Step 1
We need to solve the given differential equations using Laplace transformation.
\(a) \frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y=1,\)
\(y(0=0) , y'(0)=0\)
Now , taking Laplace transformation both of the sides.
\(L\left\{\frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y\right\}=L\left\{1\right\}\)
\(\Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+3Y(s)=\frac{1}{s}\)
\(\Rightarrow (s^2+4s+3)Y(s)=\frac{1}{s} (\text{ Using } y(0)=0 , y'(0)=0)\)
\Rightarrow Y(s)=\frac{1}{s^2+4s+3}\)
\(\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}\)
\(\Rightarrow Y(s)= \frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}\) (By the method of partial fraction)
Step 2
Now, taking inverse Laplace transformation to get the final answer.
\(\Rightarrow y(t)=L^{-1}\left\{\frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}\right\}\)
\(\Rightarrow y(t)=\frac{1}{3}L^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}L^{-1}\left\{\frac{1}{s+1}\right\}+\frac{1}{6}L^{-1}\left\{\frac{1}{s+3}\right\}\)
\(\Rightarrow y(t)=\frac{1}{3}\cdot 1 -\frac{1}{2} \cdot e^{-t}+\frac{1}{6} \cdot e^{-3t}\)
\(\Rightarrow y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}\)
Answer \(y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}\)
Step 3
\(b) \frac{d^2y}{dt^2}+4\frac{dy}{dt}=1,\)
\(y(0=0) , y'(0)=0\)
Now , taking Laplace transformation both of the sides. \(L\left\{\frac{d^2y}{dt^2}+4\frac{dy}{dt}\right\}=L\left\{1\right\}\)
\(\Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))=\frac{1}{s}\)
\(\Rightarrow (s^2+4s)Y(s)=\frac{1}{s} (\text{Using } y(0)=0 , y'(0)=0)\)
\(\Rightarrow Y(s)=\frac{1}{s(s^2+4s)}\)
\(\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}\)
\(\Rightarrow Y(s)= \frac{1}{4s^2}+\frac{1}{16(s+4)}-\frac{1}{16s}\) (By the method of partial fraction)
Step 4
Now, taking inverse Laplace transformation to get the final answer.
\(\Rightarrow y(t)=L^{-1}\left\{\frac{1}{4s^2}+\frac{1}{16(s+4)}-\frac{1}{16s}\right\}\)
\(\Rightarrow y(t)=\frac{1}{4}L^{-1}\left\{\frac{1}{s^2}\right\}+\frac{1}{16}L^{-1}\left\{\frac{1}{s+4}\right\}-\frac{1}{16}L^{-1}\left\{\frac{1}{s}\right\}\)
\(\Rightarrow y(t)=\frac{1}{4}\cdot t +\frac{1}{16} \cdot e^{-4t}-\frac{1}{16} \cdot 1\)
\(\Rightarrow y(t)=\frac{t}{4}+\frac{e^{-4t}}{16} - \frac{1}{16}\)
Answer \(y(t)=\frac{t}{4}+\frac{e^{-4t}}{16} - \frac{1}{16}\)
Step 5
For (c),
\(2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y=te^{-2t}\)
\(y(0=0) , y'(0)=-2\)
Now , taking Laplace transformation both of the sides.
\(L\left\{2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y\right\}=L\left\{te^(-2t)\right\}\)
\(\Rightarrow 2(s^2Y(s)-sy(0)-y'(0))+3(sY(s)-y(0))-2Y(s)=\frac{1}{(s+2)^2}\)
\(\Rightarrow (2s^2+3s-2)Y(s)+4=\frac{1}{(s+2)^2} (\text{ Using } y(0)=0 , y'(0)=-2)\)
\(\Rightarrow (2s^2+3s-2)Y(s)=\frac{1}{(s+2)^2-4}\)
\(\Rightarrow Y(s)=\frac{\frac{1}{(s+2)^2}-4}{2s^2+3s-2}\)
\(\Rightarrow Y(s)=\frac{-4s^2-16s-15}{(s+2)^2(2s^2+3s-2)}\)
\(\Rightarrow Y(s)=-\frac{192}{125(2s-1)}+\frac{96}{125(s+2)}-\frac{2}{25(s+2)^2}-\frac{1}{5(s+2)^3}\)
(By the method of partial fraction)
Step 6
Now, taking inverse Laplace transformation to get the final answer.
\(\Rightarrow y(t)=L^{-1}\left\{-\frac{192}{250(s-\frac{1}{2})}+\frac{96}{125(s+2)}-\frac{2}{25(s+2)^2}-\frac{1}{5(s+2)^3}\right\}\)
\(\Rightarrow y(t)=-\frac{192}{250}L^{-1}\left\{\frac{1}{s-\frac{1}{2}}\right\}+\frac{96}{125}L^{-1}\left\{\frac{1}{s+2}\right\}-\frac{2}{25}L^{-1}\left\{\frac{1}{(s+2)^2}\right\}-\frac{1}{5}L^{-1}\left\{\frac{1}{(s+2)^3}\right\}\)
\(\Rightarrow y(t)=-\frac{192}{250}\cdot e^{\frac{1}{2t}}+\frac{96}{125}\cdot e^{-2t}-\frac{2}{25}\cdot te^{-2t}-\frac{1}{5}\cdot t^2e^{-2t}\)
\(\Rightarrow y(t)=-\frac{192}{250}e^{\frac{t}{2}}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{5}t^2e^{-2t}\)
Answer: \(y(t)=-\frac{192}{250}e^{\frac{t}{2}}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{5}t^2e^{-2t}\)
\(a) \frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y=1,\)
\(y(0=0) , y'(0)=0\)
Now , taking Laplace transformation both of the sides.
