒Y\left(s\right)=\frac{2}{\left(1+{s}^{2}\right)\left({s}^{3}-7s+6\right)}$
Here, the denominator on the right hand side is factorized into:
$⇒Y\left(s\right)=\frac{2}{\left(1+{s}^{2}\right)\left(s-1\right)\left(s-2\right)\left(s+3\right)}$
Step 2 Now, find the inverse Laplace function of Y(s) to get the solution of the differential equation in the form of y(t)
$Y\left(s\right)=\frac{2}{\left(1+{s}^{2}\right)\left(s-1\right)\left(s-2\right)\left(s+3\right)}$
For this equation evaluate the partial fractions:
$\frac{2}{\left(1+{s}^{2}\right)\left(s-1\right)\left(s-2\right)\left(s+3\right)}=\frac{A}{\left(s-1\right)}+\frac{B}{\left(s-2\right)}+\frac{C}{\left(s+3\right)}+\frac{D}{\left(1+{s}^{2}\right)}$
$⇒2=A\left(1+{s}^{2}\right)\left(s-2\right)\left(s+3\right)+B\left(1+{s}^{2}\right)\left(s-1\right)\left(s+3\right)+C\left(1+{s}^{2}\right)\left(s-1\right)\left(s-2\right)+D\left(s-1\right)\left(s-2\right)\left(s+3\right)$

Equating constant terms o both sides gives D value as: $D=\frac{17}{100}$
Then, the Laplace equation turns out to be:
$Y\left(s\right)=\frac{17}{100}\left(\frac{1}{1+{s}^{2}}\right)-\frac{1}{5}\left(\frac{1}{s-1}\right)+\frac{2}{25}\left(\frac{1}{s-2}\right)+\frac{1}{100}\left(\frac{1}{s+3}\right)$
Now, apply inverse Laplace on each term of the equation.

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