91Ó°ÊÓ

Use the Laplace transform to solve the second-order initial value problems in Exercises 11-26. $$ y^{\prime \prime}+3 y^{\prime}=-3 t, \quad y(0)=-1, \quad y^{\prime}(0)=1 $$

Find the inverse Laplace transform of each function in Exercises 16-25. Create a piecewise definition for your solution that doesn't use the Heaviside function. $$ F(s)=\frac{e^{-s}}{s-2} $$

Use the Laplace transform to solve the second-order initial value problems in Exercises 11-26. $$ y^{\prime \prime}+y^{\prime}=t e^{-t}, \quad y(0)=-2, \quad y^{\prime}(0)=0 $$

Use Theorem \(7.6\) (do not use partial fraction decompositions) to find the inverse Laplace transform of the \(s\)-domain functions given in Exercises 17-24. For example, 1/ \((s(s+1))\) is the product of \(F(s)=1 / s\) and \(G(s)=1 /(s+1)\). Consequently, by Theorem 7.6, the inverse Laplace transform of \(1 /(s(s+1))\) is the convolution of \(f(t)=1\) and \(g(t)=e^{-t}\). $$ \frac{1}{s^{2}-3 s} $$

Find the inverse Laplace transform of each function in Exercises 16-25. Create a piecewise definition for your solution that doesn't use the Heaviside function. $$ F(s)=\frac{1-e^{-s}}{s^{2}} $$

Use the Laplace transform to solve the second-order initial value problems in Exercises 11-26. $$ y^{\prime \prime}-y^{\prime}-2 y=t^{2} e^{2 t}, \quad y(0)=0, \quad y^{\prime}(0)=-1 $$

Find the inverse Laplace transform of each function in Exercises 16-25. Create a piecewise definition for your solution that doesn't use the Heaviside function. $$ F(s)=\frac{2+e^{-2 s}}{s^{3}} $$

Use the Laplace transform to solve the second-order initial value problems in Exercises 11-26. $$ y^{\prime \prime}-4 y^{\prime}-5 y=e^{2 t}, \quad y(0)=-1, \quad y^{\prime}(0)=0 $$

Perform the appropriate partial fraction decomposition, and then use the result to find the inverse Laplace transform of the given function. $$ Y(s)=\frac{1}{(s+2)(s-1)} $$

Perform the appropriate partial fraction decomposition, and then use the result to find the inverse Laplace transform of the given function. $$ Y(s)=\frac{1}{(s+3)(s-4)} $$