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

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

Question
Laplace transform
asked 2021-03-09
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

Answers (1)

2021-03-10
Step 1
Given third order differential equation is: \(y"'-7y'+6y =2 \sin (t)\)
Apply Laplace transform on each term on both sides of the differential equation.
\(L\left\{y'''\right\}-7L\left\{y'\right\}+6L\left\{y\right\}=2L\left\{\sin t\right\}\)
Use the standard appropriate Laplace transforms for the higher order differentials:
\(L\left\{y'''\right\}=s^3Y(s)-s^2y(0)-sy'(0)-y''(0)\)
\(L\left\{y'\right\}=sY(s)-y(0)\)
\(L\left\{y\right\}=Y(s)\) The Laplace transform of \(\sin(t)\) is: \(L\left\{\sin t\right\}=\frac{1}{(1+s^2)}\)
Substitute the formula in the given equation:
\(\left[s^3Y(s)-s^2y(0)-sy'(0)-y''(0)\right]-7\left[sY(s)-y(0)\right]+6\left[Y(s)\right]=\frac{2}{(1+s^2)}\)
Now, substitute the given boundary conditions in the above equation.
\(\Rightarrow (s^3-7s+6)Y(s)=\frac{2}{(1+s^2)}\)
\(\Rightarrow Y(s)=\frac{2}{(1+s^2)(s^3-7s+6)}\)
Here, the denominator on the right hand side is factorized into:
\(\Rightarrow Y(s)=\frac{2}{(1+s^2)(s-1)(s-2)(s+3)}\)
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)
\(Y(s)=\frac{2}{(1+s^2)(s-1)(s-2)(s+3)}\)
For this equation evaluate the partial fractions:
\(\frac{2}{(1+s^2)(s-1)(s-2)(s+3)}=\frac{A}{(s-1)}+\frac{B}{(s-2)}+\frac{C}{(s+3)}+\frac{D}{(1+s^2)}\)
\(\Rightarrow 2=A(1+s^2)(s−2)(s+3)+B(1+s^2)(s−1)(s+3)+C(1+s^2)(s−1)(s−2)+D(s−1)(s−2)(s+3)\)
\(\text{For }s=1, \text{value of A is: }A=-\frac{1}{5}\)
\(\text{For }s=2, \text{value of B is: }B=\frac{2}{25}\)
\(\text{For }s=-3, \text{value of C is: }C=\frac{1}{100}\)
Equating constant terms o both sides gives D value as: \(D=\frac[17}{100}\)
Then, the Laplace equation turns out to be:
\(Y(s)=\frac{17}{100}\left(\frac{1}{1+s^2}\right)-\frac{1}{5}\left(\frac{1}{s-1}\right)+\frac{2}{25}\left(\frac{1}{s-2}\right)+\frac{1}{100}\left(\frac{1}{s+3}\right)\)
Now, apply inverse Laplace on each term of the equation. \(y(t)=\frac{17}{100}L^{-1}\left\{\frac{1}{1+s^2}\right\}-\frac{1}{5}L^{-1}\left\{\frac{1}{s-1}\right\}+\frac{2}{25}L^{-1}\left\{\frac{1}{s-2}\right\}+\frac{1}{100}L^{-1}\left\{\frac{1}{s+3}\right\}\)
Use the standard Laplace transforms: \(L^{-1}\left\{\frac{1}{(s-a)}\right\}=e^{at} \text{ and } L^{-1}\left\{\frac{a}{(a+s^2)}\right\}=\sin(at)\)
Which gives the solution as:
\(y(t)=\frac{17}{100}\sin t-\frac{1}{5}e^t+\frac{2}{25}e^{2t}+\frac{1}{100}e^{-3t}\)
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