Question Use the Laplace transform to solve the given equation. y'(t) = cos t + loren y(t) cos(t - t)dr, y(0) = 1
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#Given, y^{\prime}(t)=\cos t+\int_{0}^{t} y(t) \cos (t-r) d rTaking Laplace transformation on boths sides of (\theta), we get\begin{array}{c}L\left\{\dot{y}^{\prime}(t)\right\}=L\left\{\cos t+\int_{0}^{t} y(\tau) \cos (t=\tau) d \tau\right\} \\\Rightarrow s t\{y(t)\}-y(0)=L\{\cos t\}+L\left\{\int_{0}^{t} y(\tau) \cos (t-\tau) d \tau\right\} \\\Rightarrow s L\{y(t)\}-1=\frac{s}{s{ }^{2}+1}+L\{y(t)\} L\{\cos t\} \\{\left[\because L L y^{\prime}(t)\right\}=s L\{y\}-y(0)} \\L\left\{\cos =\frac{s}{s}\right. \\L\{\cos (a t)\}=\frac{s}{s^{2}+a^{2}} \\L\left\{\int_{0}^{t} f(t) g(t-\tau) d \tau\right\}=L\{f(t) \zeta L\{g(t)\}] \\\text { and yeo } y=1 \text { given }]\end{array}\begin{array}{l}\Rightarrow\left(s-\frac{s}{s^{2}+1}\right) Y(s)=\frac{s}{s^{2}+1}+1 \\\Rightarrow\left(\frac{s^{3}+s-s}{s^{2}+1}\right) Y(s)=\frac{s}{s^{2}+1}+1 \\\Rightarrow\left(\frac{s^{3}}{s^{2}+1}\right) Y(s)=\frac{s}{s^{2}+1}+1\end{array}\begin{array}{l}\Rightarrow Y(s)=\frac{s}{s^{2}+1} \times \frac{s^{2}+1}{s^{3}}+\frac{s^{2}+1}{s^{3}} \\\Rightarrow Y(s)=\frac{1}{s^{2}}+\frac{1}{s}+\frac{1}{s^{3}}\end{array}Taking inverse Laplace transfor mation,\begin{array}{l}L^{-1}\{Y(s)\}=L^{-1}\left\{\frac{1}{5^{2}}+\frac{1}{5}+\frac{1}{53}\right\} \\\Rightarrow y(t)=L^{-1}\left\{\frac{1}{52}\right\}+L^{-1}\left\{\frac{1}{5}\right\}+L^{-1}\left\{\frac{1}{53}\right\} \\\Rightarrow y(t)=\frac{t}{1 !}+1+\frac{t^{2}}{2 !} \\{\left[\because L^{+}\left\{\frac{1}{s^{n+1}}\right\}=\frac{t^{n}}{n !}\right]} \\\Rightarrow y(t)=1+t+\frac{t^{2}}{2} \\\end{array} ...