# Please solve the 2nd order differential equation by (PLEASE FOLLOW GIVEN METHOD) LAPLACE TRANSFORMATION ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE L(y) = left[frac{w}{(s^2 + a^2)(s^2+w^2)}right]*b Problem 2 Solve the differential equation frac{d^2y}{dt^2}+a^2y=b sin(omega t) where y(0)=0 and y'(0)=0

Question
Laplace transform
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right]*b$$
Problem 2 Solve the differential equation
$$\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)$$ where $$y(0)=0$$
and $$y'(0)=0$$

2021-01-28
Step 1
Given a differential equation, $$y"+a^2y=b\sin(\omega t)$$, where $$y(0)=0, y'(0)=0$$
Taking the Laplace transform of both sides of the given differential equation,
$$L(y")+a^2L(y)=bL(\sin(\omega t))$$
$$s^2L(y)−sy(0)−y'(0)+a^2L(y)=\frac{b\omega}{s^2+\omega^2}$$
$$(s^2+a^2)L(y)=\frac{b\omega}{s^2+\omega^2}$$
$$\therefore, L(y)=\frac{b\omega}{(s^2+\omega^2)}(s^2+\omega^2)$$
Step 2
Applying partial fraction,
$$\frac{b\omega}{(s^2+\omega^2)(s^2+a^2)}=\frac{A}{s^2+\omega^2}+\frac{B}{s^2+\omega^2}\dots (1)$$
$$A(s^2+a^2)+B(s^2+\omega^2)=b\omega$$
$$As^2+Aa^2+Bs^2+B\omega^2=b\omega$$
$$(A+B)s^2+Aa^2+B\omega^2=b\omega$$
Comparing coefficients,
$$A+B=0 \Rightarrow A=-B$$
$$Aa^2+B\omega^2=b\omega$$
$$\text{Substitute A in the above equation, }$$
$$-Ba^2+B\omega^2=b\omega$$
$$B=\frac{b\omega}{\omega^2-a^2}$$
Therefore, $$A=(\frac{b\omega}{a})^2-\omega^2$$
Step 3
Substitute A and B in equation (1),
$$\frac{b\omega}{(s^2+\omega^2)(s^2+a^2)}=\frac{b\omega}{(a^2-\omega^2)(s^2+\omega^2)}-\frac{b\omega}{(a^2-w^2)(s^2+a^2)}$$
$$=\frac{b\omega}{(a^2-w^2)\frac{1}{s^2+\omega^2}}-\frac{1}{(s^2+a^2)}$$
Substitute in L(y),
$$L(y)=\frac{b\omega}{a^2-\omega^2} \left(\frac{1}{s^2+\omega^2}-\frac{1}{s^2+a^2}\right)$$
$$\text{Taking inverse Laplace transform, }$$
$$y(t)=\frac{b\omega}{a^2-\omega^2}L^{-1}\left(\frac{1}{s^2+\omega^2}-\frac{1}{s^2+a^2}\right)$$
$$y(t)=\frac{b\omega}{a^2-\omega^2}\left[L^{-1}\left(\frac{1}{s^2+\omega^2}\right)-L^{-1}\left(\frac{1}{s^2+a^2}\right)\right]$$
$$=\frac{b\omega}{a^2-\omega^2}\left[\frac{1}{\omega} \sin(\omega t)-\frac{1}{a} \sin(at)\right]$$
Therefore, the required solution is,
$$y(t)=\frac{b\omega}{a^2-\omega^2}\left[\frac{1}{\omega} \sin(\omega t)-\frac{1}{a} \sin(at)\right]$$

### Relevant Questions

In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
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y(t) - ?
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Step by step
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Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
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a)$$f(t)=1+2t$$ b)$$f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}$$
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Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.$$ Solve (8.86) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{e}}}$$ can be estimated from $$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}$$
(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
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Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform
$$y"+\omega^{2}y=\sin \gamma t , y(0)=0,y'(0)=0$$
$$1)\ y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\omega^{2})^{2}}\bigg) \(2)\ y(t)=L^{-1}\bigg(\frac{\gamma}{s^{2}+\omega^{2}}\bigg) \(3)\ y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})^{2}}\bigg) \(4)\ y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})(s^{2}+\omega^{2})}\bigg)$$
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$