\(L\left\{\frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y\right\}=L\left\{1\right\}\)
\(\Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+3Y(s)=\frac{1}{s}\)
\(\Rightarrow (s^2+4s+3)Y(s)=\frac{1}{s} (\text{ Using } y(0)=0 , y'(0)=0)\)
\Rightarrow Y(s)=\frac{1}{s^2+4s+3}\)
\(\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}\)
\(\Rightarrow Y(s)= \frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}\) (By the method of partial fraction)
Step 2
Now, taking inverse Laplace transformation to get the final answer.
\(\Rightarrow y(t)=L^{-1}\left\{\frac{1}{3s}-\frac{1}{2(s+1)}+\frac{1}{6(s+3)}\right\}\)
\(\Rightarrow y(t)=\frac{1}{3}L^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}L^{-1}\left\{\frac{1}{s+1}\right\}+\frac{1}{6}L^{-1}\left\{\frac{1}{s+3}\right\}\)
\(\Rightarrow y(t)=\frac{1}{3}\cdot 1 -\frac{1}{2} \cdot e^{-t}+\frac{1}{6} \cdot e^{-3t}\)
\(\Rightarrow y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}\)
Answer \(y(t)=\frac{1}{3}-\frac{e^{-t}}{2}+\frac{e^{-3t}}{6}\)
Step 3
\(b) \frac{d^2y}{dt^2}+4\frac{dy}{dt}=1,\)
\(y(0=0) , y'(0)=0\)
Now , taking Laplace transformation both of the sides. \(L\left\{\frac{d^2y}{dt^2}+4\frac{dy}{dt}\right\}=L\left\{1\right\}\)
\(\Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))=\frac{1}{s}\)
\(\Rightarrow (s^2+4s)Y(s)=\frac{1}{s} (\text{Using } y(0)=0 , y'(0)=0)\)
\(\Rightarrow Y(s)=\frac{1}{s(s^2+4s)}\)
\(\Rightarrow Y(s)=\frac{1}{s(s+1)(s+3)}\)
\(\Rightarrow Y(s)= \frac{1}{4s^2}+\frac{1}{16(s+4)}-\frac{1}{16s}\) (By the method of partial fraction)
Step 4
Now, taking inverse Laplace transformation to get the final answer.
\(\Rightarrow y(t)=L^{-1}\left\{\frac{1}{4s^2}+\frac{1}{16(s+4)}-\frac{1}{16s}\right\}\)
\(\Rightarrow y(t)=\frac{1}{4}L^{-1}\left\{\frac{1}{s^2}\right\}+\frac{1}{16}L^{-1}\left\{\frac{1}{s+4}\right\}-\frac{1}{16}L^{-1}\left\{\frac{1}{s}\right\}\)
\(\Rightarrow y(t)=\frac{1}{4}\cdot t +\frac{1}{16} \cdot e^{-4t}-\frac{1}{16} \cdot 1\)
\(\Rightarrow y(t)=\frac{t}{4}+\frac{e^{-4t}}{16} - \frac{1}{16}\)
Answer \(y(t)=\frac{t}{4}+\frac{e^{-4t}}{16} - \frac{1}{16}\)
Step 5
For (c),
\(2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y=te^{-2t}\)
\(y(0=0) , y'(0)=-2\)
Now , taking Laplace transformation both of the sides.
\(L\left\{2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y\right\}=L\left\{te^(-2t)\right\}\)
\(\Rightarrow 2(s^2Y(s)-sy(0)-y'(0))+3(sY(s)-y(0))-2Y(s)=\frac{1}{(s+2)^2}\)
\(\Rightarrow (2s^2+3s-2)Y(s)+4=\frac{1}{(s+2)^2} (\text{ Using } y(0)=0 , y'(0)=-2)\)
\(\Rightarrow (2s^2+3s-2)Y(s)=\frac{1}{(s+2)^2-4}\)
\(\Rightarrow Y(s)=\frac{\frac{1}{(s+2)^2}-4}{2s^2+3s-2}\)
\(\Rightarrow Y(s)=\frac{-4s^2-16s-15}{(s+2)^2(2s^2+3s-2)}\)
\(\Rightarrow Y(s)=-\frac{192}{125(2s-1)}+\frac{96}{125(s+2)}-\frac{2}{25(s+2)^2}-\frac{1}{5(s+2)^3}\)
(By the method of partial fraction)
Step 6
Now, taking inverse Laplace transformation to get the final answer.
\(\Rightarrow y(t)=L^{-1}\left\{-\frac{192}{250(s-\frac{1}{2})}+\frac{96}{125(s+2)}-\frac{2}{25(s+2)^2}-\frac{1}{5(s+2)^3}\right\}\)
\(\Rightarrow y(t)=-\frac{192}{250}L^{-1}\left\{\frac{1}{s-\frac{1}{2}}\right\}+\frac{96}{125}L^{-1}\left\{\frac{1}{s+2}\right\}-\frac{2}{25}L^{-1}\left\{\frac{1}{(s+2)^2}\right\}-\frac{1}{5}L^{-1}\left\{\frac{1}{(s+2)^3}\right\}\)
\(\Rightarrow y(t)=-\frac{192}{250}\cdot e^{\frac{1}{2t}}+\frac{96}{125}\cdot e^{-2t}-\frac{2}{25}\cdot te^{-2t}-\frac{1}{5}\cdot t^2e^{-2t}\)
\(\Rightarrow y(t)=-\frac{192}{250}e^{\frac{t}{2}}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{5}t^2e^{-2t}\)
Answer: \(y(t)=-\frac{192}{250}e^{\frac{t}{2}}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{5}t^2e^{-2t}\)
